MCAT General Chemistry Review
Part I Review
Chapter 12: High-Yield Problem Solving Guide for General Chemistry
High-Yield MCAT Review
This is a High-Yield Questions section. These questions tackle the most frequently tested topics found on the MCAT. For each type of problem, you will be provided with a stepwise technique for solving the question and key directional points on how to solve for the MCAT specifically.
For each topic, you will find a “Takeaways” box, which gives a concise summary of the problem-solving approach, and a “Things to Watch Out For” box, which points out any caveats to the approach discussed above that usually lead to wrong answer choices. Finally, there is a “Similar Questions” box at the end so you can test your ability to apply the stepwise technique to analogous questions.
We’re confident that this guide, and our award-wining Kaplan teachers, can help you achieve your goals of MCAT success and admission into medical school!
Arrange the following compounds in order of increasing melting point:
1) Separate the compounds by general polarity.
In this series, we can separate the compounds into three groups of two. The alkanes (3 and 4) will be the least polar and, therefore, will melt at the lowest temperature; the alkenes (1 and 5) will be in the middle, and the aromatic compounds (2 and 6) will melt at the highest temperature.
Forces that stabilize a molecule more in the solid state than in the liquid state will cause a molecule to have a higher melting point.
Things to Watch Out For
Be careful not to confuse melting points with boiling points. Remember that in general, symmetry raises melting points, whereas branching lowers them.
2) Examine each grouping for trends in polarity and/or molecular symmetry.
For the lowest-melting-point compounds, notice that cyclohexane has a higher degree of molecular symmetry than does n–hexane; this will cause it to melt at a significantly higher temperature.
With the alkenes, the trans alkene has more symmetry than the cis alkene, because the cis alkene has a rather large “kink” in the middle of the chain that prevents it from packing together as well in the crystal, thus lowering its melting point.
Finally, acetanilide (6) is significantly more polar than aniline (2), because the amide carbonyl bond is highly polar, causing these molecules to stick together better and consequently raising their melting point.
Therefore, the ordering of the compounds’ melting points is as follows:
3 < 4 < 5 < 1 < 2 < 6
Polarity affects melting point just as it does boiling point: More polar molecules melt at a higher temperature because they tend to stick together better. Molecular symmetry also plays a more prominent role than with boiling point, because another consideration is how well molecules pack or “fit together” in the crystal. The more symmetrical a molecule is, the better it packs in the crystal, just as symmetrical puzzle pieces in a jigsaw puzzle fit together better than asymmetrical pieces.
1) For straight-chain alkanes, which do you suppose have higher melting points: alkanes with an odd number of carbons, or those with an even number of carbons?
2) Which molecule would you expect to melt at a higher temperature, n–pentane or neopentane (2,2–dimethylpropane)? Why?
3) Between phenol (hydroxybenzene) and aniline (aminobenzene), which would melt at a higher temperature and why?
Given the following five molecules, place them in order of increasing boiling point:
1) Look for unusually heavy molecules.
Remember that molecular weight is one of the key determinants of boiling point. Something that is extraordinarily heavy is going to be harder to boil than something that is lighter. In this case, all the molecules are in the same general range of molecular weight, so this factor won’t help us place the molecules in order.
2) Look for highly polar functional groups.
Compounds 3 and 5 are going to have higher-than-usual boiling points. Between compounds 3 and 5, compound 3 will boil at a higher temperature because it has a more polar functional group; also, the alcohol is capable of hydrogen bonding. Compound 3 will have hydrogen bonds that are a stronger version of dipole–dipole interactions.
The other factor that affects boiling point is the presence of polar functional groups. These groups help increase boiling point because they increase the attractions of molecules for each other.
Remember: Hydrogen bonding is the strongest type of intermolecular attraction.
Remember that only two factors affect relative boiling points between substances: molecular weight and intermolecular forces.
3) Look for the effect of dispersion forces.
Compound 4 will boil at a higher temperature than 1, 2, and 5 because it is longer (eight carbons versus five, which increases its London forces); therefore, there are more opportunities for it to attract other molecules of 4.
Although 1, 2, and 5 all have the same surface area, the polar group on 5 gives it a higher boiling point than 1 and 2.
Remember: Dispersion forces are the only kind of intermolecular attractions that cause nonpolar molecules to stick together.
Things to Watch Out For
Be sure to consider the factors in the order presented here. For example, polarity is more important than size.
4) Look for trends in the symmetry of molecules.
In this case, pentane, 1, will boil higher than neopentane, 2. This is because neopentane is more symmetrical and, therefore, a more compact molecule; thus, it has a less effective surface area. You can determine this by imagining a “bubble” around each molecule. Neopentane could very easily fit into a spherically shaped bubble, whereas pentane would require an elongated, elliptical bubble with a greater surface area.
If neopentane has a smaller surface area, then there are fewer opportunities for it to engage in dispersion-type attractions with other molecules of neopentane, making it a lower-boiling-point compound (the actual boiling points are 36.1°C for pentane and 9.4°C for neopentane).
At this point, we’re really splitting hairs. Notice that compounds 1 and 2 are merely constitutional isomers of one another. If two molecules have the same weight and are relatively nonpolar, symmetry is the factor that decides which one will boil at a higher temperature.
1) How could you easily alter the structure of compound 3 to lower its boiling point?
2) How could you easily alter the structure of compound 5 to raise its boiling point?
5) Put it all together. Order the compounds as specified by the question. The ordering of the boiling points will therefore be as follows:
2 < 1 < 5 < 4 < 3
Rate Law from Experimental Results
Consider the nitration reaction of benzene, an example of electrophilic aromatic substitution:
The rate data below were collected with the nitration of benzene carried out at 298 K. From this information, determine the rate law for this reaction.
Remember that the rate constant k depends only on temperature.
A shortcut to determine order is to use the following relation when you find two trials where one reagent’s concentration changes but all other concentrations are constant:
Change in rate = (Proportional change in concentration)x, where x = the order with respect to that reagent.
1) Write down the general form of the rate law.
Rate = k[C6H6]x [HNO2]y
Remember: The general form of the rate law must include a constant, k, that is multiplied by the concentrations of each of the reactants raised to a certain power.
2) Determine the order of the reaction with respect to each reactant. Choose two trials in which the concentration of one reagent is changing but the other is not. Take the ratio of these two trials and set up an equation.
Cancel the rate constants because they are equal to each other. Collect terms raised to the same exponent together.
Plug and chug. Substitute numbers from the rate data table into the equation.
The term raised to the y power disappears because 1 raised to any power equals 1. The only way that 4x can equal 1 is if x = 0.
To determine the order with respect to HNO2, note that there are no two trials in which the concentration of benzene stays the same. However, this does not matter, because the reaction is zero order with respect to benzene.
Plug in numbers from the table as before.
Simplify the numbers to make them easy to handle. Note that 5.40 × 10–5 is the same thing as 54.0 × 10–6.
The only way this equation can be true is if y = 2.
3) Write down the rate law with the correct orders.
Rate = k[C6H6]0[HNO2]2 = k[HNO2]2
Things to Watch Out For
Make sure that initially you select two trials where one reagent’s concentration changes but all other concentrations are constant. Otherwise, you won’t come out with the correct rate law!
1) What is the value of the rate constant k for the original reaction above? What are its units?
2) Given the data below, determine the rate law for the reaction of pyridine with methyl iodide. Find the rate constant k for this reaction and its units. Use the rate law to determine what type of reaction this is.
3) Cerium(IV) is a common inorganic oxidant. Determine the rate law for the following reaction and compute the value of the rate constant k along with its units.
Rate Law from Reaction Mechanisms
Often, changing the medium of a reaction can have a dramatic effect on its mechanism. In the gas phase, HCl reacts with propene according to the following reaction mechanism:
Step 1: HCl + HCl H2Cl2 (fast, equilibrium)
Step 2: HCl + CH3CHCH2CH3CHClCH3* (fast, equilibrium)
Step 3: CH3CHClCH3* + H2Cl2 CH3CHClCH3 + 2 HCl (slow)
where CH3CHCH2 is propene and CH3CHClCH3* represents an excited state of 2–chloropropane.
Based on these reaction steps, derive the rate law for this reaction.
1) Identify the slow step in the reaction and write down the rate law expression for that step.
Rate = k3[CH3CHClCH3*][H2Cl2]
With reaction mechanisms, the goal is to eliminate the concentrations of intermediates, because they are usually high-energy species that exist only briefly.
2) If intermediates exist in the rate law from step 1, use prior steps to solve for their concentration and eliminate them from the rate law.
k1[HCl]2 = k–1[H2Cl2]
Here, we are taking advantage of the fact that step 1 of the mechanism is in equilibrium; therefore, the rates of the forward and reverse reactions are equal.
Solve for the concentration of H2Cl2, one of the intermediates from above.
Step 2 from the mechanism is also in equilibrium, so the rates of the forward and reverse reactions are equal.
Solve for the intermediate, as above.
Plug the concentrations into the rate law for the slow step.
1) What are the units of the rate in the original question? Based on this, what must the units of kobs be for this reaction?
2) How does this rate law differ from the one that you might expect if this reaction were to be carried out in solution, instead of in the gas phase?
3) How would the key intermediates differ between this reaction in the gas phase and in solution?
Remember: Intermediates are assumed to exist for only a brief period of time because they are produced in one step and consumed in another. Therefore, their concentration cannot be measured, and they must be eliminated from the rate law.
3) Combine constants and simplify the rate law.
Combine all of the constants and concentrations.
Remember: A constant times a constant times a constant, and so on, is just another constant.
Things to Watch Out For
In this case, you may assume that the stoichiometric coefficients of each reactant are equal to the order. When you are presented with rate data, you may not make this assumption but must use the rate data to determine order.
Le Châtelier’s Principle
Le Châtelier’s principle
A chemistry student adds solid copper sulfate to water at room temperature. The resulting solution has an emerald blue color reminiscent of azulene. The student then adds a piece of aluminum foil to the solution and watches as a small hole develops in the foil. What could the student do to increase the rate at which the hole forms?
CuSO4 (s) + 5 H2O (l) CuSO4 • 5 H2O (aq) H < 0, G < 0
2 Al (s) + 3 Cu2+ (aq) 2 Al2+ (aq) + 3 Cu (s) H < 0, G < 0
1) Identify the direction of the desired reaction.
To the right of equation 2
The hole in the aluminum foil indicates that the solid is dissolving into solution. Looking at the reaction equations, solid aluminum reacts with Cu2+ to form Al2+, the desired end product.
Le Chatelier’s principle basically puts Keq and Q into words. Chemical reactions attempt to reach equilibrium. Adding more reactant, for instance, makes the reaction move to the right: More product needs to be formed to balance out the addition.
2) List the different types of stress that can be applied to any system.
Pressure, temperature, concentration
On the MCAT, liquids and solids are incompressible, so altering the pressure at which the reaction occurs should have no effect. However, we can alter the temperature or the concentration of the reactants or products. To increase the rate at which the hole forms, we are looking to push the reaction to the right.
3) Consider the effect of various system stresses: heat.
Run the reactions at a lower temperature to increase the rate of hole formation.
We are told that both reactions are exothermic; thus, we can rewrite them in this generic format: A B + , where is heat. Heat is a product of this reaction. Increasing the heat will push the reaction to the left: It is as if we’ve added more of the “product.” Alternatively, if we drop the temperature at which the reactions are conducted (i.e., remove the heat product), we push the reaction to the right and favor the formation of the hole.
MCAT Pitfall: Increasing the temperature would favor formation of the original CuSO4solid due to the exothermic nature of the two reactions.
Things to Watch Out For
Be particularly careful with ionic species and gases. Ionic species can dissociate in water, depending on their electrolytic strength, and may result in multiples of the original concentration of solid. An increase in pressure favors the side of the reaction with fewer molecules of gas.
4) Consider the effect of various system stresses: concentration.
Increase the concentration of the reactants or remove the product.
From the second reaction, it is clear that it is the Cu(II) ion that is reacting with the aluminum foil. How can we get more Cu(II) ion in solution? Add more solid copper sulfate. Alternatively, we could remove the final product. Removing Cu(II) after copper sulfate pentahydride dissociates will disable the second reaction. Removing SO42-, however, will cause more hydrate to form. Better yet, we could simply remove the solid copper that plates out when the aluminum atoms are ionized.
1) A student wants to create H2O from its natural state elements. If he plans to use gaseous hydrogen and oxygen, what type of system stress would help the reaction progress? Does it matter whether he forms water vapor or liquid water?
2) The reaction [Co(H2O)6]2+ (aq) + 4 Cl- (aq) CoCl42- (aq) + 6 H2O (l) describes the formation of a blue solution from one that is pink. A student isolates the solid of both cobalt compounds. What happens when he adds them to separate flasks of water?
3) A cold pack works by reacting NH4NO3 with water. The reaction requires energy to solvate ammonium nitrate. What is the effect of putting a cold pack on your skin? What is the effect of putting a cold pack in a 0°C freezer?
Reaction Energy Profiles
The goal of a reaction profile is to give you information about energy differences. Make sure that you identify the important differences and their significances, as above.
When chalcone (A) is subjected to reductive conditions with sodium borohydride, two products can result. The two products are the so-called “1,2-reduction” product (B), in which the carbonyl is reduced, and the “1,4-reduction” product (C), in which the conjugated alkene is reduced.
The reaction profiles leading to each reduction product are both shown in the plot below.
Based on the plot above, answer the following questions:
1) Which product is more thermodynamically stable? Which one forms faster?
2) Assume that A is in equilibrium with C. What will the ratio of C to A be at equilibrium?
3) How could the rate of the reaction of A to C be made closer to the rate of the reaction of A to B?
4) Which product would be favored if A were subjected to high temperatures for a long time? If A were subjected to low temperatures for only a brief period of time? Explain why for each situation.
Things to Watch Out For
Be careful to take note of the units of energy on the y-axis if you plan on doing any computations.
1) Look at the energy differences between the starting material and the product(s), as well as the differences between the starting material and the transition state leading to each product.
Notice that the energy of C is lower than that of B. Therefore, it is the more thermodynamically stable product.
The rate of formation of each product is determined by the difference in energy between the starting material A and the top of the “hump” leading to each product. Because this distance is lower for the formation of B, it forms faster.
2) Note that the difference in energy between the starting material and the product(s) determines the ratio of products to reactants at equilibrium.
G° = -RT ln Keq
This equation provides the relationship between Keq and G°. We need to rearrange it to solve for Keq.
Note that G° 1,000 - 3,500 = -2,500 cal mol-1, from the diagram, and R = 1.99 cal (mol K)-1 2 cal (mol K)-1 and T = 298 300 K.
Let’s say 2,500/600 is about equal to 4, and e = 2.7818 3.
Note that a negative G° gives more product than reactant, as you would expect for a spontaneous reaction.
3) Consider what role(s) the addition of a catalyst might play.
Remember: A catalyst is something that speeds up a reaction and is not consumed during a reaction. If it speeds up a reaction, it lowers the “hump” in the reaction profile. So we could increase the rate of formation of C by adding a catalyst to that reaction, making the rate closer to the rate of formation of B.
4) Consider the effects of temperature on the reaction(s).
At high temperatures for long times, A has the energy to go back and forth over and over again between B and C. Then, over time, the lowest energy product C would predominate, just as rolling a ball down a hill would cause it to fall to the lowest location.
Between the two products, B will form much faster than C because its energy of activation (height of the “hump”) is lower. At low temperatures, the products won’t have the energy to go back over the hill to get to A, so the faster-forming product will predominate (i.e., B).
1) What would be the ratio of B to A at equilibrium?
2) If a catalyst were added to the reaction of A going to C, as in step 3 above, would the energies of A and C be changed as a result? Why or why not?
3) There are actually intermediates involved in the reactions producing both B and C. These intermediates are shown below. Sketch how each reaction profile would look, including the involvement of these intermediates. Be sure to indicate which intermediate is relatively more stable.
Gibbs free energy
Equilibrium constant, Keq
Reaction quotient, Q
G = H - (kJ/mol)
G° = -RT ln Keq (kJ/mol)
G = G° + RT ln Q (kJ/mol)
The reaction 2 NO ( g) + Cl2 ( g) 2 NOCl ( g) adheres to the following thermodynamic data:
1.54 × 107
Suppose that, in equilibrium, NO exerts 0.6 atm of pressure and Cl2 adds 0.3 atm; find the partial pressure of NOCl in this equilibrium. Also, find the temperature at which the thermodynamic data in the table were reported. Keq is related to Kp by the following equation: Kp = Keq(RT)n where nis the change in number of moles of gas evolved as the reaction moves forward. (R = 8.314 J/K · mol.)
1) Find the temperature at which the thermodynamic data are true.
G = H -
(-44 kJ/mol) = (-77 kJ/mol) - T (-0.121 kJ/K × mol) T = 273 K
Being able to work with the equation G = H - is absolutely crucial for Test Day. Specifically concerning the data here, because both H and S are negative, the reaction will become “less spontaneous” as we increase temperature. This will help narrow down our answer choices on Test Day.
MCAT Pitfall: Notice that not all of the state functions were given in the same unit! Had you blindly put in entropy without changing its units, you would have obtained a temperature near absolute zero (0 K). At absolute zero, molecules no longer move, and it is unlikely that this reaction would have such a high equilibrium constant.
Two equations should get you through nearly any thermochemistry question. Remember to round your numbers and to predict the ballpark for your answers wherever possible.
2) Find n.
For the equation
2 NO ( g) + Cl2 ( g) 2 NOCl ( g),
n = 2 - (2 + 1) = -1.
Things to Watch Out For
Pay close attention to the units used.
3) Find Kp.
Kp = Keq(RT )n
Kp = (1.54 × 107)(8.314 J/K·mole)(273 K)-1 = 6,785
Use the temperature value from step #1.
You are not responsible for memorizing the equation. However, in the MCAT, you have to be able to use a brand-new equation to solve for the answer.
4) Find the partial pressure.
Plug in the given data into the reaction quotient.
1) If Keq = 7.4 × 10-3 for CH4 (g)+ 2 H2O (g)? CO2 (g)+ 4 H2 (g),which is more plentiful, the reactants or the products?
2) If pyrophosphoric acid (H4P2O7) and arsenous acid (H3AsO3) have acid dissociation constants of 3 × 10-2 and 6.6 × 10-10, respectively, at room temperature, find the Gibbs free energy of each dissociation reaction and determine if it is spontaneous. What does this mean for the H and S for these reactions?
3) A chemist is given three liquid-filled flasks, each labeled with generic thermodynamic data. She is told to put one in a cold room, to put one on a Bunsen burner, and to leave one on the benchtop—whatever conditions will best facilitate the reaction. If the flasks are labeled as follows, which flask goes where?
A H < 0, S > 0
E H < 0, S < 0
P H > 0, S > 0
Hrxn = HF(products) - HF(reactants)(kJ/mol)
Bond dissociation energy
Hrxn = Hb (reactants) - Hb (products) (kJ/mol)
Hrxn = total energy input - total energy released (kJ/mol)
An unknown compound containing only carbon and hydrogen is subjected to a combustion reaction in which 2,059 kJ of heat are released. If 3 moles of CO2 and 4 moles of steam are produced for every mole of the unknown compound reacted, find the enthalpy for a single C–H bond.
Bond Dissociation Energy (kJ/mol)
1) Write a balanced equation for this reaction.
C3H8 + 5O2 3CO2 + 4H2O
The question stem tells us a few things about the reaction: It is combustion, the carbon source has the generic structure CxHy, and the products include 10 oxygen atoms, 3 carbon atoms, and 8 H atoms. To balance the reaction, we’d need those atoms on the left side, too. Thus, we find that our unknown sample is actually propane and that we need 5 O2 molecules.
Remember: For our purposes, it is completely acceptable to have a fractional coefficient in front of a diatomic molecule. 2 C2H4 + (7/2) O23 H2O + 2 CO2is equivalent to 4 C2H4 + 7 CO26 H2O + 4 CO2.
Consider the number of bonds before applying Hess’s law. Make sure to take note of how many bonds are in a given molecule as well as how many stoichiometric equivalents of that molecule you have.
2) Determine which bonds are broken and which are formed.
C3H8: 2 C–C bonds broken, 8 C–H bonds broken
5O2: 5 O=O bonds broken
3CO2: 6 C=O bonds formed
4H2O: 8 O–H bonds formed
Combustion of C3H8 will break apart the carbon backbone and the C–H bonds. The carbon is in a straight chain (as opposed to cyclic or branched), so 2 C–C bonds and 8 C–H bonds are broken. For O2, only one O=O bond is broken. However, we have 5 moles of this reactant, and thus we have 5 O=O bonds broken. Each molecule of carbon dioxide has 2 C=O bonds, but we have 3 moles of CO2, so we have 6 C=O bonds formed. Similarly, 8 O–H bonds are formed in the 4 moles of water produced.
3) Apply Hess’s law.
Hrxn = Hb (reactants) - Hb (products)
Hrxn = total energy input - total energy released
–2,059 = [2(347) + 8x + 5(497)] - [6(805) + 8(464)]
–2,059 = [3,179 + 8 x] - [8,542]
–2,059 + 8,542 - 3,179 = 8x
x = 413 kJ/mol
Bond dissociation energy is the energy required to break a particular type of bond in one mole of gaseous molecules. Bond energies can be used to estimate the enthalpy of reaction as given by the two equations above. When we start plugging in numbers, we are given all data except for C–H bond enthalpy. We solve for this variable (x in the above equations).
Remember: The equation Hrxn = Hb(reactants) - Hb(products) is simply a restatement of Hess’s law. Bond enthalpy is for bond breaking, and enthalpy of formation, of course, is for bond making. Changing HFto Hbswitches the signs and, thus, the order of the equation. Keep in mind that it can also be written as Hrxn = Hb(bonds broken) + Hb(bonds formed), but you must remember to make the bond enthalpies for the products negative because forming bonds releases energy.
4) Use Avogadro’s number.
(413 kJ/mol) × [1 mol/(6.022 × 1023 molecules)] = 6.86 × 10-22 kJ/molecule
We see that 413 kJ are found in one mole of C–H bonds. One mole of a substance is equal to 6.022 × 1023 molecules. Here, we simply use that conversion factor. The result tells us that 6.86 × 10-22 kJ are stored in each C–H bond.
Things to Watch Out For
There are a number of ways to set up the equation for Hrxn. Whatever equation you use, keep your signs straight. Remember that forming bonds releases energy, whereas breaking bonds requires energy.
1) Ethanol metabolism in yeast consists of the conversion of ethanol (C2H5OH) to acetic acid (CH3COOH). What is the enthalpy of the reaction if 0.1 mmol of ethanol is metabolized?
2) A second metabolic process involves the net production of 2 ATP and 2 NADH from 2 ADP and 2 NAD+. If the conversion of these molecules is endothermic and adds 443.5 kJ to the overall enthalpy of the reaction, find the enthalpy for a “high-energy” phosphate bond.
3) Tristearin is oxidized in the body according to the following reaction: 2 C57H110O6 + 163 O2 114 CO2 + 110 H2O. If the standard enthalpy for this reaction is -34 MJ mol-1, find the total enthalpy for the bonds in tristearin.
Heat of Formation
Hrxn = HF (products) - HF (reactants) (kJ/mol)
Heat of formation
The heat of combustion of glucose (C6H12O6) is -2,537.3 kJ/mol. If the H°f of CO2 (g) is - 393.5 kJ/mol and the H °f of H2O (g) is -241.8 kJ/mol, what is the H°f of glucose?
1) Write a balanced equation for this reaction.
Unbalanced reaction: C6H12O6 + O2 CO2 + H2O
Balanced reaction: C6H12O6 + 6 O2 6 CO2 + 6 H2O
The unbalanced reaction above is typical of all hydrocarbon combustion reactions. (Unless otherwise noted, presume that combustion of carbohydrates is with oxygen gas.) Begin by balancing the carbons on the left side (6 CO2), then balance the hydrogens on the left side (12 H2O), and conclude by balancing the oxygen gas on the right side (6 O2).
Remember: For our purposes, it is completely acceptable to have a fractional coefficient in front of a diatomic molecule: 2 C2H4 + (7/2) O23 H2O + 2 CO2is equivalent to 4 C2H4 + 7 CO26 H2O + 4 CO2, but the math is simpler for the former.
2) Apply Hess’s law.
Hrxn = HF (products) - HF (reactants)
-2,537.3 = [6(-393.5) + 6(-241.8)] - [HF (glucose)]
Rearranging to solve for HF (glucose):
Hf (glucose) = 2,537.3 + [6(-393.5) + 6(-241.8)]
Hf (glucose) = 2,537.3 + [-2,361 + -1,450.8]
Hf (glucose) = 2,537.3 + [-3,811.8]
Hf (glucose) = -1,274.5
The heat of formation is defined as the heat absorbed or released during the formation of a pure substance from the elements at a constant pressure. Therefore, by definition, diatomic gases like oxygen have a heat of formation of zero. A negative heat of formation means that heat is released to form the product, whereas a positive heat of formation means that heat is required to form the product. The overall combustion reaction of glucose releases 2,537.3 kJ/mol of heat.
Always identify the balanced equation for the reaction before you begin to apply Hess’s law. There is a second way of thinking about Hess’s law that may be applicable in some questions as well (see previous topic). The given information in the passage and/or question stem will dictate which equation to use. Finally, recall that enthalpy is a state function, and regardless of the path you take to get from the reactants to the products, the change in enthalpy will be the same.
Things to Watch Out For
At least one of the wrong answer choices for thermochemistry questions will be a result of carelessness with signs. Organized scratchwork in a stepwise fashion will facilitate avoiding this problem, but perhaps more important is maintaining the ability to approximate the answer. Only experience (aka practice!) will breed such wisdom.
1) Given the HF of carbon dioxide and water, what other piece(s) of information must you have to calculate the Hcomb of ethane?
2) If the HF of acetylene is 226.6 kJ/mol, what is the Hcomb of acetylene?
3) If the HF of NaBr (s) is -359.9 kJ/mol, what is the sum of each HF of the following series of five reactions?
Na (s) Na (g) Na+ (g)
Br2(g) Br(g) Br- (g)
PA = XAPTotal (atm)
32 g of oxygen, 28 g of nitrogen, and 22 g of carbon dioxide are confined in a container with partial pressures of 2 atm, 2 atm, and 1 atm respectively. A student added 57 g of a halogen gas to this container and observed that the total pressure increased by 3 atm. Can you identify this gas?
1) Determine the number of moles for each gas.
This is a more complicated style of partial pressure questions, yet the first step is still the basic one of identifying the number of moles for each gas.
2) Solve for the relevant variable.
All partial pressure questions boil down to this formula. The relevant variable here is the mole fraction of the halogen gas. The partial pressure of the gas is 3 atm, and the total pressure is 8 atm.
Partial pressure questions will require manipulation of the formulas above, so the key is always to keep track of what is given to you and what the question is asking for.
3) Use the mole fraction XAto solve for the number of moles and the MW of the halogen gas.
The mole fraction of a substance is the number of moles of the substance as a fraction of the total number of moles in the container:
Rearranging the formula, we have # moles of A = (XA)(total # of moles). Note that the total number of moles is not known, but we can express it algebraically as 2.5 + nA, where nA is defined as the number of moles of A.
# moles of A = (XA)(total # of moles)
Things to Watch Out For
Dalton’s law assumes that the gases do not react with each other.
This MW corresponds to F2. Of course, on Test Day you will roughly round such that 60 g = 1.5 mol and look for the halogen gas using your calculated MW of 40 g/mol. Again, the only gas possible is F2.
1) 64 g of oxygen, 14 g of nitrogen, and 66 g of carbon dioxide are confined in a container. If the total pressure is 10 atm, what is the partial pressure of each gas?
2) The partial pressure of nitrogen is 2 atm. If its mole fraction is 2/10 and the only other gas in the container is oxygen, how many moles of oxygen are in the container?
3) An unknown substance’s mole fraction is 4/10. If its partial pressure is 5 atm, what is the sum pressure of all the other gases in the container?
Normality and Molarity
Oxidation and reduction
What volume of a 2.0 M solution of lithium aluminum hydride in ether is necessary to reduce 1 mole of methyl 5-cyanopentanoate to the corresponding amino alcohol? What if a 2.0 N (with respect to H –) solution were used instead?
1) Determine the number of equivalents of reagent necessary to accomplish the desired transformation.
There are two functional groups that need to be reduced in the molecule, the nitrile, and the ester.
The ester will require two moles of hydride to be reduced to the alcohol.
The nitrile will also require two moles of hydride because it proceeds through an imine intermediate.
The molecular formula of a molecule tells you how many equivalents of a desired reagent/atom are contained within the reagent. This is why it’s important to balance reactions and draw Lewis structures correctly.
2) Compute the necessary volume of the given solution.
Don’t forget that one mole of lithium aluminum hydride contains four moles of hydride.
Things to Watch Out For
Be careful to distinguish between equivalents and moles. With molecules containing several equivalents of reagents, the number of equivalents and moles are not equal!
So we’ll need 500 mL of the 2.0 M solution.
With the 2.0 N solution, things get a little trickier. If the solution is 2.0 N with respect to H-, that means that each liter of solution contains 2 moles of hydride, or 0.5 moles of LiAlH4.
Remember: Normality refers to equivalents per unit volume, not necessarily moles of compound per unit volume.
1) Determine the volumes necessary for the same reaction as that in the initial question if 4.0 M and 4.0 N solutions of lithium aluminum hydride were used instead.
2) If the following molecule were subjected to LiAlH4 reduction as well, what would the product be? How much of the 2.0 M solution would be necessary? The 2.0 N solution?
3) Diisobutylaluminum hydride (DIBAL) is a common alternate hydride-reducing agent. Its structure is shown below. How much of a 2.5 M solution would be necessary to carry out the same reaction described above? A 2.5 N solution?
Common ion effect
Le Châtelier’s principle
Ksp = [A+]xsat [B - ]ysat
The molar solubility of iron(III) hydroxide in pure water at 25°C is 9.94 × 10–10 mol/L. How would the substance’s molar solubility change if placed in an aqueous solution of pH 10.0 at 25°C?
1) Identify the balanced equation for the dissociation reaction.
The generic dissociation reaction may be expressed as follows:
AxBy (s) xA+ (aq) + yB- (aq)
Plugging in for iron(III) hydroxide, the reaction expression is:
Fe(OH)3(s) Fe+3(aq) + 30H-(aq)
This step allows us to see how many moles of ions are added to the solution per mole dissolved.
The value of Ksp does not change when a common ion is present; it is a constant that is dependent on temperature. The molar solubility of the salt, however, does change if a common ion is present. To find the change in molar solubility due to the common ion effect, you must find the Ksp of the substance first.
2) Find the Kspexpression for the dissociation reaction.
Generic: Ksp = [xA+]xsat [yB-]ysat
Fe(OH)3: Ksp = [Fe+3] [OH-]3
Ksp is merely an equilibrium constant, just like K, and it is given the special name “solubility product” because it tells us how soluble a solid is.
Recall that the concentrations are those at equilibrium; thus, the solution is saturated. A saturated solution contains the maximum concentration of dissolved solute.
Remember: Like all Ks, a substance’s Kspvaries only with temperature.
Things to Watch Out For
Be careful when applying Le Châtelier’s principle in cases of precipitation and solvation. For a solution at equilibrium (i.e., saturated), adding more solid would not shift the equilibrium to the right. More solid does not dissociate to raise the ion concentrations; the solid just piles up at the bottom.
3) Calculate molar solubility of each product by assuming that you’re starting with x mols of reactant.
x mol Fe(OH)3x mol Fe +3 + 3x mol OH-
Use the balanced equation from step 1 to determine the appropriate coefficients. For each mole of iron hydroxide dissolved, four ions are created: one Fe+3 and three OH-.
4) Plug the molar solubility for each product into the Kspequation:
Fe(OH)3: Ksp = [Fe+3] [OH-]3
Ksp = [x] [3x]3 = 27x4 (x = 9.94 × 10-10)
Ksp = 27(9.94 × 10-10)4 = 2.64 × 10-35
Simply plug in the coefficients from step 3 into the Ksp equation. The molar solubility, x, was given in the question stem. Thus, to calculate Ksp, plug in 9.94 × 10-10. Now that we are armed with the Ksp, we can find the change in molar solubility due to the common ion.
5) If a common ion is present in a solution, it must also be accounted for in the Kspequation:
Fe(OH)3: Ksp = [Fe+3] [OH-]3
2.64 × 10-35 = x(3x + 10-4)3x(10-4)3
2.64 × 10-35 = x10-12
x = 2.64 × 10-23
A solution of pH 10.0 has an OH- concentration of 10-4. Although iron hydroxide will also contribute to the solution’s total concentration of OH–, its contribution will be negligible relative to the 10-4 already present in solution; thus, when plugging in for the [OH-], we can approximate that it equals 10-4. For the pH 10.0 solution, the molar solubility is on the order of 10-23 mol/L, a steep decrease from the 10-9 in pure water. This is due to the common ion effect. Look at it from the perspective of Le Châtelier’s principle: The addition of more OH- will shift the reaction to the left, so less iron hydroxide will dissociate.
1) Given a substance’s Ksp, how would you solve for its molar solubility in pure water? What if a common ion were also present in solution?
2) Given a table listing substances and their solubility constants, how would you determine which substance was most soluble in pure water?
3) Given that the sulfate ion can react with acid to form hydrogen sulfate, how would the molar solubility of sulfate salts be affected by varying a solution’s pH?
pH and pKa
Acids and bases
What is the pH of the resulting solution if 4 g of sodium acetate (CH3CO2Na) is dissolved in 0.5 L of water? (The pKa of acetic acid, CH3CO2H, is 4.74.)
1) Convert masses to concentrations.
Molecular weight of sodium acetate in g mol-1: 23 + 2(12) + 3(1) + 2(16) = 82. The concentration of a solution is usually expressed in units of moles per liter.
Choose numbers that are easy to work with (i.e., 80 g mol-1 instead of 82 g mol-1). Remember, you won’t have a calculator on Test Day! Moles of sodium acetate:
Remember: Moles, not grams, are the “common currency” of chemistry problems dealing with reactions. Concentration, usually in units of moles per liter, is the essential quantity for these acid-base problems.
2) Choose the constant that will make most sense for the reaction in question and compute its value.
Now we need to decide which constant we will use, Ka or Kb. Here’s where a little common sense goes a long way. If you think about sodium acetate, you should realize that it is the conjugate base of acetic acid. Therefore, we need the value of Kb for sodium acetate. Even though we are given the pKa of acetic acid, getting the Kb from this information is no sweat.
In water, Ka × Kb is always equal to Kw, or 10-14. Take the negative logarithm of both sides.
-log(Ka × Kb) = - log(10-14)
Remember that log(a × b) = log a + log b and that log 10x = x.
-log Ka + -log Kb = - log 10-14 = 14
pKa + pKb = 14
Don’t forget that -log(whatever) = p(whatever). We will want to find Kb because sodium acetate is the conjugate base of acetic acid.
Plug in the pKa from above (to make our lives easier, let’s say 4.74 5).
pKb = 14 - pKa = 9
Kb = 10-pKb = 10-9
Keep in mind that Kb is nothing more than an equilibrium constant for the reaction of a base picking up a proton from water. So all of the things that are true for Keq are true for Kb, especially that Keq only depends on temperature. At constant temperature, it never changes (as the name suggests), even if the concentrations of the species in solution change.
Remember: The p-scale is a hugely important value for acid-base problems. Remember that p(something) = -log(something).
3) Write down the appropriate chemical reaction and set up a table.
Here we need to set up a table to reflect the data we’ve collected. This is the “putting it all together” step and is crucial.
Our table will be as follows:
H2O (l ) + CH3CO2-(aq) CH3CO2H (aq) + OH-(aq)
0.1 - x
Initial: The idea is that we’re going to take some sodium acetate, dump it into water, and see what happens. Our initial row in the table shows the concentrations that we have before any reaction takes place. That means that we’ll start with the amount of sodium acetate we computed, 0.1 M. There’s no acetic acid or hydroxide because no reaction has happened yet.
Change: Here’s where all the action happens. As our acetate reacts with water, the concentration is going to decrease by some amount. We don’t know what that will be yet, so let’s just call it x. If the acetate concentration goes down by x, the concentrations of acetic acid and hydroxide must goup by the same amount, so we put x’s in their columns.
Equilibrium: This is the easy part. Just add up all of the columns above.
Remember: Always, always, always make sure that any chemical reaction you write down is balanced. This means to make sure that mass is balanced (the number of atoms on either side of the reaction) and that charge is balanced. Also, the concentration of pure liquids (e.g., water) and pure solids is never taken into account in equilibria.
Things to Watch Out For
Remember that the cardinal principle of handling computation on the MCAT is to avoid it whenever possible because it is so time consuming and drastically increases the chances of making a mistake. If you can’t avoid doing computation, choose numbers that are easy to work with. So, for example, you wouldn’t want to use 199.9999; you would just use 200.
4) Plug the equilibrium concentrations from the table into the appropriate acidity or basicity expression.
We know that Kb is just the Keq for the reaction in our table above. Plug in the numbers from the table.
5) Simplify the expression from step 4 and solve it.
Let’s make our lives easier (and save ourselves time on Test Day) by assuming that x is much, much smaller than 0.1, so that 0.1 - x 0.1.
Now why would we want to assume that x is very, very small? Well, remember that sodium acetate is a weak base because it has a Kb value. So if it’s a weak base, it won’t react much with water, thus making x a very small number.
Remember: Don’t forget to check the assumption we made to simplify our equation. Because x = 10-5, which is indeed much less than 0.1 (by a factor of 10,000), our assumption holds.
1) What would be the pH if the initial concentration of sodium acetate in the opening question was halved? If it were doubled?
2) Sodium acetate is a buffer, a species that resists changes in pH. To see why this is true, compute the pH of the resulting solution if 0.1 mol of pure NaOH were added to 1 L of water. How does this pH compare to that of the solution with sodium acetate?
3) What would the pH of the solution be if just enough HCl were added to the solution in the original problem to consume all of the sodium acetate?
6) Answer the question.
-log[OH-] = -log[10-5] = 5 = pOH
pH = 14 - 5 = 9
This step is trickier than it sounds and is where many, many mistakes are committed. Here’s where attention to detail counts. You don’t want to slog through all of the work above and then mess up at the end, when 99 percent of the work is done!
Think about what you’ve solved for. What is x? Well, if we look at the table, we see that x is the concentration of hydroxide. So if we take the negative log of the hydroxide ion concentration, we get the pOH.
Remember that pH + pOH = 14. The actual pH is 8.89. So all of our assumptions and roundings didn’t affect the answer much, but they saved us a lot of time in computation!
Remember: Always ask yourself whether your final answer makes sense. The MCAT isn’t a computation test; it’s a test of critical thinking. Here, we have a base being dissolved in water, so at the end of the day, the pH better be above 7, which it is.
Acids and bases
Hydrazoic acid, HN3, is a highly toxic compound that can cause death in minutes if inhaled in concentrated form. 100 mL of 0.2 M aqueous solution of HN3 (pKa = 4.72) is to be titrated with a 0.5 M solution of NaOH.
a) What is the pH of the HN3 solution before any NaOH is added?
b) The half-equivalence point of a titration is where half the titrant necessary to get to the equivalence point has been added. How much of the NaOH solution will be needed to get to the half-equivalence point? What is the pH at the half-equivalence point?
c) What is the pH at the equivalence point?
1) Determine the pH before the titration.
What you need to ask yourself in each stage of this problem is which species is present, H+ or OH-, and where is it coming from? Before the titration begins, we have H+ around because, as the name of the compound suggests, hydrazoic acid is acidic. The major source of H+ is from the hydrazoic acid itself, so we can set up a table as follows:
Setting up the tables as shown makes quick work of titration pH questions. Remember to make approximations and use numbers that are easy to work with in order to minimize the computation necessary to get to the answer.
Plug in the concentrations from the table. We know the pKa is around 5, so the Ka must be 10-5. Whenever exponents or logarithms are involved, use numbers that are easy to work with. Make the approximation that 0.2 - x 0.2 because HN3 is a weak acid.
If you must take a square root, try to get the power of 10 to be even to make matters simple. Remember from the table that x is the hydronium ion concentration.
Remember that -log[a × 10-b] = b - log[a] = somewhere between b - 1 and b. 2 < pH < 3
2) Find the equivalence point and half-equivalence point.
Compute the number of moles of HN3 that you start with.
Each mole of HN3 will react with one mole of OH-. Therefore, the equivalence point is reached when 40 mL of the NaOH solution are added and the half-equivalence point is at 40/2 = 20 mL. We could go through the whole rigamarole of setting up another table to figure out the pH at the half-equivalence point, or we could use a little common sense to avoid computation. At the half-equivalence point, half of the HN3 has been consumed and converted to N3-. Therefore, the HN3 and N3 concentrations are equal.
Because [HN3] = [N3-], Ka = [H3O+], and pH = pKa = 4.72.
Remember: Whenever possible, avoid computation!
Things to Watch Out For
Be careful in choosing whether you will use Ka or Kb to determine the pH. Make this decision based on whether the dominant species in solution is acidic or basic, respectively.
3) Determine the reactive species at the equivalence point to find the pH.
At the equivalence point, all of the HN3 has been consumed, leaving only N3- behind. Because N3- is a Brønsted-Lowry base, we need to worry about OH-, not H3O+.
Remember that Ka × Kb = Kw = 10-14. Now we can set up our table:
Remember that the volume of our solution has increased by 40 mL, so the concentration of N - 3 is 0.02 mol (0.1 + 0.04) L 0.15 M.
Make the approximation that 0.15 - x 0.15.
1) What is the pH after the equivalence point has been exceeded by 5 mL of the NaOH solution in the opening question?
2) What would be the pH if the same amount of NaOH solution necessary to get to the half-equivalence point in this titration were added to pure water? How does the pH of each situation compare? This demonstrates how weak acids can serve as buffers, solutions that resist changes in pH.
3) What would be the pH of a solution that was 0.2 M in HN3 and 0.10 M in N3–?
Here, let’s say that 1.6 is close to 1.5 to make the square root computation trivial.
Note that pH + pOH = 14 in water. Remember to ask yourself whether or not a result makes sense. Here, because we have a basic species (N3-), the pH should be above 7, which it is.
Balancing Redox Reactions
Balancing electrochemical half-reactions
Balance the following reaction that takes place in basic solution.
ZrO(OH)2 (s) + SO32- (aq) Zr (s) + SO42- (aq)
1) Separate the overall reaction into two half-reactions.
Break the reactions up by looking at atoms other than hydrogen and oxygen.
2) Balance the oxygens in each reaction by adding the necessary number of moles of water to the appropriate side.
ZrO(OH)2Zr (s) + 3 H2O
H2O + SO32-SO42-
3) Balance hydrogen by adding the necessary number of H+ions to the appropriate side of each reaction.
4 H+ + ZrO(OH)2Zr + 3 H2O
H2O + SO 32-SO42- + 2 H+
Don’t fall into the trap of simply balancing mass in these reactions. If oxidation and reduction are occurring, you must go through this procedure to balance the reaction.
4) If the reaction is carried out in basic solution, “neutralize” each equivalent of H+with one equivalent of OH-.
4 OH- + 4 H+ + ZrO(OH)2Zr + 3 H2O + 4 OH-
2 OH- + H2O + SO32- SO42- + SO42- + 2 H+ + 2 OH-
4 H2O + ZrO(OH)2Zr + 3 H2O + 4 OH-
2 OH- + H2O + SO32-SO42- + 2 H2O
Combine each mole of H+ and OH- into one mole of water and simplify each reaction.
Remember: Don’t forget to add OH-to each side of both reactions!
Things to Watch Out For
These kinds of problems can be extremely tedious. You must take extra care to avoid careless addition and subtraction errors!
5) Balance the overall charge in each reaction using electrons.
4 e- + 4 H2O + ZrO(OH)2 Zr + 3 H2O + 4 OH-
2 OH- + H2O + SO32- SO42- + 2 H2O + 2 e-
The top equation has a total charge of -4 on the right from the 4 moles of hydroxide, so 4 electrons need to be added to the left side of the equation.
In the bottom equation, there is a total charge of -4 on the left, -2 from the 2 moles of hydroxide and -2 from the 2 moles of sulfite anion (SO32-).
Remember: Don’t forget to account for all charges in this step, including the charge contributed by molecules other than H+and OH-.
6) Multiply each reaction by the necessary integer to ensure that equal numbers of electrons are present in each reaction.
4 e- + 4 H2O + ZrO(OH)2 Zr + 3 H2O + 4 OH-
4 OH- + 2 H2O + 2 SO32- 2 SO42- + 4 H2O + 4 e-
Here, the lowest common multiple among the four electrons in the top reaction and the two in the bottom is four electrons, so we must multiply everything in the bottom reaction by two.
7) Combine both reactions and simplify by eliminating redundant molecules on each side of the reaction.
4 e- + 4 H2O (l ) + ZrO(OH)2 (s) + 4 OH- (aq) + 2 H2O (l ) + 2 SO32- (aq)
Zr (s) + 3 H2O (l) + 4 OH- (aq) + 2 SO42- (aq) + 4 H2O (l) + 4 e-
Combine common terms on each side of the net reaction.
4 e- + 6 H2O (l) + ZrO(OH)2 (s) + 4 OH- (aq) + 2 SO32- (aq)
Zr (s) + 7 H2O (l) + 4 OH- (aq) + 2 SO42- (aq) + 4 e-
Eliminate the redundant water molecules, as well as the electrons and excess hydroxide equivalents.
ZrO(OH)2 (s) + 2 SO32- (aq) Zr (s) + 2 SO42- (aq) + H2O (l)
Check to make sure that the reaction is balanced, in terms of both mass (number of atoms on each side) and overall charge.
Remember: This last step is extremely important. If mass and charge aren’t balanced, then you made an error in one of the previous steps.
1) Which atom is being oxidized in the original equation? Which is being reduced? Identify the oxidizing and reducing agents.
2) A disproportionation is a redox reaction in which the same species is both oxidized and reduced during the course of the reaction. One such reaction is shown below. Balance the reaction, assuming that it takes place in acidic solution:
PbSO4 (s) Pb (s) +
PbO2 (s) + SO42- (aq)
3) Dentists often use zinc amalgams to make temporary crowns for their patients. It is absolutely vital that they keep the zinc amalgam dry. Any exposure to water would cause pain to the patient and might even crack a tooth. The reaction of zinc metal with water is shown below: Zn (s) + H2O (l) Zn2+ (aq) + H2 (g)
Balance this reaction, assuming that it takes place in basic solution. Why would exposure to water cause the crown, and perhaps the tooth, to crack?
E°cell = E°cathode–E°anode (V)
G° =–nFE°cell (kJ/mol)
A galvanic cell is to be constructed using the
MnO4- | Mn2+ (E°red = 1.49 V) and Zn2+ | Zn
(E°red = -0.76 V) couples placed in an acidic solution. Assume that all potentials given are measured against the standard hydrogen electrode at 298 K and that all reagents are present in 1 M concentration (their standard states). What is the maximum possible work output of this cell per mole of reactant if it is used to run an electric motor for one hour at room temperature (298 K)? During this amount of time, how much Zn metal would be necessary to run the cell, given a current of 5 A?
1) Determine which half-reaction is occurring at the anode and which is occurring at the cathode of the cell.
MnO4- (aq) + 5 e- Mn2+ (aq) E°red = 1.49 V
Zn2+ (aq) + 2 e- Zn (s) E°red = -0.76 V
Compare the standard reduction potentials for both reactions. The permanganate reduction potential is greater than the zinc potential, so it would prefer to be reduced and zinc oxidized. Therefore, the zinc is being oxidized at the anode, and the manganese is being reduced at the cathode.
Remember: Oxidation occurs at the anode. (Hint: They both start with a vowel.)
Double-check your work when you balance the cell equation to make sure that you haven’t made any arithmetic errors. One small addition or subtraction mistake can have drastic consequences!
2) Write a balanced reaction for the cell.
Balance the reactions one at a time.
Balance oxygen with water, then hydrogen with acid (H+).
8 H+ + MnO4- Mn2+ + 4 H2O
Balance overall charge with electrons.
8 H+ + MnO4- + 5 e- Mn2+ + 4 H2O
This one is easy; all you have to do is balance electrons.
Zn Zn2+ + 2 e-
To combine both equations, we need to multiply each by the appropriate integer to get to the lowest common multiple of 2 and 5, which is 10.
16 H+ + 2 MnO4- + 10 e- 2 Mn2+ + 8 H2O
5 Zn 5 Zn2+ + 10 e-
Things to Watch Out For
Remember to balance the electron flow in order to figure out the maximum amount of work that the cell can perform.
Now add the equations up to get the balanced cell equation, and you’re golden.
16 H+ (aq) + 2 MnO4- (aq) + 5 Zn (s) 2 Mn2+ (aq)
+ 5 Zn 2+ (aq) + 8 H2O (l)
3) Calculate the standard potential for the cell as a whole.
E°cell = 1.49 V - (-0.76 V) = 2.25 V
Use the equation E°cell = E°cathode - E°anode. Because the standard potential for the cell is positive, this confirms that this is a galvanic (or voltaic) cell—once you hook up the electrodes and immerse them in the designated solutions, current will start to flow on its own.
4) Compute G° for the cell.
G° = -(10 mol e-)(105 C mol-1)(2.25 V) = -2.25 × 106 J mol-1
maximum work output per mole of reactant = 2.25 × 103 KJ mol-1
Use the equation G° = -nFE°cell. The upper limit on the amount of work a reaction can perform is the same thing as G°.
Remember: Power is work over time, and 1 h = 3,600 s 4 × 103s.
5) Use Faraday’s constant to determine the number of moles of electrons transferred and to do any stoichiometric calculations.
4 × 103 s (5 C s-1) (10-5 mol e- C-1) = 0.2 mol e-
Remember that current is charge passing though a point per unit of time, and Faraday’s constant tells us how many coulombs of charge make up one mole of electrons.
The balanced half-reaction is used to determine the necessary mole ratio.
1) How could you alter the cell setup to reverse the direction of current flow?
2) What would the cell potential be if Mn2+ and Zn2+ were at 2 M concentration and the MnO4- concentration remained at 1 M? Would changing the amount of zinc metal present in the cell change this potential? Why or why not?
3) Compute the minimum mass of potassium permanganate (KMnO4) necessary to run the cell for the same amount of time as specified above.
The Nernst Equation
Oxidation and reduction
E°cell = E°cathode - E°anode (V)
A galvanic cell is created at 298 K using the following net reaction:
2 H+ (aq) + Ca (s) Ca2+ (aq) + H2 (g)
Fluoride anions are added to the anode section of the cell only until precipitation is observed. Right at this point, the concentration of fluoride is 1.4 × 10-2 M, the pH is measured to be 0, the pressure of hydrogen gas is 1 atm, and the measured cell voltage is 2.96 V. Given this information, compute the Ksp of CaF2 at 298 K.
R = 8.314 J (mol K)-1
Ca2+ (aq) + 2 e- Ca (s) E°red = -2.76 V
2 H+ (aq) + 2 e- H2 (g) E°red = 0.00 V
F = 96,485 C mol-1
1) Write down the expression for the Ksp.
The first part of this problem begins as with any other solubility problem: We need to write down the expression for the Ksp.
CaF2 (s) Ca2+ (aq) + 2 F- (aq)
We’re given the concentration of fluoride right when precipitation begins, so we can plug that right into the Ksp expression above. All we need is the concentration of Ca2+ ions, and we’re golden.
Keq = [Ca2+][F-]2 = Ksp
Remember what the superscript ° means: that a reaction is at standard conditions. This means that reagents are at 1.0 M or 1 atm, depending on their phase. If you are working with an electrochemical cell where the concentrations are nonstandard, you must apply the Nernst equation to determine what the effect on the cell voltage will be.
2) Separate the net cell reaction into half-reactions and find E°cell.
Ca (s) Ca2+ (aq) + 2 e-E ° = 2.76 V
2 H+ (aq) + 2 e- H2 ( g) E° = 0.00 V
NET: 2 H+ (aq) + Ca (s) Ca2+ (aq) + H2 (g)
E°cell = 0.00 V - (-2.76 V) = 2.76 V
Now we know what the standard potential for the cell is. Our only problem is that, in the situation given in the problem, we are in nonstandard conditions, because the concentration of fluoride is not 1 M.
Remember: E°cell = E°cathode- E°anode.
Things to Watch Out For
Don’t forget to check signs during problems that require computation. When logarithms and exponents are involved, one small sign error can have a massive impact on the answer!
3) Apply the Nernst equation.
The reaction quotient (Q) in this case is of the cell reaction. Recall that the pressure of hydrogen gas is 1 atmosphere. As the pH = 0, [H+] = 10-0 = 1.0 M.
Rearrange the Nernst equation to solve for ln[Ca2+].
Start plugging in numbers. Here, n = 2 mol e-, from the cell equation; F = 96,485 100,000 C mol-1; T = 298 300 K; R 8 J (mol K)-1; and 2.96 - 2.76 = 0.2 = 2 × 10-1 V.
Here, assume–16.7 -20.
–20 = In [Ca2+]
–20 = 2.3 log[Ca2+]
Recall that ln x = 2.3 log x. Assume that 2.3 2.5 so that - –8.
–8 = log[Ca2+ ]
[Ca2+] = 10-8
1) If the Ksp of copper(I) bromide is 4.2 × 10-8, compute the concentration of bromide necessary to cause precipitation in an electrochemical cell with the Cu|Cu+ and H+|H2 couples. Assume the conditions are as follows: E°red of Cu+ = 0.521 V; pH = 0; PH2 (g) = 1 atm; T = 298 K; Ecell when precipitation begins = 0.82 V.
2) Compute the equilibrium constant at 298 K for the cell comprised of the Zn2+ | Zn (Ered° = -0.76 V) and MnO4- | Mn2+ (Ered° = 1.49 V) couples. Given this number, comment on the oxidizing ability of the permanganate anion.
3) A buffer solution is prepared that is 0.15 M in acetic acid and 0.05 M in sodium acetate. If oxidation is occuring at a platinum wire with 1 atm of H2 bubbling over it that is submerged in the buffer solution, and the wire is connected to a standard Cu2+ | Cu half-cell (Ered° = 0.34 V), the measured cell voltage is 0.592 V. Based on this information, compute the pKa of acetic acid.
Remember: When you absolutely must do computation, choose numbers that are easy to work with.
4) Plug the concentrations into the Kspexpression and solve.
Ksp = [Ca2+][F-]2
Ksp = (10-8)(10-2)2 = 10-12
The “actual” value for the Ksp is 3.9 × 10-11, so we are quite close.
Art Credits for General Chemistry
Figure 1.1—Image credited to Slim Films. From The Coming Revolutions in Particle Physics by Chris Quigg. Copyright © 2008 by Scientific American, Inc. All rights reserved.
Figure 1.2—Image credited to Jared Schneidman Designs. From The Earth’s Elements by Robert P. Kirshner. Copyright © 1994 by Scientific American, Inc. All rights reserved.
Figure 6.1 (Intact egg)—Image credited to Richard Drury/Getty Images. From The Cosmic Origins of Time’s Arrow by Sean M. Carroll. Copyright © 2008 by Scientific American, Inc. All rights reserved.
Figure 6.1 (Slightly cracked egg)—Image credited to Graeme Montgomery/Getty Images. From The Cosmic Origins of Time’s Arrow by Sean M. Carroll. Copyright © 2008 by Scientific American, Inc. All rights reserved.
Figure 6.1 (Egg cracked in half)—Image credited to Jan Stromme/Getty Images. From The Cosmic Origins of Time’s Arrow by Sean M. Carroll. Copyright © 2008 by Scientific American, Inc. All rights reserved.
Figure 6.1 (Egg half with yolk)—Image credited to Michael Rosenfeld/Getty Images. From The Cosmic Origins of Time’s Arrow by Sean M. Carroll. Copyright © 2008 by Scientific American, Inc. All rights reserved.
Figure 6.1 (Smashed egg with seeping yolk)—Image credited to Jonathan Kantor/Getty Images. From The Cosmic Origins of Time’s Arrow by Sean M. Carroll. Copyright © 2008 by Scientific American, Inc. All rights reserved.
Figure 6.1 (Over easy egg)—Image credited to Diamond Sky Images/Getty Images. From The Cosmic Origins of Time’s Arrow by Sean M. Carroll. Copyright © 2008 by Scientific American, Inc. All rights reserved.
Figure 8.1—Image credited to Don Foley. From The Ice of Life by David F. Blake and Peter Jenniskens. Copyright © 2001 by Scientific American, Inc. All rights reserved.
Figure 8.6—Image credited to Daniels and Daniels. From The Amateur Scientist: Tackling the Triple Point by Shawn Carlson. Copyright © 1999 by Scientific American, Inc. All rights reserved.