## SAT Subject Test Chemistry

__PART 2__

__REVIEW OF MAJOR TOPICS__

__CHAPTER 10__

__CHAPTER 10__

__Chemical Equilibrium__

__Chemical Equilibrium__

**These skills are usually tested on the SAT Subject Test in Chemistry. You should be able to …**

• Explain the development of an equilibrium condition and how it is expressed as an equilibrium constant, and use it mathematically.

• Describe Le Châtelier’s Principle and how changes in temperature, pressure, and concentrations affect an equilibrium.

• Solve problems dealing with ionization of water, finding the pH, solubility products, and the common ion effect.

• Explain the relationship of enthalpy and entropy as driving forces in a reaction and how they are combined in the Gibbs equation.

**This chapter will review and strengthen these skills. Be sure to do the Practice Exercises at the end of the chapter.**

**I**n some reactions no product is formed to allow the reaction to go to completion; that is, the reactants and products can still interact in both directions. This can be shown as follows:

A + B C + D

The double arrow indicates that C and D can react to form A and B, while A and B react to form C and D.

__REVERSIBLE REACTIONS AND EQUILIBRIUM__

__REVERSIBLE REACTIONS AND EQUILIBRIUM__

The reaction is said to have reached **equilibrium **when the forward reaction rate is equal to the reverse reaction rate. Notice that this is a dynamic condition, NOT a static one, although in appearance the reaction *seems *to have stopped. An example of an equilibrium is a crystal of copper sulfate in a saturated solution of copper sulfate. Although to the observer the crystal seems to remain unchanged, there is actually an equal exchange of crystal material with the copper sulfate in solution. As some solute comes out of solution, an equal amount is going into solution.

**TIP **

**Equilibrium **is reached when the forward and reverse reaction rate are equal.

To express the rate of reaction in numerical terms, we can use the **Law of Mass Action**, discussed in Chapter 9, which states: The rate of a chemical reaction is proportional to the product of the concentrations of the reacting substances. The concentrations are expressed in moles of gas per liter of volume or moles of solute per liter of solution. Suppose, for example, that 1 mole/liter of gas A_{2} (diatomic molecule) is allowed to react with 1 mole/liter of another diatomic gas, B_{2}, and they form gas AB; let *R *be the rate for the forward reaction forming AB. The bracketed symbols [A_{2}] and [B_{2}] represent the concentrations in moles per liter for these diatomic molecules. Then A_{2} + B_{2} → 2AB has the rate expression

*R *[A_{2}]×[B_{2}]

where is the symbol for “proportional to.” When [A_{2}] and [B_{2}] are both 1 mole/liter, the reaction rate is a certain constant value (*k*_{1}) at a fixed temperature.

*R *= *k*_{1} (*k*_{1} is called the rate constant)

For any concentrations of A and B, the reaction rate is

*R*=*k** _{1}* ×[A

_{2}]×[B

_{2}]

If [A_{2}] is 3 moles/liter and [B_{2}] is 2 moles/liter, the equation becomes

*R *= *k*_{1} × 3 × 2 = 6*k*_{1}

The reaction rate is six times the value for a 1 mole/liter concentration of each reactant.

At the fixed temperature of the forward reaction, AB molecules are also decomposing. If we designate this reverse reaction as *R*´, then, since

2AB (or AB + AB)→A_{2}+B_{2}

two molecules of AB must decompose to form one molecule of A_{2} and one of B_{2}. Thus the reverse reaction in this equation is proportional to the square of the molecular concentration of AB:

where *k*_{2} represents the rate of decomposition of AB at the fixed temperature. Both reactions can be shown in this manner:

A_{2} + B_{2} 2AB (note double arrow)

When the first reaction begins to produce AB, some AB is available for the reverse reaction. If the initial condition is only the presence of A_{2} and B_{2} gases, then the forward reaction will occur rapidly to produce AB. As the concentration of AB increases, the reverse reaction will increase. At the same time, the concentrations of A_{2} and B_{2} will be decreasing and consequently the forward reaction rate will decrease. Eventually the two rates will be equal, that is, *R *= *R*´. At this point, equilibrium has been established, and

*k*_{1}[A_{2}]×[B_{2}]=k_{2}[AB]^{2}

or

The convention is that *k*_{1} (forward reaction) is placed over *k*_{2} (reverse reaction) to get this expression. Then *k*_{1}/*k*_{2} can be replaced by *K** _{eq}*, which is called the

**equilibrium constant**for this reaction under the particular conditions.

In another general example:

*a*A + *b*B *c*C + *d*D

the reaction rates can be expressed as

Note that the values of *k*_{1} and *k*_{2} are different, but that each is a constant for the conditions of the reaction. At the start of the reaction, [A] and [B] will be at their greatest values, and *R *will be large; [C], [D], and *R*′ will be zero. Gradually *R *will decrease and *R*′ will become equal. At this point the reverse reaction is forming the original reactants just as rapidly as they are being used up by the forward reaction. Therefore no further change in *R*, *R*´, or any of the concentrations will occur.

If we set *R*´ equal to *R*, we have:

*k** _{2}*×[C]

*×[D]*

^{c}*=k*

^{d}_{1}×[A]

*×[B]*

^{a}

^{b}or

This process of two substances, A and B, reacting to form products C and D and the reverse can be shown graphically to represent what happens as equilibrium is established. The hypothetical equilibrium reaction is described by the following general equation:

*a*A + *b*B *c*C + *d*D

At the beginning (time *t *_{0}), the concentrations of C and D are zero and those of A and B are maximum. The graph below shows that over time the rate of the forward reaction decreases as A and B are used up. Meanwhile, the rate of the reverse reaction increases as C and D are formed. When these two reaction rates become equal (at time *t*_{1}), equilibrium is established. The individual concentrations of A, B, C, and D no longer change if conditions remain the same. To an observer, it appears that all reaction has stopped; in fact, however, both reactions are occurring at the same rate.

**TIP **

Notice that as the forward reaction rate decreases, the reverse reaction rate increases until they are at “equilibrium.”

We see that, for the given reaction and the given conditions, *K*_{eq}* *is a constant, called the **equilibrium constant**. If *K*_{eq}* *is large, this means that equilibrium does not occur until the concentrations of the original reactants are small and those of the products large. A small value of *K*_{eq}* *means that equilibrium occurs almost at once and relatively little product is produced.

The equilibrium constant, *K** _{eq}*, has been determined experimentally for many reactions, and the values are given in chemical handbooks.

Suppose we find *K*_{eq}* *for reacting H_{2} and I_{2} at 490°C to be equal to 45.9. Then the equilibrium constant for this reaction

H_{2}(g) + I_{2}(g) 2HI(g) at 490°C

is

**SAMPLE PROBLEM:** Three moles of H_{2}(g) and 3.00 moles of I_{2}(g) are introduced into a 1-liter box at a temperature of 490°C. Find the concentration of each substance in the box when equilibrium is established.

Initial conditions:

[H_{2}]=3.00 mol/L

[I_{2}]=3.00 mol/L

[HI]=0 mol/L

The reaction proceeds to equilibrium and

At equilibrium, then,

[H |
(where |

[I |
(the same |

[HI] = 2*x *mol/L

Then

If

then taking the square root of each side gives

Solving for *x*:

*x *= 2.32

**Note: On the SAT Subject Test in Chemistry, calculators are not permitted. So answers to questions like this would be easily calculated “perfect squares.”**

Substituting this *x *value into the concentration expressions at equilibrium we have

[H_{2}]=(3.00−*x*)=0.68 mol/L

[I_{2}]=(3.00−*x*)=0.68 mol/L

[HI]=2*x*=4.64 mol/L

The crucial step in this type of problem is setting up your concentration expressions from your knowledge of the equation. Suppose that this problem had been:

**SECOND SAMPLE PROBLEM:** Find the concentrations at equilibrium for the same conditions as in the preceding example except that only 2.00 moles of HI are injected into the box.

[H_{2}]=0 mol/L

[I_{2}]=0 mol/L

[HI]=2.00 mol/L

At equilibrium

[HI]=(2.00−*x*)mol/L

(For every mole of HI that decomposes, only 0.5 mole of H_{2} and 0.5 mole of I_{2} are formed.)

[H_{2}]=0.5*x*

[I_{2}]=0.5*x*

Solving for *x *gives:

*x*=0.456

Then, substituting into the equilibrium conditions,