Cracking the AP Chemistry Exam

Part IV

Content Review for the AP Chemistry Exam

Chapter 7

Big Idea #5: Laws of Thermodynamics and Changes in Matter


Heat capacityCP , is a measure of how much the temperature of an object is raised when it absorbs heat.

An object with a large heat capacity can absorb a lot of heat without undergoing much of a change in temperature, whereas an object with a small heat capacity shows a large increase in temperature even if only a small amount of heat is absorbed.

Specific heat is the amount of heat required to raise the temperature of one gram of a substance one degree Celsius.

Specific Heat

q = mcΔT

q    = heat added (J or cal)

m   = mass of the substance (g or kg)

c    = specific heat

T = temperature change (K or °C)

Calorimetry is the measurement of heat changes during chemical reactions, and is frequently accomplished via the equation above. Determining the amount of heat transfer in a reaction can lead directly to determining the enthalpy for that reaction, as shown in the example below.

H+ + OH → H2O(l)

25.0 mL of 1.5 M HCl and 30.0 mL of 2.0 M NaOH are mixed together in a styrofoam cup and the reaction above occurs. The temperature of the reaction rises from 23.00°C to 31.60°C over the course of the reaction. Assuming the density of the solutions is 1.0 g/mL and the specific heat of the mixture is 4.18 J/g°C, calculate the enthalpy of the reaction.

To solve this, we first need to determine the amount of heat released during the reaction. The mass of the final solution can be determined by taking the total volume of the solution (55.0 mL) and multiplying that by the density, yielding 55.0g. The temperature change is 31.60 − 23.00 = 8.60°C. So:

q = mcT

q = (55.0g)(4.18J/g°×C)(8.60°C)

q = 1970 J

From here, we need to determine how many moles of product are formed in this reaction. Looking at the two reactants, we can see that there is going to be less of the HCl. As everything here is in a 1:1 ratio, that means the HCl will be limiting and we can use it to determine how many moles of product will form.

Molarity = mol/volume

1.5 M = n/0.025 L

n = 0.038 mol HCl = 0.038 mol H2O formed

The calculated heat gain from earlier (1970 J) was the heat gained by the water. Due to conservation of energy, this is also the heat lost by the reaction itself. To calculate the enthalpy of reaction, we have to flip that sign to indicate heat is lost, and then divide that value by the number of moles.

H = −1970 J/0.038 mol = −52,000 J/mol = −52 kJ/mol

It is important to emphasize that ∆H is always measured in joules per mole (or kilojoules per mole). Enthalpy is not just heat; it’s the amount of heat given per mole of product that is created. Limiting reagent calculations can be done in situations where it is not easy to determine which reactant is limiting by inspection, as we did above.