AP Chemistry 6.4 Heat Capacity and Calorimetry Questions

Practice AP Chemistry Unit 6.4 questions on calculating heat, comparing specific heats, interpreting temperature changes, and evaluating calorimetry measurements.

Syllabus
Effective Fall 2025
Course
AP Chemistry

Exam points

  • Use reaction equations and stoichiometry to analyze aqueous reactions, limiting reagents, yields, and composition.
  • Calculate concentration, mass, moles, and gas quantities from balanced reactions and measured experimental data.

Question 1

[Maximum number: 2]

White phosphorus is composed of P4\mathrm{P}_{4} molecules with a tetrahedral structure, as shown in the diagram on the left. Each P atom is bonded to the other three P atoms by single bonds, as shown in the incomplete Lewis diagram on the right.

Figure for Question 1 — AP Chemistry
Figure for Question 1 — AP Chemistry

Question (a)

(a)

The chemist carries out the calorimetry experiment and records the following information.

Table for Question (a) — AP Chemistry
[ 1 ]

Question (i)

(i)

Calculate the amount of heat, q, released during the experiment, in kJ. Assume that the specific heat of the solution is the same as that of water.

[ 1 ]

Question (b)

(b)

The chemist weighed out 0.100 gP4O100.100 \mathrm{~g} \mathrm{P}_{4} \mathrm{O}_{10} of and 100.0 g of H2O\mathrm{H}_{2} \mathrm{O} to perform a second trial. In the second trial, some of the solid P4O10\mathrm{P}_{4} \mathrm{O}_{10} stuck to the weighing paper and was not transferred to the calorimeter. Given that P4O10\mathrm{P}_{4} \mathrm{O}_{10} is the limiting reactant, would ΔT\Delta T for the second trial be greater than, less than, or equal to the value in the first trial? Justify your answer.

P4(s)\mathrm{P}_{4}(s) also reacts readily with Cl2(g)\mathrm{Cl}_{2}(g) to produce phosphorus trichloride, PCl3(g)\mathrm{PCl}_{3}(g), which in turn reacts with Cl2(g)\mathrm{Cl}_{2}(g) in an equilibrium process to produce PCl5(g)\mathrm{PCl}_{5}(g). The reactions are represented by equations 3 and 4.

 Equation 3: P4(s)+6Cl2(g)4PCl3(g)ΔH1=1148 kJ/molrxn Equation 4: PCl3(g)+Cl2(g)PCl5(g)ΔH2=88 kJ/molrxn\begin{array}{ll} \text { Equation 3: } \mathrm{P}_{4}(s)+6 \mathrm{Cl}_{2}(g) \rightarrow 4 \mathrm{PCl}_{3}(g) & \Delta H_{1}^{\circ}=-1148 \mathrm{~kJ} / \mathrm{mol}_{r x n} \\ \text { Equation 4: } \mathrm{PCl}_{3}(g)+\mathrm{Cl}_{2}(g) \rightleftarrows \mathrm{PCl}_{5}(g) & \Delta H_{2}^{\circ}=-88 \mathrm{~kJ} / \mathrm{mol}_{r x n} \end{array}
[ 1 ]
All question bank results loaded