AP Chemistry Unit 6.4: Calorimetry

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

[Maximum number: 2]

White phosphorus is composed of $\mathrm{P}_{4}$ 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 6.4 Heat Capacity and Calorimetry question 1 — AP Chemistry

(a)

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

[ 1 ]

(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 ]

C (i) For the correct calculated value reported with the correct number of significant figures:

Point 04

$\begin{aligned} & q=m c \Delta T=(100.1 \mathrm{~g})\left(4.18 \mathrm{~J} /\left(\mathrm{g} \cdot{ }^{\circ} \mathrm{C}\right)\right)\left(22.38^{\circ} \mathrm{C}-22.00^{\circ} \mathrm{C}\right) \\ & q=160 \mathrm{~J}=0.16 \mathrm{~kJ} \end{aligned}$

(b)

The chemist weighed out $0.100 \mathrm{~g} \mathrm{P}_{4} \mathrm{O}_{10}$ of and 100.0 g of $\mathrm{H}_{2} \mathrm{O}$ to perform a second trial. In the

second trial, some of the solid $\mathrm{P}_{4} \mathrm{O}_{10}$ stuck to the weighing paper and was not transferred to

the calorimeter. Given that $\mathrm{P}_{4} \mathrm{O}_{10}$ is the limiting reactant, would $\Delta T$ for the second trial be

greater than, less than, or equal to the value in the first trial? Justify your answer.

$\mathrm{P}_{4}(s)$ also reacts readily with $\mathrm{Cl}_{2}(g)$ to produce phosphorus trichloride, $\mathrm{PCl}_{3}(g)$ , which in turn

reacts with $\mathrm{Cl}_{2}(g)$ in an equilibrium process to produce $\mathrm{PCl}_{5}(g)$ . The reactions are represented

by equations 3 and 4.

\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}

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D For the correct answer and a valid justification:

Point 07

Less than. If less $\mathrm{P}_{4} \mathrm{O}_{10}$ is present, less thermal energy will be transferred to the water

during the reaction, causing the temperature increase to be less than it was with 0.100 g

of $\mathrm{P}_{4} \mathrm{O}_{10}$ .

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AP Chemistry Unit 6.4: Calorimetry

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6.4 Heat Capacity and Calorimetry question 1

Question (a)

Question (i)

Question (b)