AP Chemistry Unit 6: Thermochemistry

Explore AP Chemistry Unit 6 questions on endothermic and exothermic changes, calorimetry, phase-change energy, reaction enthalpy, and Hess law.

Covered syllabus points

6.1 Endothermic and Exothermic Processes 6.2 Energy Diagrams 6.3 Heat Transfer and Thermal Equilibrium 6.4 Heat Capacity and Calorimetry 6.5 Energy of Phase Changes 6.6 Introduction to Enthalpy of Reaction 6.7 Bond Enthalpies 6.8 Enthalpy of Formation 6.9 Hess’s Law

[Maximum number: 3]

Answer the following questions about an experiment in which $\mathrm{CaCO}_{3}(s)$ is combined with HCl(a q), represented by the following balanced equation.

\mathrm{CaCO}_{3}(s)+2 \mathrm{HCl}(a q) \rightarrow \mathrm{CaCl}_{2}(a q)+\mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l)

(a)

Is the reaction endothermic or exothermic? Justify your answer using the information in the table.

[ 1 ]

(b)

Based on the experimental data, the mass of the system is 51.0 g, and the specific heat of the reaction mixture is $4.0 \mathrm{~J} /\left(\mathrm{g} \cdot{ }^{\circ} \mathrm{C}\right)$ .

[ 2 ]

(i)

(ii)

For the correct calculated value (sign not required):
$q_{\text {surr }}=m c \Delta T=(51.0 \mathrm{~g})\left(4.0 \frac{\mathrm{~J}}{\mathrm{~g} \cdot{ }^{\circ} \mathrm{C}}\right)\left(21.90^{\circ} \mathrm{C}-21.20^{\circ} \mathrm{C}\right)=140 \mathrm{~J}$

1 point

(i)

Calculate the magnitude of heat transfer, q, in joules.

[ 1 ]

(ii)

For the correct calculated value (sign not required):
$q_{\text {surr }}=m c \Delta T=(51.0 \mathrm{~g})\left(4.0 \frac{\mathrm{~J}}{\mathrm{~g} \cdot{ }^{\circ} \mathrm{C}}\right)\left(21.90^{\circ} \mathrm{C}-21.20^{\circ} \mathrm{C}\right)=140 \mathrm{~J}$

1 point

(ii)

Calculate the enthalpy of reaction in units of $\mathrm{kJ} / \mathrm{mol}_{r x n}$ . Include the algebraic sign on your answer.

[ 1 ]

$\begin{aligned} \text { For the correct calculated value, consistent with ( } \mathrm{g} \text { )(i), and the cor } \text { (f): } q_{s y s}=-q_{s u r r}=-140 \mathrm{~J}=-0.14 \mathrm{~kJ} 1.00 \mathrm{~g} \mathrm{CaCO}_{3} \times \frac{1 \mathrm{~mol} \mathrm{CaCO}_{3}}{100.09 \mathrm{~g} \mathrm{CaCO}_{3}} \times \frac{1 \mathrm{~mol}_{r x n}}{1 \mathrm{~mol} \mathrm{CaCO}_{3}}=0.00999 \mathrm{~mol}_{r x n} \Delta H_{r x n}^{\circ}=\frac{-0.14 \mathrm{~kJ}}{0.00999 \mathrm{~mol}_{r x n}}=-14 \mathrm{~kJ} / \mathrm{mol}_{r x n} \end{aligned}$

1 point

Total for part (g)

for question 3

10 points

[Maximum number: 1]

Which of the following diagrams represents a potential energy diagram of an endothermic reaction?

Figure for Question 6 Thermochemistry question 2 — AP Chemistry — Option A

Figure for Question 6 Thermochemistry question 2 — AP Chemistry — Option B

Figure for Question 6 Thermochemistry question 2 — AP Chemistry — Option C

Figure for Question 6 Thermochemistry question 2 — AP Chemistry — Option D

[Maximum number: 4]

A student performs an experiment to determine the specific heat capacity of a metal. The student places a cube of the metal in boiling water so its temperature will be 100.0°C. The student then places the metal cube into a calorimeter that contains water and records the highest temperature of the water. A data table and a diagram of the thermometer at the highest temperature are shown.

Figure for Question 6 Thermochemistry question 3 — AP Chemistry

(a)

What should the student report as the highest temperature of the water?

[ 1 ]

For the correct answer, reported to the correct decimal place: 1 point $38.5^{\circ} \mathrm{C}$

(b)

A particle-level representation of water molecules in the calorimeter before and after the metal cube was added is shown. The length of the arrows in the Before diagram represents the speed of the water molecules in the system. In the After diagram, draw an arrow for each molecule to indicate how the speed of each of the molecules changes after the metal cube is added.

[ 1 ]

(c)

Assuming the metal transfers 2940 J of thermal energy to the water, calculate the specific heat of the metal in $\mathrm{J} /\left(\mathrm{g} \cdot{ }^{\circ} \mathrm{C}\right)$ .

[ 1 ]

For the correct calculated value, consistent with part (a): 1 point

\begin{aligned} & q=m c \Delta T \\ & c_{\text {metal }}=\frac{q_{\text {metal }}}{m_{\text {metal }} \Delta T_{\text {metal }}}=\frac{-2940 \mathrm{~J}}{(98.1 \mathrm{~g})\left(38.5^{\circ} \mathrm{C}-100.0^{\circ} \mathrm{C}\right)}=0.487 \frac{\mathrm{~J}}{\mathrm{~g} \cdot{ }^{\circ} \mathrm{C}} \end{aligned}

(d)

In a second experiment, 2940 J of thermal energy is transferred from 98.1 g of aluminum, which has a specific heat capacity of $0.897 \mathrm{~J} /\left(\mathrm{g} \cdot{ }^{\circ} \mathrm{C}\right)$ . Explain how the magnitude of the temperature change of the aluminum, $\Delta T_{\mathrm{Al}}$ , compares with the magnitude of the temperature change of the metal in the original experiment.

[ 1 ]

For a valid explanation, consistent with part (c): 1 point
Accept one of the following:

- The value of $\Delta T_{\mathrm{Al}}$ will be smaller because Al has a greater specific heat capacity than the metal in the original experiment. Therefore, the same thermal energy transfer applied to the same mass will result in a smaller change in temperature, according to the equation $q=m c \Delta T$ .

- $q=m c \Delta T$

\begin{aligned} & \left|\Delta T_{\mathrm{Al}}\right|=\left|\frac{q_{\mathrm{Al}}}{m_{\mathrm{Al}} c_{\mathrm{Al}}}\right|=\left|\frac{-2940 \mathrm{~J}}{(98.1 \mathrm{~g})\left(0.897 \frac{\mathrm{~J}}{\mathrm{~g} \cdot \mathrm{C}}\right)}\right|=33.4^{\circ} \mathrm{C} \\ & \left|\Delta T_{\text {metal }}\right|=\left|38.5^{\circ} \mathrm{C}-100.0^{\circ} \mathrm{C}\right|=61.5^{\circ} \mathrm{C} \\ & \text { Thus, } \Delta T_{\mathrm{Al}}<\Delta T_{\text {metal }} \end{aligned}

[Maximum number: 5]

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 Thermochemistry question 4 — AP Chemistry

(a)

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

[ 3 ]

(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

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

(ii)

Calculate the value of $\Delta H_{\text {rxn }}^{\circ}$ for equation 2 in $\mathrm{kJ} / \mathrm{mol}_{\text {rxn }}$ . Include the sign in your answer.

[ 2 ]

(ii) For the correct calculated value, consistent with part C (i):

Point 05

$\begin{aligned} & q_{r x n}=-q_{s u r r}=-0.16 \mathrm{~kJ} \\ & \Delta H_{r x n}^{\circ}=\frac{-0.16 \mathrm{~kJ}}{0.100 \mathrm{~g} \mathrm{P}_{4} \mathrm{O}_{10}} \times \frac{283.9 \mathrm{~g} \mathrm{P}_{4} \mathrm{O}_{10}}{1 \mathrm{~mol} \mathrm{P}_{4} \mathrm{O}_{10}} \times \frac{1 \mathrm{~mol} \mathrm{P}_{4} \mathrm{O}_{10}}{1 \mathrm{~mol}_{r x n}}=-450 \mathrm{~kJ} / \mathrm{mol}_{r x n} \end{aligned}$

For the correct sign:

Point 06

$-450 \mathrm{~kJ} / \mathrm{mol}_{r x n}$

(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}

[ 1 ]

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

(c)

Calculate the standard enthalpy of formation of $\mathrm{PCl}_{5}(g)$ represented by equation 5.

\text { Equation 5: } \frac{1}{4} \mathrm{P}_{4}(s)+\frac{5}{2} \mathrm{Cl}_{2}(g) \rightarrow \mathrm{PCl}_{5}(g) \quad \Delta H_{f}^{\circ}=\text { ? }

The following particle-level diagram represents the contents of the vessel in an equilibrium

mixture at 546 K involving equation 4.

[ 1 ]

E For the correct calculated value:

Using Hess's law:

$\begin{aligned} \Delta H_{f, \mathrm{PCl}_{5}(g)}^{\circ} & =\frac{1}{4} \Delta H_{1}^{\circ}+\Delta H_{2}^{\circ} \\ \Delta H_{f, \mathrm{PCl}_{5}(g)}^{\circ} & =\frac{1}{4}(-1148)+(-88)=-375 \mathrm{~kJ} / \mathrm{mol} \end{aligned}$

All question bank results loaded

Covered syllabus points

6 Thermochemistry question 1

Question (a)

Question (b)

Question (i)

Question (ii)

6 Thermochemistry question 2

6 Thermochemistry question 3

Question (a)

Question (b)

Question (c)

Question (d)

6 Thermochemistry question 4

Question (a)

Question (i)

Question (ii)

Question (b)

Question (c)