AP Chemistry Unit 9: Thermodynamics and Electrochemistry

Explore AP Chemistry Unit 9 questions on entropy, Gibbs free energy, thermodynamic favorability, equilibrium, electrochemical cells, cell potential, and electrolysis.

Covered syllabus points

9.1 Introduction to Entropy 9.2 Absolute Entropy and Entropy Change 9.3 Gibbs Free Energy and Thermodynamic Favorability 9.4 Thermodynamic and Kinetic Control 9.5 Free Energy and Equilibrium 9.6 Free Energy of Dissolution 9.7 Coupled Reactions 9.8 Galvanic (Voltaic) and Electrolytic Cells 9.9 Cell Potential and Free Energy 9.10 Cell Potential Under Nonstandard Conditions 9.11 Electrolysis and Faraday’s Law

[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 9 Thermodynamics and Electrochemistry question 1 — AP Chemistry

(a)

The reaction of white phosphorus with oxygen to form $\mathrm{P}_{4} \mathrm{O}_{10}(s)$ is thermodynamically

favorable at 298 K. The reaction is represented by equation 1.

[ 2 ]

(i)

The entropy change of the reaction, $\Delta S^{\circ}$ , is negative. Using particle-level reasoning,

explain why the entropy decreases as the reaction progresses.

[ 1 ]

(ii)

The enthalpy change of the reaction, $\Delta H^{\circ}$ , is also negative. A student claims that the

favorability of the reaction is driven by enthalpy and not by entropy. Is the student's

claim correct? Justify your answer by using the relationship between $\Delta G^{\circ}, \Delta H^{\circ}$ , and $\Delta S^{\circ}$ .

$\mathrm{P}_{4} \mathrm{O}_{10}(s)$ reacts exothermically with water to form phosphoric acid, as represented by

equation 2.

A chemist uses a calorimetry experiment to determine the enthalpy change for the reaction, as

represented by the following diagram.

[ 1 ]

Point 03

Yes. Given that $\Delta G_{r x n}^{\circ}=\Delta H_{r x n}^{\circ}-T \Delta S_{r x n}^{\circ}$ , the reaction must have $\Delta G_{r x n}^{\circ}<0$ to be

favorable. Because the reaction is exothermic, $\Delta H_{\text {rxn }}^{\circ}<0$ and enthalpy contributes to

favorability. $\Delta S_{\text {rxn }}^{\circ}<0$ , so entropy does not contribute to favorability.

[Maximum number: 3]

Methanol vapor decomposes to form carbon monoxide gas and hydrogen gas at high temperatures in the presence of a platinum catalyst, as represented by the balanced chemical equation given.

(a)

The values of the standard molar entropies of the compounds involved in the reaction are given in the following table.

[ 3 ]

(i)

Use the data in the table to calculate the value of the standard entropy change, $\Delta S^{\circ}$ , in $\mathrm{J} /\left(\mathrm{K} \cdot \operatorname{mol}_{r x n}\right)$ , for the reaction.

[ 2 ]

(i) For the correct stoichiometry (may be implicit): 1 point

\begin{aligned} & \Delta S^{\circ}{ }_{r x n}=\Sigma \Delta S^{\circ}{ }_{\text {products }}-\Sigma \Delta S^{\circ}{ }_{\text {reactants }} \\ & \Delta S^{\circ}{ }_{r x n}=\left(\Delta S^{\circ}{ }_{\mathrm{CO}(g)}+2\left(\Delta S^{\circ}{ }_{\mathrm{H}_{2}(g)}\right)\right)-\left(\Delta S^{\circ}{ }_{\mathrm{CH}_{3} \mathrm{OH}(g)}\right) \end{aligned}

For the correct calculated value: 1 point

\Delta S^{\circ}{ }_{r x n}=198+2(131)-240 .=220 \cdot \frac{\mathrm{~J}}{\mathrm{~K} \cdot \mathrm{~mol}_{r x n}}

(ii)

Calculate the value of $\Delta G^{\circ}$ , in $\mathrm{kJ} / \mathrm{mol}_{r x n}$ , for the reaction at 375 K. Assume that $\Delta H^{\circ}$ and $\Delta S^{\circ}$ are independent of temperature.

The following particle-level diagram shows a representative sample of the equilibrium mixture represented by the equation given.

[ 1 ]

For the correct calculated value: 1 point

\begin{aligned} & \Delta G^{\circ}=\Delta H^{\circ}-T \Delta S^{\circ} \\ & \Delta G^{\circ}=90.0 \frac{\mathrm{~kJ}}{\mathrm{~mol}_{r x n}}-(375 \mathrm{~K})\left(0.220 \frac{\mathrm{~kJ}}{\mathrm{~K} \cdot \mathrm{~mol}_{r x n}}\right)=+7.5 \mathrm{~kJ} / \mathrm{mol}_{r x n} \end{aligned}

Total for part (c) 3 points

[Maximum number: 3]

Answer the following questions about the element Si and some of its compounds.

(a)

Explain why the absolute molar entropy of Si(s) is less than that of $\mathrm{H}_{2}(g)$ .

[ 1 ]

For a correct explanation:

The $H_{2}(g)$ molecules are more highly dispersed than the Si(s) atoms and, therefore, have a higher absolute molar entropy. Silicon is a solid; therefore, its atoms are in fixed positions, are less dispersed, and have a lower absolute molar entropy.

1 point

(b)

Calculate the value, in $\mathrm{J} /(\mathrm{mol} \cdot \mathrm{K})$ , of $\Delta S^{\circ}$ for the reaction.

[ 1 ]

For the correct calculated value:

$\Delta S_{\text {rxn }}^{\circ}=(18+2(131))-205=+75 \mathrm{~J} /\left(\mathrm{mol}_{\text {rxn }} \cdot \mathrm{K}\right)$

1 point

(c)

The reaction is thermodynamically favorable at all temperatures. Explain why the reaction occurs only at high temperatures.

[ 1 ]

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Covered syllabus points

9 Thermodynamics and Electrochemistry question 1

Question (a)

Question (i)

Question (ii)

9 Thermodynamics and Electrochemistry question 2

Question (a)

Question (i)

Question (ii)

9 Thermodynamics and Electrochemistry question 3

Question (a)

Question (b)

Question (c)