AP Chemistry Unit 9.3: Gibbs Free Energy

Practice AP Chemistry Unit 9.3 questions on calculating ΔG°, judging thermodynamic favorability, and explaining enthalpy, entropy, and temperature effects.

[Maximum number: 1]

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 9.3 Gibbs Free Energy and Thermodynamic Favorability question 1 — AP Chemistry

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.

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.

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.

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Covered syllabus points

9.3 Gibbs Free Energy and Thermodynamic Favorability question 1