AP Chemistry: 9.5 Free Energy and Equilibrium – Exam Style questions with Answer- MCQ

Question

    

▶️Answer/Explanation

Ans:A

The graph for reaction 1 shows that when equilibrium is reached (no observable changes in the concentrations of reactant and product), the concentration of the product is much larger than the concentration of the reactant, hence, \(\frac{[Product]}{[Reactant]}\)>>1. ΔG°\(_{rxn}=−RTlnK\) and \(K=\frac{[Product]}{[Reactant]}\)>>1, hence, ΔG°\(_{rxn}<0\) because the natural logarithm of a number greater than 1 is positive. ΔG°\(_{rxn}<0\) is the criterion that determines if a process is thermodynamically favorable at constant T and P. Applying similar reasoning to the graph for reaction 2, \(\frac{[Product]}{[Reactant]}\)<<1 and ΔG°\(_{rxn}>0\).

Question

                                          \(N_2(g)+O_2(g)→2NO(g)\)        \(K=1.0\times 10^{−30}\)

At 720°C , the reaction represented above has an equilibrium constant of  \(1.0\times 10^{−30}\). Which of the following mathematical expressions can be used to support the claim that the reaction is not thermodynamically favored under these conditions?
A ΔG°\(_{rxn}=\frac{−8.314×(720+273)×ln(1.0\times 10^{−30})}{1,000}>0\)

B ΔG°\(_{rxn}\)=−0.08206×(720+273)×ln\(1.0\times 10^{−30}\)×1,000>0

C ΔG°\(_{rxn}=\frac{8.314×(720)×ln(1.0\times 10^{−30})}{1,000}<0\)

D ΔG°\(_{rxn}\)=0.08206×(720+273)×ln\(1.0\times 10^{−30}\)×1,000<0

▶️Answer/Explanation

Ans:A

ΔG°\(_{rxn}=−RTlnK\) where T is the temperature in Kelvin and R=8.314 J/(mol⋅K). Based on the information given and after substituting the appropriate values,ΔG°\(_{rxn}=\frac{−8.314×(720+273)×ln(1.0\times 10^{−30})}{1,000}\) . A process is not thermodynamically favored when ΔG°\(_{rxn}>0\). Because K<<1, the term lnK is negative, resulting in ΔG°\(_{rxn}>0\).

Question

                                                                  W(s)→W(l)

                                              \( \Delta H^{\circ}_{fusion}=35 kJ/mol\) ; \(\Delta S^{\circ}_{fusion}=10J/(mol.K)\)

The chemical equation and thermodynamic data for the melting of tungsten are given above. Based on this information, which of the following provides the best prediction with correct justification about whether a sample of pure tungsten will melt at 3723K ?
A The sample will not melt because \(T>\frac{\Delta H^{\circ}_{fusion}}{\Delta S^{\circ}_{fusion}}\)
B The sample will not melt because \(T<\frac{\Delta H^{\circ}_{fusion}}{\Delta S^{\circ}_{fusion}}\) .
C The sample will melt because     \(T>\frac{\Delta H^{\circ}_{fusion}}{\Delta S^{\circ}_{fusion}}\) .
D The sample will melt because    \(T<\frac{\Delta H^{\circ}_{fusion}}{\Delta S^{\circ}_{fusion}}\) .

▶️Answer/Explanation

Ans:C

At the melting point, the solid and liquid phases are in equilibrium and \(\Delta S^{\circ}_{fusion}=\Delta H^{\circ}_{fusion}−Tm\Delta S^{\circ}_{fusion}=0\) , where \(T_m\) is the melting point in kelvins. Rearranging the equation to solve for \(T_m\) gives \(T_m=\frac{\Delta H^{\circ}_{fusion}}{\Delta S^{\circ}_{fusion}}=\frac{350}{.010}=3500K\) for tungsten. Because T(3723K) is greater than \(T_m\), the sample will melt.

Refer to the following information.

Two molecules of the amino acid glycine join through the formation of a peptide bond, as shown above. The thermodynamic data for the reaction are listed in the following table.

Question

Based on the bond energies listed in the table above, which of the following is closest to the bond energy of the C-N bond?

(A) 200 kJ/mol

(B) 300 kJ/mol

(C) 400 kJ/mol

(D) 500 kJ/mol

▶️Answer/Explanation

Ans:C

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