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Chapter 16: Problem 98

Consider the reaction: $$\mathrm{H}_{2} \mathrm{S}(g)+\mathrm{SO}_{2}(g) \longrightarrow 3 \mathrm{S}(g)+2 \mathrm{H}_{2} \mathrm{O}(l)$$ for which \(\Delta H\) is \(-233 \mathrm{kJ}\) and \(\Delta S\) is \(-424 \mathrm{J} / \mathrm{K}.\) a. Calculate the free energy change for the reaction \((\Delta G)\) at \(393 \mathrm{K}.\) b. Assuming \(\Delta H\) and \(\Delta S\) do not depend on temperature, at what temperatures is this reaction spontaneous?

Short Answer

Expert verified
(a) The Gibbs free energy change (ΔG) at 393 K is -66,368 J. (b) The reaction is spontaneous at temperatures greater than 549.06 K.

Step by step solution

01

Identify the Given Variables

Enthalpy Change (ΔH) = -233 kJ Entropy Change (ΔS) = -424 J/K Temperature (T) = 393 K Note: Enthalpy and entropy changes should be in the same unit, so we will need to convert ΔH to J (joules).
02

Convert ΔH to J

1 kJ = 1000 J ΔH = -233 kJ × 1000 J/kJ = -233,000 J
03

Calculate ΔG using the Gibbs Free Energy Formula

The formula for Gibbs free energy change (ΔG) is: ΔG = ΔH - TΔS Now, plug in the values for ΔH, T, and ΔS: ΔG = -233,000 J - (393 K × -424 J/K)
04

Calculate ΔG

ΔG = -233,000 J + (393 K × 424 J/K) ΔG ≈ -233,000 J + 166,632 J ΔG ≈ -66,368 J (a). Therefore, the Gibbs free energy change (ΔG) at 393 K is -66,368 J.
05

Determine the Temperature at which the Reaction is Spontaneous (ΔG < 0)

For the reaction to be spontaneous, ΔG should be negative (ΔG < 0). Using the formula for ΔG, we can set up an inequality: ΔG = ΔH - TΔS < 0 T > ΔH/ΔS We will keep the ΔH and ΔS in their original units (kJ and J/K): T > -233 kJ / (-0.424 kJ/K)
06

Calculate the Temperature

T > 549.06 K (b). So, the reaction is spontaneous at temperatures greater than 549.06 K.

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Most popular questions from this chapter

Gas \(\mathrm{A}_{2}\) reacts with gas \(\mathrm{B}_{2}\) to form gas \(\mathrm{AB}\) at a constant temperature. The bond energy of AB is much greater than that of either reactant. What can be said about the sign of \(\Delta H ? \Delta S_{\text {surr }}\) ? \(\Delta S ?\) Explain how potential energy changes for this process. Explain how random kinetic energy changes during the process.

Some nonelectrolyte solute (molar mass \(=142 \mathrm{g} / \mathrm{mol}\) ) was dissolved in \(150 . \mathrm{mL}\) of a solvent (density \(=0.879 \mathrm{g} / \mathrm{cm}^{3}\) ). The elevated boiling point of the solution was \(355.4 \mathrm{K} .\) What mass of solute was dissolved in the solvent? For the solvent, the enthalpy of vaporization is \(33.90 \mathrm{kJ} / \mathrm{mol},\) the entropy of vaporization is \(95.95 \mathrm{J} / \mathrm{K} \cdot \mathrm{mol},\) and the boiling-point elevation constant is \(2.5 \mathrm{K} \cdot \mathrm{kg} / \mathrm{mol}.\)

Given the values of \(\Delta H\) and \(\Delta S,\) which of the following changes will be spontaneous at constant \(T\) and \(P ?\) a. \(\Delta H=+25 \mathrm{kJ}, \Delta S=+5.0 \mathrm{J} / \mathrm{K}, T=300 . \mathrm{K}\) b. \(\Delta H=+25 \mathrm{kJ}, \Delta S=+100 . \mathrm{J} / \mathrm{K}, T=300 . \mathrm{K}\) c. \(\Delta H=-10 . \mathrm{kJ}, \Delta S=+5.0 \mathrm{J} / \mathrm{K}, T=298 \mathrm{K}\) d. \(\Delta H=-10 . \mathrm{kJ}, \Delta S=-40 . \mathrm{J} / \mathrm{K}, T=200 . \mathrm{K}\)

The standard free energies of formation and the standard enthalpies of formation at \(298 \mathrm{K}\) for diffuoroacetylene \(\left(\mathrm{C}_{2} \mathrm{F}_{2}\right)\) and hexafluorobenzene \(\left(\mathrm{C}_{6} \mathrm{F}_{6}\right)\) are $$\begin{array}{ccc} & \Delta G_{f}^{\circ}(\mathrm{kJ} / \mathrm{mol}) & \Delta H_{f}^{\circ}(\mathrm{kJ} / \mathrm{mol}) \\ \hline \mathrm{C}_{2} \mathrm{F}_{2}(g) & 191.2 & 241.3 \\\ \mathrm{C}_{6} \mathrm{F}_{6}(g) & 78.2 & 132.8 \end{array}$$ For the following reaction: $$\mathrm{C}_{6} \mathrm{F}_{6}(g) \rightleftharpoons 3 \mathrm{C}_{2} \mathrm{F}_{2}(g)$$ a. calculate \(\Delta S^{\circ}\) at \(298 \mathrm{K}\). b. calculate \(K\) at 298 K. c. estimate \(K\) at \(3000 .\) K, assuming \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) do not depend on temperature.

Given the following data: $$2 \mathrm{C}_{6} \mathrm{H}_{6}(l)+15 \mathrm{O}_{2}(g) \longrightarrow 12 \mathrm{CO}_{2}(g)+6 \mathrm{H}_{2} \mathrm{O}(l)$$ $$\Delta G^{\circ}=-6399 \mathrm{kJ}$$ $$\mathrm{C}(s)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g) \quad \Delta G^{\circ}=-394 \mathrm{kJ}$$ $$\mathrm{H}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l) \quad \Delta G^{\circ}=-237 \mathrm{kJ}$$ calculate \(\Delta G^{\circ}\) for the reaction $$6 \mathrm{C}(s)+3 \mathrm{H}_{2}(g) \longrightarrow \mathrm{C}_{6} \mathrm{H}_{6}(l)$$
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