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Chapter 17: Problem 80

The equilibrium constant \(K\) for the reaction $$ 2 \mathrm{Cl}(g) \rightleftharpoons \mathrm{Cl}_{2}(g) $$ was measured as a function of temperature (Kelvin). A graph of \(\ln (K)\) versus \(1 / T\) for this reaction gives a straight line with a slope of \(1.352 \times 10^{4} \mathrm{~K}\) and a \(y\) -intercept of \(-14.51\). Determine the values of \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) for this reaction. See Exercise 79

Short Answer

Expert verified
In conclusion, the standard enthalpy change (ΔH°) for the given reaction is approximately -112428 J/mol, and the standard entropy change (ΔS°) is approximately -120.61 J/mol·K.

Step by step solution

01

Determine the ΔH°

To find the value of ΔH°, we can use the given slope of the graph (1.352 × 10^4 K) and the relationship between the slope of the graph and ΔH°: Slope = -ΔH°/R We can now solve for ΔH°: ΔH° = -Slope × R ΔH° = -(1.352 × 10^4 K) × (8.314 J/mol·K) ΔH° ≈ -112428 J/mol So, the standard enthalpy change (ΔH°) for the reaction is approximately -112428 J/mol.
02

Determine the ΔS°

To find the value of ΔS°, we can use the given y-intercept of the graph (-14.51) and the relationship between the y-intercept of the graph and ΔS°: y-intercept = ΔS°/R We can now solve for ΔS°: ΔS° = y-intercept × R ΔS° = (-14.51) × (8.314 J/mol·K) ΔS° ≈ -120.61 J/mol·K So, the standard entropy change (ΔS°) for the reaction is approximately -120.61 J/mol·K. In conclusion, the standard enthalpy change (ΔH°) for the given reaction is approximately -112428 J/mol, and the standard entropy change (ΔS°) is approximately -120.61 J/mol·K.

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Most popular questions from this chapter

Cells use the hydrolysis of adenosine triphosphate, abbreviated as ATP, as a source of energy. Symbolically, this reaction can be written as $$ \mathrm{ATP}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{ADP}(a q)+\mathrm{H}_{2} \mathrm{PO}_{4}^{-}(a q) $$ where ADP represents adenosine diphosphate. For this reaction, \(\Delta G^{\circ}=-30.5 \mathrm{~kJ} / \mathrm{mol}\). a. Calculate \(K\) at \(25^{\circ} \mathrm{C}\). b. If all the free energy from the metabolism of glucose $$ \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(s)+6 \mathrm{O}_{2}(g) \longrightarrow 6 \mathrm{CO}_{2}(g)+6 \mathrm{H}_{2} \mathrm{O}(l) $$ goes into forming ATP from ADP, how many ATP molecules can be produced for every molecule of glucose?

Using Appendix 4 and the following data, determine \(S^{\circ}\) for \(\mathrm{Fe}(\mathrm{CO})_{5}(g)\) $$ \begin{aligned} \mathrm{Fe}(s)+5 \mathrm{CO}(g) & \longrightarrow \mathrm{Fe}(\mathrm{CO})_{5}(g) & & \Delta S^{\circ}=? \\ \mathrm{Fe}(\mathrm{CO})_{5}(l) & \longrightarrow \mathrm{Fe}(\mathrm{CO})_{5}(g) & & \Delta S^{\circ}=107 \mathrm{~J} / \mathrm{K} \\ \mathrm{Fe}(s)+5 \mathrm{CO}(g) & \longrightarrow \mathrm{Fe}(\mathrm{CO})_{5}(l) & & \Delta S^{\circ}=-677 \mathrm{~J} / \mathrm{K} \end{aligned} $$

Predict the sign of \(\Delta S^{\circ}\) for each of the following changes. a. \(\mathrm{K}(s)+\frac{1}{2} \mathrm{Br}_{2}(g) \longrightarrow \mathrm{KBr}(s)\) b. \(\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{NH}_{3}(g)\) c. \(\mathrm{KBr}(s) \longrightarrow \mathrm{K}^{+}(a q)+\mathrm{Br}^{-}(a q)\) d. \(\mathrm{KBr}(s) \longrightarrow \mathrm{KBr}(l)\)

For a liquid, which would you expect to be larger, \(\Delta S_{\text {fusion }}\) or \(\Delta S_{\text {evaporation }}\) ? Why?

Consider the reaction $$ 2 \mathrm{POCl}_{3}(g) \longrightarrow 2 \mathrm{PCl}_{3}(g)+\mathrm{O}_{2}(g) $$ a. Calculate \(\Delta G^{\circ}\) for this reaction. The \(\Delta G_{\mathrm{f}}^{\circ}\) values for \(\mathrm{POCl}_{3}(g)\) and \(\mathrm{PCl}_{3}(g)\) are \(-502 \mathrm{~kJ} / \mathrm{mol}\) and \(-270 . \mathrm{kJ} / \mathrm{mol}\), respectively. b. Is this reaction spontaneous under standard conditions at \(298 \mathrm{~K}\) ? c. The value of \(\Delta S^{\circ}\) for this reaction is \(179 \mathrm{~J} / \mathrm{K} \cdot\) mol. At what temperatures is this reaction spontaneous at standard conditions? Assume that \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) do not depend on temperature.
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