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Chapter 17: Problem 72

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 71 .

Short Answer

Expert verified
The values of \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) for this reaction are approximately -112400 J/mol and -120.6 J/K·mol, respectively.

Step by step solution

01

Recall the Van 't Hoff equation

Recall the Van 't Hoff equation, which relates the equilibrium constant K, temperature (T), and the Gibbs free energy change ∆G°: \( \ln K = -\frac{\Delta G^\circ}{R \cdot T} \) where R is the gas constant (approximately 8.314 J/K·mol) and T is the temperature in Kelvin.
02

Relate ∆G° to ∆H° and ∆S°

Now we need to establish a connection between ∆G°, ∆H°, and ∆S°. The relevant relationship is given by the Gibbs-Helmholtz equation: \( \Delta G^\circ = \Delta H^\circ - T \cdot \Delta S^\circ \) Now we can substitute this expression for ∆G° into the Van 't Hoff equation: \( \ln K = -\frac{\Delta H^\circ - T \cdot \Delta S^\circ}{R \cdot T} \)
03

Rearrange the equation for ln(K) vs. 1/T

We need to rewrite this equation to match the given information about the graph: \( \ln K = -\frac{\Delta H^\circ}{R} \cdot \frac{1}{T} + \frac{\Delta S^\circ}{R} \) This equation is in the form of a linear equation y = mx + b, where y = ln(K), x = 1/T, m = -∆H°/R, and b = ∆S°/R. We are given the slope (m = -∆H°/R) and y-intercept (b = ∆S°/R) as 13520 K and -14.51, respectively.
04

Calculate ∆H° and ∆S°

Now that we have the slope and y-intercept values, we can calculate ∆H° and ∆S°: For the slope, m = -∆H°/R, we have: \( -\Delta H^\circ = m \cdot R \) \( \Delta H^\circ = -m \cdot R = -(1.352 \times 10^4 \,\text{K}) \cdot (8.314 \,\text{J/K·mol}) \) \( \Delta H^\circ \approx -112400 \, \text{J/mol} \) For the y-intercept, b = ∆S°/R, we have: \( \Delta S^\circ = b \cdot R = (-14.51) \cdot (8.314 \, \text{J/K·mol}) \) \( \Delta S^\circ \approx -120.6 \, \text{J/K·mol} \)
05

Final Answer

The values of ∆H° and ∆S° for this reaction are approximately -112400 J/mol and -120.6 J/K·mol, respectively.

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Most popular questions from this chapter

Consider the following reaction at \(298 \mathrm{~K}\) : $$2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{SO}_{3}(g)$$ An equilibrium mixture contains \(\mathrm{O}_{2}(\mathrm{~g})\) and \(\mathrm{SO}_{3}(g)\) at partial pressures of \(0.50\) atm and \(2.0\) atm, respectively. Using data from Appendix 4 , determine the equilibrium partial pressure of \(\mathrm{SO}_{2}\) in the mixture. Will this reaction be most favored at a high or a low temperature, assuming standard conditions?

For the process \(\mathrm{A}(l) \longrightarrow \mathrm{A}(g)\), which direction is favored by changes in energy probability? Positional probability? Explain your answers. If you wanted to favor the process as written, would you raise or lower the temperature of the system? Explain.

A green plant synthesizes glucose by photosynthesis, as shown in the reaction $$6 \mathrm{CO}_{2}(g)+6 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(s)+6 \mathrm{O}_{2}(g)$$ Animals use glucose as a source of energy: $$\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(s)+6 \mathrm{O}_{2}(g) \longrightarrow 6 \mathrm{CO}_{2}(g)+6 \mathrm{H}_{2} \mathrm{O}(l)$$ If we were to assume that both these processes occur to the same extent in a cyclic process, what thermodynamic property must have a nonzero value?

It is quite common for a solid to change from one structure to another at a temperature below its melting point. For example, sulfur undergoes a phase change from the rhombic crystal structure to the monoclinic crystal form at temperatures above \(95^{\circ} \mathrm{C}\). a. Predict the signs of \(\Delta H\) and \(\Delta S\) for the process \(S_{\text {rhcmbic }} \longrightarrow\) \(\mathrm{S}_{\text {monoclinic }}\) b. Which form of sulfur has the more ordered crystalline structure (has the smaller positional probability)?

Hydrogen cyanide is produced industrially by the following exothermic reaction: $$2 \mathrm{NH}_{3}(g)+3 \mathrm{O}_{2}(g)+2 \mathrm{CH}_{4}(g) \stackrel{\mathrm{low}^{\circ} \mathrm{c}}{\mathrm{Pt} \cdot \mathrm{Rh}} 2 \mathrm{HCN}(g)+6 \mathrm{H}_{2} \mathrm{O}(g)$$ Is the high temperature needed for thermodynamic or kinetic reasons?
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