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Chapter 17: Problem 65

Consider the following reaction at \(25.0^{\circ} \mathrm{C}\) : $$2 \mathrm{NO}_{2}(g) \rightleftharpoons \mathrm{N}_{2} \mathrm{O}_{4}(g)$$ The values of \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) are \(-58.03 \mathrm{~kJ} / \mathrm{mol}\) and \(-176.6 \mathrm{~J} / \mathrm{K}\). mol, respectively. Calculate the value of \(K\) at \(25.0^{\circ} \mathrm{C}\). Assuming \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) are temperature independent, estimate the value of \(K\) at \(100.0^{\circ} \mathrm{C}\).

Short Answer

Expert verified
The equilibrium constant (\(K\)) at \(25.0^{\circ}C\) (298.15 K) is approximately \(2.576\), and at \(100.0^{\circ}C\) (373.15 K) is approximately \(1.176\).

Step by step solution

01

(Step 1: Convert temperatures to Kelvin)

Convert the given temperatures to Kelvin, as the Van't Hoff equation requires temperatures to be in Kelvin: $$T_1 = 25.0^{\circ}C + 273.15K = 298.15 \, K$$ $$T_2 = 100.0^{\circ}C + 273.15K = 373.15 \, K$$
02

(Step 2: Calculate the initial equilibrium constant)

Use the Van't Hoff equation to calculate the equilibrium constant (\(K_1\)) for \(25.0^{\circ}C\) (\(298.15K\)), with the given values of \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\): $$\ln K_1 = -\frac{-58.03 \, kJ/mol}{8.314 \, J/mol \cdot K} \cdot \frac{1}{298.15 \, K} + \frac{-176.6 \, J/K}{8.314 \, J/mol \cdot K}$$ $$\ln K_1 = -\frac{-58.03 \times 10^3 \, J/mol}{8.314 \, J/mol \cdot K} \cdot \frac{1}{298.15 \, K} + \frac{-176.6 \, J/K}{8.314 \, J/mol \cdot K}$$ After calculating the values, we get: $$\ln K_1 = 0.94662$$ Convert the natural logarithm value to the actual value of \(K_1\): $$K_1 = e^{0.94662} \approx 2.576$$
03

(Step 3: Calculate the equilibrium constant at the new temperature)

Use the Van't Hoff equation again to calculate the equilibrium constant (\(K_2\)) for \(100.0^{\circ}C\) (\(373.15K\)), with the same values of \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\): $$\ln K_2 = -\frac{-58.03 \, kJ/mol}{8.314 \, J/mol \cdot K} \cdot \frac{1}{373.15 \, K} + \frac{-176.6 \, J/K}{8.314 \, J/mol \cdot K}$$ $$\ln K_2 = -\frac{-58.03 \times 10^3 \, J/mol}{8.314 \, J/mol \cdot K} \cdot \frac{1}{373.15 \, K} + \frac{-176.6 \, J/K}{8.314 \, J/mol \cdot K}$$ After calculating the values, we get: $$\ln K_2 = 0.16241$$ Convert the natural logarithm value to the actual value of \(K_2\): $$K_2 = e^{0.16241} \approx 1.176$$ So, the values of \(K\) at \(25.0^{\circ}C\) and \(100.0^{\circ}C\) are approximately \(2.576\) and \(1.176\), respectively.

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Most popular questions from this chapter

When the environment is contaminated by a toxic or potentially toxic substance (for example, from a chemical spill or the use of insecticides), the substance tends to disperse. How is this consistent with the second law of thermodynamics? In terms of the second law, which requires the least work: cleaning the environment after it has been contaminated or trying to prevent the contamination before it occurs? Explain.

As \(\mathrm{O}_{2}(I)\) is cooled at \(1 \mathrm{~atm}\), it freezes at \(54.5 \mathrm{~K}\) to form solid \(\mathrm{I}\). At a lower temperature, solid I rearranges to solid II, which has a different crystal structure. Thermal measurements show that \(\Delta H\) for the \(\mathrm{I} \rightarrow\) II phase transition is \(-743.1 \mathrm{~J} / \mathrm{mol}\), and \(\Delta S\) for the same transition is \(-17.0 \mathrm{~J} / \mathrm{K} \cdot \mathrm{mol}\). At what temperature are solids I and II in equilibrium?

Predict the sign of \(\Delta S_{\text {surr }}\) for the following processes. a. \(\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{H}_{2} \mathrm{O}(g)\) b. \(I_{2}(g) \longrightarrow I_{2}(s)\)

Calculate \(\Delta G^{\circ}\) for \(\mathrm{H}_{2} \mathrm{O}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \rightleftharpoons \mathrm{H}_{2} \mathrm{O}_{2}(g)\) at \(600 . \mathrm{K}\) using the following data: \(\mathrm{H}_{2}(\mathrm{~g})+\mathrm{O}_{2}(g) \rightleftharpoons \mathrm{H}_{2} \mathrm{O}_{2}(g) \quad K=2.3 \times 10^{6}\) at \(600 . \mathrm{K}\) \(2 \mathrm{H}_{2}(\mathrm{~g})+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{H}_{2} \mathrm{O}(g) \quad K=1.8 \times 10^{37}\) at 600. \(\mathrm{K}\)

Consider the following reaction: $$\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g)$$Calculate \(\Delta G\) for this reaction under the following conditions (assume an uncertainty of \(\pm 1\) in all quantities): a. \(T=298 \mathrm{~K}, P_{\mathrm{N}_{2}}=P_{\mathrm{H}_{2}}=200 \mathrm{~atm}, P_{\mathrm{NH}_{3}}=50 \mathrm{~atm}\) b. \(T=298 \mathrm{~K}, P_{\mathrm{N}_{2}}=200 \mathrm{~atm}, P_{\mathrm{H}_{2}}=600 \mathrm{~atm}, P_{\mathrm{NH}_{3}}=200 \mathrm{~atm}\)
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