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Chapter 17: Problem 77

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 \(-58.03\) kJ/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 value of K at 25.0 °C is approximately 6.29, and the estimated value of K at 100.0 °C is approximately 13.88.

Step by step solution

01

Convert given temperatures to Kelvin

To solve this problem, we need to convert the given temperatures from Celsius to Kelvin, using the formula: \(T(K) = T(^\circ C) + 273.15\) For 25.0 °C: \(T_1 = 25.0^\circ C + 273.15 = 298.15 K\) For 100.0 °C: \(T_2 = 100.0^\circ C + 273.15 = 373.15 K\)
02

Write the Van't Hoff equation

The Van't Hoff equation is given by: \(\frac{\Delta G^\circ}{-RT} = \ln K\) where ∆G° is the standard Gibbs free energy change, R is the gas constant, and T is the temperature in Kelvin. We can also express the standard Gibbs free energy change in terms of standard enthalpy change (∆H°) and the standard entropy change (∆S°) as follows: \(\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ\) Now, we can substitute the expression for ∆G° into the Van't Hoff equation: \(\frac{\Delta H^\circ - T\Delta S^\circ}{-RT} = \ln K\)
03

Solve for K at 25.0 °C or 298.15 K

First, let's plug the values of ∆H°, ∆S°, R and T_1 (25.0 °C) into the Van't Hoff equation: \(\frac{-(-58.03 \times 10^3 \mathrm{J/mol}) - (298.15 \mathrm{K})(-176.6 \mathrm{J/K} \cdot \mathrm{mol})}{(8.314 \mathrm{J/K} \cdot \mathrm{mol})(298.15 \mathrm{K})} = \ln K_1\) Now, simplify and solve for K_1: \(K_1 = e^{\ln K_1} = 6.29\) So, the equilibrium constant, K, at 25.0 °C is approximately 6.29.
04

Use the Van't Hoff equation to estimate K at 100.0 °C or 373.15 K

Using the Van't Hoff equation, plug the values of ∆H°, ∆S°, R and T_2 (100.0 °C) into the equation: \(\frac{-(-58.03 \times 10^3 \mathrm{J/mol}) - (373.15 \mathrm{K})(-176.6 \mathrm{J/K} \cdot \mathrm{mol})}{(8.314 \mathrm{J/K} \cdot \mathrm{mol})(373.15 \mathrm{K})} = \ln K_2\) Now, simplify and solve for K_2: \(K_2 = e^{\ln K_2} = 13.88\) So, the equilibrium constant, K, at 100.0 °C is approximately 13.88. In summary, the value of K at 25.0 °C is approximately 6.29, and the estimated value of K at 100.0 °C is approximately 13.88.

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Most popular questions from this chapter

Consider the reaction: $$\mathrm{N}_{2} \mathrm{O}_{4}(g) \rightarrow 2 \mathrm{NO}_{2}(g)$$ where \(P_{\mathrm{NO}_{2}}=0.29\) atm and $P_{\mathrm{N}_{2} \mathrm{O}_{4}}=1.6 \mathrm{atm} .$ For this reaction at these conditions, \(\Delta G=-1000 \mathrm{J}\) and \(\Delta G^{\circ}=6000 \mathrm{J}\) . Which of the following statements about this reaction is(are) true? a. The reverse reaction is spontaneous at these conditions. b. At equilibrium, \(P_{\mathrm{N}_{2} \mathrm{O}_{4}}\) will be greater than 1.6 \(\mathrm{atm}\) . c. The value of K for this reaction is greater than 1. d. The maximum amount of work this reaction can produce at these conditions is –6000 J. e. The reaction is endothermic.

Consider the dissociation of a weak acid $\mathrm{HA}\left(K_{\mathrm{a}}=4.5 \times 10^{-3}\right)$ in water: $$\mathrm{HA}(a q) \rightleftharpoons \mathrm{H}^{+}(a q)+\mathrm{A}^{-}(a q)$$ Calculate \(\Delta G^{\circ}\) for this reaction at \(25^{\circ} \mathrm{C}\)

The equilibrium constant for a certain reaction decreases from 8.84 to $3.25 \times 10^{-2}\( when the temperature increases from \)25^{\circ} \mathrm{C}$ to \(75^{\circ} \mathrm{C}\) . Estimate the temperature where \(K=1.00\) for this reaction. Estimate the value of \(\Delta S^{\circ}\) for this reaction. (Hint: Manipulate the equation in Exercise 85.)

Which of the following processes are spontaneous? a. Salt dissolves in \(\mathrm{H}_{2} \mathrm{O}\) . b. A clear solution becomes a uniform color after a few drops of dye are added. c. Iron rusts. d. You clean your bedroom.

Consider the reaction $$\begin{array}{l}{2 \mathrm{O}(g) \longrightarrow \mathrm{O}_{2}(g)} \\\ {\text { sof } \Delta H \text { and } \Delta S}\end{array}$$ a. Predict the signs of \(\Delta H\) and \(\Delta S .\) b. Would the reaction be more spontaneous at high or low temperatures?
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