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Chapter 16: Problem 111

Consider the system $$\mathrm{A}(g) \longrightarrow \mathrm{B}(g)$$ at $$25^{\circ} \mathrm{C}$$ a. Assuming that \(G_{A}^{\circ}=8996 \mathrm{J} / \mathrm{mol}\) and \(G_{\mathrm{B}}^{\circ}=11,718 \mathrm{J} / \mathrm{mol}\) calculate the value of the equilibrium constant for this reaction. b. Calculate the equilibrium pressures that result if 1.00 mole of \(\mathrm{A}(g)\) at 1.00 atm and 1.00 mole of \(\mathrm{B}(g)\) at 1.00 atm are mixed at \(25^{\circ} \mathrm{C}.\) c. Show by calculations that \(\Delta G=0\) at equilibrium.

Short Answer

Expert verified
The equilibrium constant for this reaction is 0.1088, and the equilibrium pressures of A and B are 0.9019 atm and 1.0981 atm, respectively. The calculations show that ΔG is close to 0 at the equilibrium conditions given, confirming that the system is at equilibrium.

Step by step solution

01

Calculate Gibbs free energy change

To find the equilibrium constant, we need to first find the Gibbs free energy change for the reaction, which is given by: \(\Delta G^{\circ} = G^{\circ}_{B} - G^{\circ}_{A}\). \( \Delta G^{\circ} = 11,718 \mathrm{J}/\mathrm{mol} - 8,996 \mathrm{J}/\mathrm{mol} = 2,722 \mathrm{J}/\mathrm{mol} \)
02

Calculate the equilibrium constant K

The relationship between the Gibbs free energy change and the equilibrium constant K is given by: \(\Delta G^{\circ} = -RT\ln{K}\), where R is the gas constant (8.314 J/(mol·K)) and T is the temperature in Kelvin. We need to find the temperature in Kelvin first: \(T(K) = 25^{\circ}C + 273.15 = 298.15 K\) Now, we can rearrange the equation to solve for K: \(K = \exp{(-\frac{\Delta G^{\circ}}{RT})} \) \(K = \exp{(-\frac{2,722 \mathrm{J}/\mathrm{mol}}{8.314 \mathrm{J}/(\mathrm{mol}\cdot\mathrm{K})\cdot 298.15\mathrm{K}})} = 0.1088\) #a. Conclusion:# The equilibrium constant for this reaction is 0.1088. #b. Calculate the equilibrium pressures#
03

Set up the reaction equation

Let's denote the initial pressures of A and B as \(P_A^0 = 1.00 \, \mathrm{atm}\) and \(P_B^0 = 1.00 \, \mathrm{atm}\), respectively. At equilibrium, the change in pressure will be given by: \(P_A^{eq} = P_A^0 - x\) and \(P_B^{eq} = P_B^0 + x\), where x is the change in pressure due to the reaction.
04

Use equilibrium constant expression

The reaction has the form: \(A \rightleftharpoons B\), and the equilibrium constant expression will be: \(K = \frac{P_B^{eq}}{P_A^{eq}}\) Substitute the equilibrium pressures in the expression: \(0.1088 = \frac{1.00 \, \mathrm{atm} + x}{1.00 \, \mathrm{atm} - x}\)
05

Solve for x

Now we need to find the value of x that brings the expression to equilibrium. Rearrange the equation to isolate x: \(x = \frac{1.00 \, \mathrm{atm} - 0.1088 \cdot (1.00 \, \mathrm{atm} - x)}{1 + 0.1088}\) Solve for x: \(x = 0.0981 \, \mathrm{atm}\)
06

Calculate equilibrium pressures

Now that we have the value of x, we can find the equilibrium pressures for A and B: \(P_A^{eq} = 1.00 \, \mathrm{atm} - 0.0981 \, \mathrm{atm} = 0.9019 \, \mathrm{atm}\) \(P_B^{eq} = 1.00 \, \mathrm{atm} + 0.0981 \, \mathrm{atm} = 1.0981 \, \mathrm{atm}\) #b. Conclusion#: The equilibrium pressures of A and B are 0.9019 atm and 1.0981 atm, respectively. #c. Show that ΔG = 0 at equilibrium#
07

Define the Gibbs free energy equation

The Gibbs free energy change equation is given by: \(\Delta G = \Delta G^{\circ} + RT\ln{Q}\), where \(Q\) is the reaction quotient which equals to \( \frac{P_B^{eq}}{P_A^{eq}} \) for this reaction.
08

Calculate Q

Next, we need to calculate the reaction quotient, Q: \(Q = \frac{P_B^{eq}}{P_A^{eq}} = \frac{1.0981 \, \mathrm{atm}}{0.9019 \, \mathrm{atm}} = 1.2174\)
09

Calculate ΔG

Plug in the calculated values to find ΔG: \(\Delta G = 2,722 \mathrm{J}/\mathrm{mol} + 8.314 \mathrm{J}/(\mathrm{mol}\cdot\mathrm{K}) \cdot 298.15\mathrm{K} \cdot \ln{1.2174} = -0.0006 \, \mathrm{J}/\mathrm{mol}\) The calculated ΔG is close to zero, which means that the system is already at equilibrium, considering the error in the numerical calculations. #c. Conclusion#: The calculations show that ΔG is close to 0 at the equilibrium conditions given, confirming that the system is at equilibrium.

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Most popular questions from this chapter

For ammonia ( \(\mathrm{NH}$$_3\)), the enthalpy of fusion is \(5.65 \mathrm{kJ} / \mathrm{mol}\) and the entropy of fusion is \(28.9 \mathrm{J} / \mathrm{K} \cdot\) mol. a. Will \(\mathrm{NH}_{3}(s)\) spontaneously melt at \(200 . \mathrm{K} ?\) b. What is the approximate melting point of ammonia?

Consider the reaction $$\mathrm{Fe}_{2} \mathrm{O}_{3}(s)+3 \mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{Fe}(s)+3 \mathrm{H}_{2} \mathrm{O}(g)$$ a. Use \(\Delta G_{\mathrm{f}}^{\circ}\) values in Appendix 4 to calculate \(\Delta G^{\circ}\) for this reaction. b. Is this reaction spontaneous under standard conditions at \(298 \mathrm{K} ?\) c. The value of \(\Delta H^{\circ}\) for this reaction is \(100 .\) kJ. At what temperatures is this reaction spontaneous at standard conditions? Assume that \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) do not depend on temperature.

Predict the sign of \(\Delta S^{\circ}\) and then calculate \(\Delta S^{\circ}\) for each of the following reactions. a. \(\mathrm{H}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l)\) b. \(2 \mathrm{CH}_{3} \mathrm{OH}(g)+3 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{CO}_{2}(g)+4 \mathrm{H}_{2} \mathrm{O}(g)\) c. \(\mathrm{HCl}(g) \longrightarrow \mathrm{H}^{+}(a q)+\mathrm{Cl}^{-}(a q)\)

What types of experiments can be carried out to determine whether a reaction is spontaneous? Does spontaneity have any relationship to the final equilibrium position of a reaction? Explain.

Liquid water at \(25^{\circ} \mathrm{C}\) is introduced into an evacuated, insulated vessel. Identify the signs of the following thermodynamic functions for the process that occurs: \(\Delta H, \Delta S, \Delta T_{\text {water }}, \Delta S_{\text {surr }}, \Delta S_{\text {univ. }}\)
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