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Chapter 16: Problem 119

For the equilibrium $$\mathrm{A}(g)+2 \mathrm{B}(g) \rightleftharpoons \mathrm{C}(g)$$ the initial concentrations are \([\mathrm{A}]=[\mathrm{B}]=[\mathrm{C}]=0.100 \mathrm{atm}\) Once equilibrium has been established, it is found that \([\mathrm{C}]=\) 0.040 atm. What is \(\Delta G^{\circ}\) for this reaction at \(25^{\circ} \mathrm{C} ?\)

Short Answer

Expert verified
The standard Gibbs free energy change (ΔG°) for the given reaction at 25°C is approximately -6.73 kJ/mol.

Step by step solution

01

Determine the equilibrium concentrations

Given the initial concentrations of A, B, and C are all 0.100 atm, we can write the change in concentrations as they approach equilibrium as follows: A: 0.100 - x B: 0.100 - 2x C: 0.040 + x At equilibrium, we're given that the concentration of C is 0.040 atm. So, we can set up the following equation to solve for x: 0.040 + x = 0.100
02

Solve for x

Solving for x, we find that x = 0.060. Now, we can determine the equilibrium concentrations of A and B: A: 0.100 - x = 0.100 - 0.060 = 0.040 atm B: 0.100 - 2x = 0.100 - 2(0.060) = 0.020 atm
03

Calculate the reaction quotient Q and equilibrium constant K

Using the equilibrium concentrations, we can calculate the reaction quotient Q: Q = [C] / ([A][B]^2) = (0.040) / ((0.040)(0.020)^2) = 50 Since the system is at equilibrium, Q is equal to the equilibrium constant K: K = 50
04

Determine ΔG° using K

We can find the standard Gibbs free energy change (ΔG°) using the relationship between K and ΔG°: ΔG° = -RT ln(K) Where R is the gas constant, 8.314 J/(mol K), and T is the temperature in Kelvin, which is 25°C + 273.15 = 298.15 K. So, ΔG° = - (8.314 J/(mol K)) (298.15 K) ln(50)
05

Calculate ΔG°

Now, we calculate the value of ΔG°: ΔG° ≈ - (8.314 J/(mol K)) (298.15 K) ln(50) ≈ -6733 J/mol We can convert this to kJ/mol: ΔG° ≈ -6.73 kJ/mol Thus, the standard Gibbs free energy change (ΔG°) for this reaction at 25°C is approximately -6.73 kJ/mol.

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Most popular questions from this chapter

At \(1500 \mathrm{K},\) the process $$\begin{aligned} &\mathbf{I}_{2}(g) \longrightarrow 2 \mathbf{I}(g)\\\ &10 \mathrm{atm} \quad 10 \mathrm{atm} \end{aligned}$$ is not spontaneous. However, the process $$\begin{aligned} &\mathbf{I}_{2}(g) \longrightarrow 2 \mathbf{I}(g)\\\ &0.10 \mathrm{atm} \quad 0.10 \mathrm{atm} \end{aligned}$$ is spontaneous at \(1500 \mathrm{K}\). Explain.

Hydrogen sulfide can be removed from natural gas by the reaction $$2 \mathrm{H}_{2} \mathrm{S}(g)+\mathrm{SO}_{2}(g) \rightleftharpoons 3 \mathrm{S}(s)+2 \mathrm{H}_{2} \mathrm{O}(g)$$ Calculate \(\Delta G^{\circ}\) and \(K\) (at 298 K) for this reaction. Would this reaction be favored at a high or low temperature?

The standard enthalpy of formation of \(\mathrm{H}_{2} \mathrm{O}(l)\) at \(298 \mathrm{K}\) is \(-285.8 \mathrm{kJ} / \mathrm{mol} .\) Calculate the change in internal energy for the following process at \(298 \mathrm{K}\) and 1 atm: $$\mathrm{H}_{2} \mathrm{O}(l) \rightarrow \mathrm{H}_{2}(g)+1 / 2 \mathrm{O}_{2}(g) \quad \Delta E^{\circ}=?$$ (Hint: Using the ideal gas equation, derive an expression for work in terms of \(n, R,\) and \(T .\) )

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: \(\Delta H, \Delta S, \Delta T_{\text {water }}, \Delta S_{\text {surr }}, \Delta S_{\text {univ. }}\)

At \(100 .^{\circ} \mathrm{C}\) and \(1.00 \mathrm{atm}, \Delta H^{\circ}=40.6 \mathrm{kJ} / \mathrm{mol}\) for the vaporiza- tion of water. Estimate \(\Delta G^{\circ}\) for the vaporization of water at \(90 .^{\circ} \mathrm{C}\) and \(110 .^{\circ} \mathrm{C} .\) Assume \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) at \(100 .^{\circ} \mathrm{C}\) and 1.00 atm do not depend on temperature.
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