Consider the system $$\mathrm{A}(g) \rightleftharpoons \mathrm{B}(g)$$ at \(25^{\circ} \mathrm{C}\) a. Assuming that \(G_{\mathrm{A}}^{\circ}=8996 \mathrm{J} / \mathrm{mol}\) 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.

Given the standard Gibbs free energy of reactant A, \(G_A^\circ = 8,996 \;J/mol\), and the standard Gibbs free energy of product B, \(G_B^\circ = 11,718 \;J/mol\), we can determine the standard change in Gibbs free energy for the reaction \(\Delta G^\circ\): \(\Delta G^\circ = G_B^\circ - G_A^\circ = 11,718 \;J/mol - 8,996 \;J/mol = 2,722 \;J/mol\). Now, we can use the temperature (in Kelvin) and \(\Delta G^\circ\) to calculate the equilibrium constant: First, convert the temperature from Celsius to Kelvin: \(T = 25^\circ C + 273.15 = 298.15 K\) Using the equation \(\Delta G^\circ = -RT\ln K\), we can rearrange for the equilibrium constant: \(K = e^{(-\Delta G^\circ) / (RT)} = e^{(-2,722 \;J/mol) / (8.314 \;J/(mol \cdot K) \cdot 298.15 \;K)} = 0.104\) Thus, the equilibrium constant, \(K\), for the given reaction is 0.104.

Initially, we have 1 mole of A at 1 atm and 1 mole of B at 1 atm. Let's denote the change in pressure for A as \(x\) and for B as \(y\). The resulting equilibrium pressures will be \(P_A = (1-x) atm\) for A and \(P_B = (1+y) atm\) for B. At equilibrium, the equilibrium expression would be: \(K = \dfrac{P_B}{P_A} = \dfrac{1+y}{1-x}\) Using the calculated value of the equilibrium constant, \(K = 0.104\), we get: \(0.104 = \dfrac{1+y}{1-x}\) Now, let's solve the linear equation for x and y: From the equation, we can find \(y = 0.104(1-x) - 1\), which gives us \(x = \dfrac{1-0.104(1+y)}{0.104}\). Substituting \(x\) in terms of \(y\) in the initial equation, we get: \(y = 0.104 \cdot \dfrac{1-0.104(1+y)}{0.104} - 1\) Solving for \(y\), we find that \(y = -0.329 atm\). Now, substituting the value of \(y\) back into the equation for \(x\): \(x = \dfrac{1-0.104(1-0.329)}{0.104} = 0.467 atm\) The equilibrium pressures for reactant A and product B are: \(P_A = 1 - x = 1 - 0.467 = 0.533 atm\) \(P_B = 1 + y = 1 - 0.329 = 0.671 atm\) So, the equilibrium pressures are 0.533 atm for A and 0.671 atm for B.

We have previously calculated the standard change in Gibbs free energy for the reaction, \(\Delta G^\circ\). Now, to show that the change in Gibbs free energy \(\Delta G\) at equilibrium is zero, we will use the following equation: \(\Delta G = \Delta G^\circ + RT\ln Q\) The reaction quotient, \(Q\), can be represented as: \(Q=\dfrac{P_B}{P_A}\) We know the equilibrium pressures for both reactants A and B are \(P_A = 0.533 atm\) and \(P_B = 0.671 atm\), respectively. Therefore, the reaction quotient \(Q\) can be calculated: \(Q = \dfrac{0.671}{0.533} = 1.259\) Now, plugging the values into the equation for \(\Delta G\): \(\Delta G = 2,722 \;J/mol + (8.314 \;J/(mol \cdot K) \cdot 298.15 \;K) \cdot \ln(1.259) \approx 0 \;J/mol\) Thus, as the result of \(\Delta G\) is approximately zero, we have shown that \(\Delta G = 0\) at equilibrium.