Consider the reaction $$ 2 \mathrm{HI}(g) \rightleftharpoons \mathrm{H}_{2}(g)+\mathrm{I}_{2}(g) $$ At \(500^{\circ} \mathrm{C}\), a flask initially has all three gases, each at a partial pressure of \(0.200\) atm. When equilibrium is established, the partial pressure of HI is determined to be \(0.48\) atm. What is \(\Delta G^{\circ}\) for the reaction at \(500^{\circ} \mathrm{C}\) ?

We know the initial pressures of the three gases, and the equilibrium pressure of HI. Using the balanced equation, we can find the equilibrium pressures of H2 and I2. For every 2 moles of HI, 1 mole of H2 and 1 mole of I2 are produced. So, we can see the stoichiometry as: $$ 2HI → H_2 + I_2 $$ Initial pressures: $$ [HI]_0 = 0.2\,\mathrm{atm}, [H_2]_0 = [I_2]_0 = 0.2\,\mathrm{atm} $$ Equilibrium pressures: $$ [HI]_{eq} = 0.48\,\mathrm{atm}, [H_2]_{eq} = [I_2]_{eq} = 0.2\,\mathrm{atm} - x $$ As 2 moles of HI are consumed to produce 1 mole of H2 and 1 mole of I2, therefore: $$ 2x = (0.2\,\mathrm{atm} - [HI]_{eq}) = (0.2\,\mathrm{atm} - 0.48\,\mathrm{atm}) $$ Solving for x: $$ x = \frac{-0.28\,\mathrm{atm}}{2} = -0.14\,\mathrm{atm} $$ Thus, equilibrium pressures are: $$ [H_2]_{eq} = [I_2]_{eq} = 0.2\,\mathrm{atm} - (-0.14\,\mathrm{atm}) = 0.34\,\mathrm{atm} $$

Using the equilibrium concentrations, we can find the equilibrium constant (K). The reaction quotient, Q, is defined as: $$ Q = \frac{[H_2]_{eq}[I_2]_{eq}}{[HI]_{eq}^2} $$ Since, at equilibrium, Q = K: $$ K = \frac{0.34 \times 0.34}{0.48^2} = 0.4993 $$

The van't Hoff equation relates the equilibrium constant, temperature, and standard Gibbs free energy change and is given by: $$ \Delta G^{\circ} = -RT\ln K $$ Where \(\Delta G^{\circ}\) is the standard Gibbs free energy change, R is the gas constant (8.314 J/mol K), T is the temperature in Kelvin, and ln K is the natural logarithm of the equilibrium constant. First, convert the temperature in Celsius to Kelvin: $$ T = 500°C + 273.15 = 773.15\,\mathrm{K} $$ Now, calculate the standard Gibbs free energy change: $$ \Delta G^{\circ} = - (8.314\,\mathrm{\frac{J}{mol\cdot K}}) (773.15\,\mathrm{K}) \ln(0.4993) = 3731.85\,\mathrm{J/mol} $$

Since the standard Gibbs free energy change is usually reported in kJ/mol, we need to convert the value obtained: $$ \Delta G^{\circ} = 3731.85\,\mathrm{J/mol} \times \frac{1\,\mathrm{kJ}}{1000\,\mathrm{J}} = 3.73\,\mathrm{kJ/mol} $$ The standard Gibbs free energy change for the reaction at 500°C is 3.73 kJ/mol.