Question

For a given reversible reaction at \(300\,K\): \[ 2\text{N}_2\text{O}_5(g) \rightleftharpoons 4\text{NO}_2(g) + \text{O}_2(g) \] If \( \text{N}_2\text{O}_5 \) is 50% dissociated at 5 bar and 300 K at equilibrium, then find \( \Delta G^\circ \) for the reaction? [Given: \( \log 2 = 0.3,\ \log 7 = 0.84 \)]

• A. \(-4\,\text{kJ}\)
• B. \(-8\,\text{kJ}\)
• C. \(-6\,\text{kJ}\)
• D. \(-10\,\text{kJ}\)

Answer 1

The Correct Option is B

Concept: For gaseous equilibria involving dissociation, the equilibrium constant \(K_p\) can be calculated using mole fractions and total pressure. The relation between standard Gibbs free energy change and equilibrium constant is: \[ \Delta G^\circ = -RT \ln K_p \] where \(R = 8.314\,\text{J mol}^{-1}\text{K}^{-1}\) and \(T = 300\,K\).
Step 1: Write the equilibrium reaction. \[ 2\text{N}_2\text{O}_5(g) \rightleftharpoons 4\text{NO}_2(g) + \text{O}_2(g) \] Assume initially \(1\) mole of \( \text{N}_2\text{O}_5 \). Degree of dissociation \( \alpha = 0.5 \). \[ \begin{array}{c|c|c|c} & \text{N}_2\text{O}_5 & \text{NO}_2 & \text{O}_2
\hline \text{Initial} & 1 & 0 & 0
\text{Change} & -\alpha & +2\alpha & +\alpha/2
\text{Equilibrium} & 1-\alpha & 2\alpha & \alpha/2 \end{array} \] Substitute \( \alpha = 0.5 \) \[ \text{N}_2\text{O}_5 = 0.5,\quad \text{NO}_2 = 1,\quad \text{O}_2 = 0.25 \] Total moles: \[ n_T = 0.5 + 1 + 0.25 = 1.75 \]
Step 2: Write the expression for \(K_p\). \[ K_p = \frac{(P_{\text{NO}_2})^4 (P_{\text{O}_2})}{(P_{\text{N}_2\text{O}_5})^2} \] Using mole fraction relation: \[ K_p = \frac{(n_{\text{NO}_2})^4 (n_{\text{O}_2})}{(n_{\text{N}_2\text{O}_5})^2} \left(\frac{P_T}{n_T}\right)^{\Delta n} \] Here \[ \Delta n = (4+1)-2 = 3 \]
Step 3: Substitute the values. \[ K_p = \frac{(1)^4(0.25)}{(0.5)^2} \left(\frac{5}{1.75}\right)^3 \] \[ K_p = \left(\frac{20}{7}\right)^3 \]
Step 4: Calculate \( \Delta G^\circ \). \[ \Delta G^\circ = -RT\ln K_p \] \[ = -8.314 \times 300 \ln\left(\frac{20}{7}\right)^3 \] \[ = -7.926\,\text{kJ} \] \[ \Delta G^\circ \approx -8\,\text{kJ} \]