Question

For the following reaction at 50 \(^\circ\)C and at 2 atm pressure,
\(2N_2O_5(g) \rightleftharpoons 2N_2O_4(g) + O_2(g)\)
\(N_2O_5\) is 50% dissociated. The magnitude of standard free energy change at this temperature is x.
x = _______ J \(\cdot mol^{-1}\) [Nearest integer].
Given : R = 8.314 J \(mol^{-1} K^{-1}\), \(\log 2 = 0.30, \log 3 = 0.48, \ln 10 = 2.303, ^\circ C + 273 = K\)

Answer 1

Correct Answer: 2470

Step 1: Write equilibrium moles
For reaction \[ 2N_2O_5 \rightleftharpoons 2N_2O_4 + O_2 \] Assume initially $2$ moles of $N_2O_5$. Degree of dissociation: \[ \alpha=0.5 \] Hence equilibrium moles are \[ N_2O_5 = 2(1-\alpha)=2(1-0.5)=1 \] \[ N_2O_4 = 2\alpha = 1 \] \[ O_2=\alpha=0.5 \] Total moles: \[ n_{total}=1+1+0.5=2.5 \]
Step 2: Find partial pressures
Given total pressure \[ P=2\ \text{atm} \] Using \[ P_i = X_iP \] For $N_2O_5$: \[ P_{N_2O_5}=\frac{1}{2.5}\times 2=0.8\ \text{atm} \] For $N_2O_4$: \[ P_{N_2O_4}=\frac{1}{2.5}\times 2=0.8\ \text{atm} \] For $O_2$: \[ P_{O_2}=\frac{0.5}{2.5}\times 2=0.4\ \text{atm} \]
Step 3: Calculate $K_p$
\[ K_p= \frac{(P_{N_2O_4})^2(P_{O_2})} {(P_{N_2O_5})^2} \] \[ = \frac{(0.8)^2(0.4)}{(0.8)^2} \] \[ =0.4 \]
Step 4: Use free energy relation
Formula: \[ \Delta G^\circ=-RT\ln K_p \] Temperature: \[ T=50+273=323\ \text{K} \] So, \[ \Delta G^\circ = -8.314\times 323\times \ln(0.4) \] Now \[ \ln(0.4)=\ln\left(\frac{4}{10}\right) \] \[ =\ln 4-\ln 10 \] Using \[ \ln 4 = 2\ln 2 \] and \[ \ln 2 = 2.303\log 2 = 2.303(0.30)=0.6909 \] \[ \ln 4=2(0.6909)=1.3818 \] Hence \[ \ln(0.4)=1.3818-2.303=-0.9212 \] Therefore \[ \Delta G^\circ = -8.314\times 323\times (-0.9212) \] \[ =2470\ \text{J mol}^{-1} \] Final Answer: \[ \boxed{2470\ \text{J mol}^{-1}} \]