The standard free energy change (ΔG°) for 50% dissociation of N2O4 into NO2 at 27 °C and 1 atm pressure is −x J mol−1 . The value of x

To find the value of $x$ corresponding to the standard free energy change ($\Delta G°$) for the dissociation of $\text{N}_2\text{O}_4$ into $\text{NO}_2$ at a given temperature, we can utilize the relationship between free energy change and equilibrium constant, along with the ideal gas law.

1. Break Down the Problem

• The reaction of interest is: $\text{N}_2\text{O}_4 (g) \rightleftharpoons 2 \text{NO}_2 (g)$
• We know that at 50% dissociation, the concentrations at equilibrium can be considered:
  • Initial moles: 1 mole of $\text{N}_2\text{O}_4$
  • Moles at Equilibrium: 0.5 moles of $\text{N}_2\text{O}_4$ and 1 moles of $\text{NO}_2$

2. Relevant Concepts

• The equilibrium constant $K$ at 50% dissociation can be calculated as: $K = \frac{[\text{NO}_2]^2}{[\text{N}_2\text{O}_4]} = \frac{(1)^2}{0.5} = 2$
• The relationship between $\Delta G°$ and $K$ is given by the formula: $\Delta G° = -RT \ln K$ Where:
  • $R$ = 8.314 J/(mol·K) (universal gas constant)
  • $T$ = 27 °C + 273.15 = 300.15 K
  • $K = 2$

3. Analysis and Detail

• Substituting the values into the formula: $\Delta G° = - (8.314) (300.15) \ln(2)$
• Calculate $\ln(2)$: $\ln(2) \approx 0.693$
• Now substitute this back into the equation: $\Delta G° \approx - (8.314) (300.15) (0.693)$
• Performing the calculations step-by-step:
  • Calculate $8.314 \times 300.15$: $8.314 \times 300.15 \approx 2498.3 \text{ J}$
  • Now multiply by $-0.693$: $\Delta G° \approx -2498.3 \times 0.693 \approx -1730.3 \text{ J}$

4. Verify and Summarize

• After performing the calculations and verifying each step:
  • The calculated $x$ is therefore approximately $1730.3$ J mol$^{-1}$.

Final Answer

The value of $x$ is $1730.3$ J mol$^{-1}$.