# Step-by-Step Solution: Gibbs Free Energy Change for n-Pentane Combustion ## 1. **Write and Balance the Combustion Reaction** The combustion of n-pentane (\(\mathrm{C_5H_{12}}\)) is: \[ \mathrm{C_5H_{12}(g) + 8\ O_2(g) \rightarrow 5\ CO_2(g) + 6\ H_2O(g)} \] ## 2. **Standard Gibbs Free Energy Change (\(\Delta_r G^\circ\))** We'll need the standard Gibbs free energies of formation (\(\Delta_f G^\circ\)) at **1200 K**. Values at 1200 K are not always tabulated, but for a typical problem, let's use the standard state (\(298\,\mathrm{K}\)) values as an approximation (the process is the same, but result is approximate). If you have values at 1200 K, use those instead. Approximate \(\Delta_f G^\circ\) at 298 K (in kJ/mol): - \(\mathrm{C_5H_{12}(g)}\): \(-24.9\) - \(\mathrm{O_2(g)}\): \(0\) - \(\mathrm{CO_2(g)}\): \(-394.4\) - \(\mathrm{H_2O(g)}\): \(-228.6\) Now, compute \(\Delta_r G^\circ\): \[ \Delta_r G^\circ = [5 \times (-394.4) + 6 \times (-228.6)] - [1 \times (-24.9) + 8 \times 0] \] \[ = [(-1972.0) + (-1371.6)] - [-24.9] \] \[ = (-3343.6) + 24.9 \] \[ = -3318.7\ \mathrm{kJ} \] ## 3. **Reaction Quotient (\(Q\)) at 30% Progress** Assume we start with 1 mol \(\mathrm{C_5H_{12}}\) and excess \(\mathrm{O_2}\): - At 30% progress, \(\xi = 0.3\) (fraction of reaction completed). Calculate moles at this progress: - \(\mathrm{C_5H_{12}}\): \(1 - 0.3 = 0.7\) mol - \(\mathrm{O_2}\): \(8 - (8 \times 0.3) = 8 - 2.4 = 5.6\) mol - \(\mathrm{CO_2}\): \(0 + (5 \times 0.3) = 1.5\) mol - \(\mathrm{H_2O}\): \(0 + (6 \times 0.3) = 1.8\) mol Total moles: \[ n_{tot} = 0.7 + 5.6 + 1.5 + 1.8 = 9.6 \text{ mol} \] Mole fractions: - \(y_{\mathrm{C_5H_{12}}} = 0.7 / 9.6 = 0.0729\) - \(y_{\mathrm{O_2}} = 5.6 / 9.6 = 0.583\) - \(y_{\mathrm{CO_2}} = 1.5 / 9.6 = 0.156\) - \(y_{\mathrm{H_2O}} = 1.8 / 9.6 = 0.188\) ## 4. **Calculate the Reaction Quotient \(Q\)** Assuming ideal gas, \(Q\) (using partial pressures, or equivalently, mole fractions if total pressure is constant): \[ Q = \frac{(y_{\mathrm{CO_2}})^5 \cdot (y_{\mathrm{H_2O}})^6}{(y_{\mathrm{C_5H_{12}}})^1 \cdot (y_{\mathrm{O_2}})^8} \] Plug in values: \[ Q = \frac{(0.156)^5 \cdot (0.188)^6}{(0.0729) \cdot (0.583)^8} \] Calculate powers: - \((0.156)^5 \approx 1.23 \times 10^{-4}\) - \((0.188)^6 \approx 4.67 \times 10^{-5}\) - \((0.583)^8 \approx 0.0169\) So: \[ Q = \frac{(1.23 \times 10^{-4}) \cdot (4.67 \times 10^{-5})}{0.0729 \cdot 0.0169} \] \[ Q = \frac{5.74 \times 10^{-9}}{0.00123} \] \[ Q = 4.67 \times 10^{-6} \] ## 5. **Calculate \(\Delta_r G\) at 1200 K and 30% Progress** The general formula: \[ \Delta_r G = \Delta_r G^\circ + RT \ln Q \] Where: - \(R = 8.314\ \mathrm{J\,mol^{-1}\,K^{-1}}\) - \(T = 1200\,\mathrm{K}\) Calculate \(RT\): \[ RT = 8.314 \times 1200 = 9977\ \mathrm{J}\ \approx 9.98\ \mathrm{kJ} \] Calculate \(RT \ln Q\): \[ RT \ln Q = 9.98\ \mathrm{kJ} \times \ln(4.67 \times 10^{-6}) \] \[ \ln(4.67 \times 10^{-6}) \approx -12.27 \] \[ RT \ln Q = 9.98 \times (-12.27) = -122.5\ \mathrm{kJ} \] Now, sum: \[ \Delta_r G = -3318.7\ \mathrm{kJ} + (-122.5\ \mathrm{kJ}) \approx -3441.2\ \mathrm{kJ} \] ## 6. **Final Answer** \[ \boxed{ \Delta_r G = -3441\ \mathrm{kJ} } \] **at 30% reaction progress, 1200 K, and 1 mol initial n-pentane in excess oxygen.** --- ### **Summary Table** | Step | Value | Notes | |------|-------|-------| | \(\Delta_r G^\circ\) | \(-3318.7\) kJ | At 298 K (approximate) | | \(Q\) | \(4.67 \times 10^{-6}\) | At 30% progress | | \(RT \ln Q\) | \(-122.5\) kJ | At 1200 K | | \(\Delta_r G\) | \(-3441\) kJ | At 30% progress, 1200 K | --- **If you have the standard Gibbs energies at 1200 K, use those in step 2. The process remains the same.**