Steam decomposes at high temperature: 2(g) ⇌ 2(g) + (g); ΔH° = 240 kJ/mol and ΔS° = 50 J K^-1 mol^-1 — Chemical Equilibrium Chemistry Question

💡 Solution & Explanation

Step 1 - Relate standard Gibbs free energy change to the equilibrium constant When Kp = 1.0, \(\ln K_p = \ln(1.0) = 0\), so: \[\Delta G^\circ = -RT \ln K_p = 0\] Step 2 - Apply the Gibbs-Helmholtz equation \[\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ\] Setting \(\Delta G^\circ = 0\): \[0 = \Delta H^\circ - T\Delta S^\circ\] \[T = \frac{\Delta H^\circ}{\Delta S^\circ}\] Step 3 - Convert units and calculate \[\Delta H^\circ = 240 \text{ kJ mol}^{-1} = 240000 \text{ J mol}^{-1}\] \[\Delta S^\circ = 50 \text{ J K}^{-1}\text{ mol}^{-1}\] \[T = \frac{240000}{50} = \boxed{4800 \text{ K}}\] Step 4 - Evaluate each option * **Option (A) 4.8 K:** Incorrect. Unit error — dividing 240 kJ directly by 50 J without converting kJ to J. * **Option (B) 4800 K:** Correct. Proper unit conversion gives T = 240000/50 = 4800 K. * **Option (C) 480 K:** Incorrect. Decimal scaling error. * **Option (D) Impossible:** Incorrect. 4800 K is thermodynamically reachable. \[\boxed{\text{B}}\]