ΔrG° for the formation of (g) from its gaseous elements is -2.303 kcal/mol at 500 K. When the partia — Chemical Equilibrium Chemistry Question

💡 Solution & Explanation

Step 1 - Write the reaction for H2 + I2 → 2HI and find ΔrG°(rxn) The formation of \ce{HI}(g) is: \[\ce{H2(g) + I2(g) <=> 2HI(g)}\] Given \(\Delta_f G°(\ce{HI}) = -2.303\text{ kcal/mol}\), the standard Gibbs energy for the overall reaction (producing 2 mol HI): \[\Delta_r G°_{\text{rxn}} = 2 \times (-2.303) = -4.606\text{ kcal/mol}\] Step 2 - Calculate Kp from ΔrG° Using \(\Delta G° = -2.303 RT \log K_p\) with \(R = 2 \times 10^{-3}\text{ kcal/(mol·K)}\) and \(T = 500\text{ K}\): \[-4.606 = -2.303 \times (2 \times 10^{-3}) \times 500 \times \log K_p\] \[-4.606 = -2.303 \times 1 \times \log K_p\] \[\log K_p = \frac{4.606}{2.303} = 2 \implies K_p = 10^2 = 100\] Step 3 - Apply condition ΔG = 0 means Q = Kp When \(\Delta G = 0\), the system is at equilibrium, so the reaction quotient equals \(K_p\): \[Q_p = K_p = 100\] \[K_p = \frac{p_{\ce{HI}}^2}{p_{\ce{H2}} \cdot p_{\ce{I2}}}\] Substituting \(p_{\ce{HI}} = 10\text{ atm}\), \(p_{\ce{I2}} = 0.001\text{ atm}\): \[100 = \frac{(10)^2}{p_{\ce{H2}} \times 0.001} = \frac{100}{p_{\ce{H2}} \times 0.001}\] \[p_{\ce{H2}} \times 0.001 = \frac{100}{100} = 1\] \[p_{\ce{H2}} = \boxed{1000\text{ atm}}\] Step 4 - Evaluate all options - **Option (A) 1000 atm**: Correct. With Kp=100 and Qp=Kp at ΔG=0, P_H2 = P_HI²/(Kp × P_I2) = 100/(100 × 0.001) = 1000 atm. - **Option (B) 10000 atm**: Incorrect. Arithmetic error; off by factor of 10. - **Option (C) 100 atm**: Incorrect. Misses the 0.001 factor for iodine pressure. - **Option (D) 31.63 atm**: Incorrect. Different mathematical error.