For .3(s) ⇌ .(s) + 2(g), dissociation pressure is 7×10^-3 atm at 27°C. Dissociation pressure at 127° — Chemical Equilibrium Chemistry Question

💡 Solution & Explanation

Step 1 - Express Kp for the heterogeneous reaction For \(\ce{CuSO4.3H2O(s) <=> CuSO4.H2O(s) + 2H2O(g)}\): Solids have unit activity, so: \[K_p = (p_{\ce{H2O}})^2 = P^2\] where \(P\) is the total dissociation pressure. Step 2 - Calculate Kp1 at T1 = 27°C = 300 K \[K_{p1} = (P_1)^2 = (7 \times 10^{-3})^2 = 4.9 \times 10^{-5} \text{ atm}^2\] Step 3 - Apply Van't Hoff equation from 300 K to 400 K \[\ln\left(\frac{K_{p2}}{K_{p1}}\right) = \frac{\Delta H^\circ}{R}\left(\frac{1}{T_1} - \frac{1}{T_2}\right) = \frac{3360}{2}\left(\frac{1}{300} - \frac{1}{400}\right) = 1680 \times \frac{100}{120000} = 1.4\] Since \(\ln 2 = 0.7\): \(\quad 1.4 = 2 \times 0.7 = 2\ln 2 = \ln 4\) \[\therefore \frac{K_{p2}}{K_{p1}} = 4 \implies K_{p2} = 4 \times 4.9 \times 10^{-5} = 1.96 \times 10^{-4} \text{ atm}^2\] Step 4 - Find P2 \[P_2 = \sqrt{K_{p2}} = \sqrt{1.96 \times 10^{-4}} = 1.4 \times 10^{-2} \text{ atm}\] \[\boxed{P_2 = 1.4 \times 10^{-2} \text{ atm}}\] Step 5 - Explain all options * **Option (A) 9.8×10^-3 atm**: Incorrect. This is 1.4×P1 without accounting for the P^2 relationship in Kp. * **Option (B) 1.4×10^-2 atm**: Correct. Van't Hoff gives Kp2 = 4Kp1; taking square root gives 1.4×10^-2 atm. * **Option (C) 1.4×10^-3 atm**: Incorrect. Off by a factor of 10 — power-of-ten arithmetic error. * **Option (D) 9.8×10^-2 atm**: Incorrect. Arithmetic error in the final calculation.