Consider the reaction X₂(g) → 2X(g). When a vessel initially containing 755 torr of X2 comes to equilibrium at 298 K, the equilibrium partial pressure of X is 103 torr. The same reaction is repeated with an initial partial pressure of 748 torr of X₂ at 755 K; the equilibrium partial pressure of X is 532 torr. Find ΔH° for the reaction.

Verified step by step guidance

1. <insert step 1> Start by writing the expression for the equilibrium constant (Kp) for the reaction X2(g) → 2X(g). The expression is Kp = (P_X)^2 / P_X2, where P_X is the partial pressure of X and P_X2 is the partial pressure of X2.>
2. <insert step 2> For the first condition at 298 K, calculate the equilibrium partial pressure of X2 using the initial pressure and the change in pressure due to the formation of X. The change in pressure is half of the equilibrium pressure of X, so P_X2 = initial pressure - (1/2) * P_X.>
3. <insert step 3> Calculate Kp at 298 K using the equilibrium pressures found in step 2.>
4. <insert step 4> Repeat steps 2 and 3 for the second condition at 755 K to find the equilibrium constant Kp at this temperature.>
5. <insert step 5> Use the van 't Hoff equation, ln(K2/K1) = -ΔH°/R * (1/T2 - 1/T1), where K1 and K2 are the equilibrium constants at temperatures T1 and T2, respectively, and R is the gas constant, to solve for ΔH°.>