Consider the following aquarium, for which Kp = 7.49 × 10-2 at 500 degree C 2 Cl

Question

Consider the following aquarium, for which Kp = 7.49 × 10-2 at 500 degree C 2 Cl2(g) + 2 H2O(g) = 4 HCl(g) + O2(g) What a the value of Kp for the reaction 4 HCl(g) + O2(g) rlhar2 2 Cl2(g) + 2 H2O(g)? ANSWER: What is the value of Kp for the reaction Cl2(g) + H2O(g) rlhar2 2 HCl(g) + 1/2O2(g)? ANSWER: What is the value of Kc for the reaction in part B? ANSWER

Explanation / Answer

part A. Kp = 1 / 7.49*10^-2 = 13.35. Part B. As all moles are divided by two thus Kp new = (Kp old)^1/2 -----> Kp new = (7.49*10^-2)^1/2 = 0.274 Part C. Kc = Kp / (RT) ^(delta n) where delta n = sum of mole numbers of product - sum of mole number of reactants. thus delta n = 2.5-2 = 0.5 R = 0.082 L atm K-1 mol-1 and T = 273 + 500 = 773 K. Kc = 0.274 / (0.082 * 773)^(delta n) ----> Kc = 3.44*10^-2

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