For the following chemical equilibrium,K_p = 4.6 × 10^-14 at 25 °C,find the value of K_c for this reaction at 25 °C. 2 Cl₂(g)+ 2 H₂O(g)⇌ 4 HCl(g)+ O₂(g) A)K_c = 1.9 × 10^-15 B)K_c = 2.2 × 10^-14 C)K_c = 1.1 × 10^-12 D)K_c = 9.4 × 10^-14 E)K_c = 4.6 × 10^-14

Question

Quizplus · Accepted Answer

The relationship between Kp and Kc is given by the equation Kp = Kc(RT)^(Δn), where Δn is the change in moles of gas. For this reaction, Δn = (4+1)-(2+2) = 1. At 25°C (298 K), R = 0.0821 L·atm/mol·K. Solving for Kc gives Kc = Kp / (RT)^(Δn) = 4.6 × 10^-14 / (0.0821 × 298)^1 = 1.9 × 10^-15.

For the Following Chemical Equilibrium,Kp = 4

For the Following Chemical Equilibrium,K_p = 4