Given the following reaction at equilibrium,if K_c = 1.90 × 10¹⁹at 25.0 °C,K_p = ________. H₂ (g)+ Br₂ (g) 2 HBr (g)

Question

Quizplus · Accepted Answer

From the given equilibrium constant expression, we can write: K_c = [H2₂][Br2₂]/[HBr]2 The equilibrium constant expression for the reverse reaction will be: K_p = [HBr]2/[H2₂][Br2₂] Since the equilibrium constant for the reverse reaction is equal to 1/Kc, we have: K_p = 1/K_c Substituting the given value for K_c, we get: K_p = 1/(1.90 × 10¹⁹) K_p = 5.26 × 10^-20 Therefore, the answer is (D) 5.26 × 10^-20.

Given the Following Reaction at Equilibrium,if Kc = 1

Given the Following Reaction at Equilibrium,if K_c = 1