The K_eq for the equilibrium below is 50. H₂ (g) + I₂ (g) 2HI (g)
What is the value of K_eq for the following reaction? H₂ (g) + I₂(g) HI (g)

Question

The K_eq for the equilibrium below is 50. H₂ (g)+ I₂ (g) 2HI (g) What is the value of K_eq for the following reaction? H₂ (g)+ I₂(g) HI (g)

Quizplus · Accepted Answer

The value of K_eq for a reaction is equal to the ratio of the product concentrations to the reactant concentrations at equilibrium, with each concentration raised to the appropriate stoichiometric coefficient. In the given reaction, the stoichiometric coefficients for each species are 1, so K_eq = [HI]^2 / ([HS12S]^1[IS12S]^1). Plugging in the equilibrium concentrations from the given K_eq value gives: 50 = ([HI]^2) / ([HS12S][IS12S]). Solving for [HI]^2 gives [HI]^2 = 50[HS12S][IS12S]. Taking the square root of both sides gives [HI] = sqrt(50[HS12S][IS12S]) = sqrt(50*0.1*0.1) = 0.707 M. Thus, K_eq = [HI]^2 / ([HS12S][IS12S]) = (0.707)^2 / (0.1*0.1) = 7.07.

The Keq for the Equilibrium Below Is 50

The K_eq for the Equilibrium Below Is 50