At 400ºC, K_c = 64 for the equilibrium H₂(g)+ I₂(g) 2HI(g). If 3.00 mol H₂ and 3.00 mol I₂ are introduced into an empty 4.0 L vessel, find the equilibrium concentration of HI at 400ºC.

Question

Quizplus · Accepted Answer

First, we need to write the reaction equation and the expression for the equilibrium constant: H₂(g) + I₂(g) <=> 2HI(g) K_c = [HI]^2 / ([H₂][I₂]) At equilibrium, let's assume that x mol of HI is formed. Thus, the concentration of H₂ and I₂ will be (3.00 - x) mol/L. Substituting these values into the expression for K_c, we get: 64 = x^2 / [(3.00 - x)(3.00 - x)] Solving for x, we get x = 1.20 mol/L. Therefore, the equilibrium concentration of HI is 1.20 M, which is answer choice B.

At 400ºC, Kc = 64 for the Equilibrium H2(g)+ I2(g)

At 400ºC, K_c = 64 for the Equilibrium H₂(g)+ I₂(g)