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, what is the equilibrium concentration of HI at 400ºC?

Question

Quizplus · Accepted Answer

First, use the equation K_c = 64 to find the equilibrium constant, Kc. Kc = [HI]^2/([H₂] [I₂]) Plug in the given values: 64 = [HI]^2/((3.00 mol/L) (3.00 mol/L)) Simplify: [HI]^2 = (64) (9.00 mol^2/L^2) [HI] = 4.8 M Therefore, the equilibrium concentration of HI is 4.8 M. However, we must check if this is a valid answer by making sure that the initial amounts of H₂ and I₂ are not significantly depleted at equilibrium. The initial concentration of H₂ and I₂ in the 4.0-L vessel is: [H₂] = 3.00 mol / 4.0 L = 0.75 M [I₂] = 3.00 mol / 4.0 L = 0.75 M The equilibrium concentration of H₂ and I₂ can be calculated using the expression: Kc = [HI]^2 / ([H₂] [I₂]) Substituting the values obtained: 64 = (4.8 M)^2 / ([H₂] [I₂]) [H₂] [I₂] = (4.8 M)^2 / 64 = 0.36 M^2 At equilibrium, the concentration of H₂ and I₂ will decrease by x, which will be very small compared to their initial concentrations because Kc is large. Therefore, we can ignore x in the denominator: 0.36 M^2 = (0.75 M - x)(0.75 M - x) Solving for x: x = 0.135 M The change in concentration is only 0.135 M, which is small compared to the initial concentration of 0.75 M. Therefore, we can assume that the initial concentration of H₂ and I₂ are not significantly depleted at equilibrium. Therefore, the equilibrium concentration of HI is 4.8 M, answer choice (B).

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

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