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

Using the given equilibrium constant, we can set up the following reaction quotient expression at equilibrium:

Qc = [HI]² / [HSI2][I2]

At equilibrium, Qc = Kc = 64. We can use an ICE table to determine the equilibrium concentrations of the reactants and products.

Initial:    3.00 M      3.00 M          0
Change:   -x           -x              +2x
Equilibrium: 3.00 - x  3.00 - x       2x

Substituting these equilibrium concentrations into the Qc expression and simplifying, we get:

64 = [HI]² / (3.00 - x)²

Solving for x, we get:

x = 1.20 M

Therefore, the equilibrium concentration of HI is:

[HI] = 2x = 2(1.20) = 2.40 M

So the best choice is B.

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

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