The equilibrium constant, Kc, for the reaction given below is 794. A 5.1 L vesse

Question

The equilibrium constant, Kc, for the reaction given below is 794. A 5.1 L vessel is filled with 0.03 mol H2, 0.09 mol I2, and 0.02 mol HI.H2(g) +I2(g) - 2 HI(g) Is the mixture at equilibrium? Yes no If not, in which direction will the reaction proceed? The reaction will proceed in the backwards direction, producing more reactant. The reaction will proceed in the forward direction, producing more product. The reaction is in equilibrium. What are the equilibrium concentrations of each of the gases?

Explanation / Answer

Kc = [HI]^2/[H2][I2] = 794

[H2] = 0.03/5 = 6 x 10^-3 M

[I2] = 0.09/5 = 0.018 M

[HI] = 0.02/5 = 4 x 10^-3 M

Feed the values in Kc equation,

Kc = (4 x 10^-3)^2/(6 x 10^-3)(0.018) = 0.148

the given Kc does not match with calculated Kc so the reaction is not at equilibrium

Answer would be,

No

The reaction is not at equlibrium, it would proceed,

The reaction will proceed in the forward direction,producing more product.

Equilibrium concentrations of each gases,

let x be the change at equilibrium then,

749 = ((4 x 10^-3 + 2x)^2/(6 x 10^-3-x)(0.018-x)

0.081 - 17.98x + 749x^2 = 1.6 x 10^-5 + 0.016x + 4x^2

745x^2 - 17.996x + 0.081 = 0

x = 6 x 10^-3 M

Equilibrium concentrations thus become,

[H2] = 6 x 10^-3 - 6 x 10^-3 = 0 M

[I2] = 0.018 - 6 x 10^-3 = 0.012 M

[HI] = 4 x 10^-3 + 2 x 6 x 10^-3 = 0.016 M

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