Considering H2(g) + I2(g) ↔ 2HI and a temperature of 731K; 2.40 mol of H2 and 2.40 mol of I2 are placed in a 1.00 L vessel. What is the equilibrium concentration of H2 in the gaseous mixture? The equilibrium constant is K = 49.0

1 Answer
Dec 13, 2014

The answer is 0.53M.

Starting from the balanced chemical equation

H_(2(g)) + I_(2(g)) rightleftharpoons2HI_((g))

Since the reaction's equlibrium constant, K_(eq), greater than 1, the reaction will favor the formation of the product, HI, so we would expct the equilibrium concentrations of H_2 and I_2 to be smaller than the concentration of HI.

From the data given we can determine the starting concentrations of both H_2 and I_2 to be

C_(H_2) = n_(H_2)/V = (2.40 mol es)/(1.00L) = 2.40M
C_(I_2) = n_(I_2)/V = (2.40 mol es)/(1.00L) = 2.40M

We can now determine the equilibrium concentrations for this reaction by using the ICE method (more here: http://en.wikipedia.org/wiki/RICE_chart)

...H_(2(g)) + I_(2(g))rightleftharpoons2HI_((g))
I: 2.40........2.40...........0
C: (-x).............(-x)..........(+2x)
E: 2.40-x.....2.40-x........2x

We know that K_(eq) = ([HI]^2)/([H_2]*[I_2]), so we get

K_(eq) = (2x)^2/((2.40-x)(2.40-x)) = (4x^2)/(2.40-x)^2 = 49.0

Rearranging this equation will give us

45x^2 - 235.2x + 282.24 = 0, which produces two values for x, x_1 = 3.36 and x_2 = 1.87; we cannot choose x_1, since that would imply negative concentration values at equilibrium for both H_2 and I_2 (2.40 -3.36 = -0.96);

Therefore, x = 1.87, which means that, at equilibrium,

[H_2] = 2.40 - 1.87 = 0.53M
[I_2] = 2.40 - 1.87 = 0.53M
{HI] = 2 * 1.87 = 3.74M

Notice how the final concentrations match the estimate derived from K_(eq)'s value - the reaction indeed favors the product ,HI.

A quick word on the ICE table...the starting concentration of HI is 0 M because only H_2 and I_2 are present in the vessel; x simply represents the change in concentrations in accordance to the balanced chemical equation -> 1 mole of H_2 and 1 mole of I_2 combine to form 2 moles of HI - that is where the +2x comes from...