Question

For the reaction H2 + I2 (g)2HI (g) with Kc = 54.3 at 698 K, if the initial amounts were 0.800 mole H2 and 0.500 mole I2 in a 5.25-L vessel at 698 K, write the ICE table, and what will be the amounts of reactants and products (in mole(s)) when equilibrium is attained?

Answer 1

initially

[H2] = 0.8/5.25 = 0.15238

[I2] = 0.5/5.25 = 0.09523

[HI] = 0

the change

[H2] = +x

[I2] = -x

[HI] = +2x

in equilbirium

[H2] = 0.15238 - x

[I2] = 0.09523 - x

[HI] = 0 + 2x

substitute in Kc

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

54.3 = (2x)^2 /((0.15238 - x)(0.09523 - x))

4/54.3 x^2 = 0.15238 *0.09523 - (0.15238 +0.09523 )x + x^2

0.07366x^2 = 0.01451 - 0.24761x + x^2

0.92634x^2 - 0.24761x + 0.01451 = 0

x = 0.0867

[H2] = 0.15238 - 0.0867 = 0.06568

[I2] = 0.09523 - 0.0867= 0.00853

[HI] = 0 + 2*0.0867= 0.1734

in mol: mol = MV

[H2] = 0.06568*5.25 = 0.34482 mol

[I2] = 0.00853*5.25  = 0.0447825 mol

[HI] = 0.1734*5.25  = 0.91035mol