Question

The equilibrium constant, Kc, for the following reaction is 1.80×10-2 at 698 K.

2HI(g)------> H2(g) + I2(g)

Calculate the equilibrium concentrations of reactant and products when 0.311 moles of HI are introduced into a 1.00 L vessel at 698 K.

[HI]= ___ M

[H2]= ___M

[I2]= ____M

Answer 1

ICE Table:

[HI] [H2] [I2]

initial 0.311 0 0

change -2x +1x +1x

equilibrium 0.311-2x +1x +1x

Equilibrium constant expression is

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

0.018 = (1*x)^2/(0.311-2*x)^2

sqrt(0.018) = (1*x)/(0.311-2*x)

0.1341640786499874 = (1*x)/(0.311-2*x)

4.173*10^-2-0.2683*x = 1*x

4.173*10^-2-1.268*x = 0

x = 0.0329

At equilibrium:

[HI] = 0.311-2x = 0.311-2*0.0329 = 0.2452 M

[H2] = +1x = +1*0.0329 = 0.0329 M

[I2] = +1x = +1*0.0329 = 0.0329 M

[HI] = 0.245 M

[H2] = 0.0329 M

[I2] = 0.0329 M