Question

The equilibrium constant, K_c, 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 0.249 moles of HI are introduced into a 1.00 L vessel at 698 K.

[HI]	=	M
[H2]	=	M
[I2]	=	M

Answer 1

2 HI (g)    <—> H2 (g)   +   I2 (g)

0.249           0           0       (initial)

0.249-2x       x           x       (at equilibrium)

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

1.80*10^-2 = x^2 / (0.249-2x)^2

sqrt(1.80*10^-2) = x / (0.249-2x)

x / (0.249-2x) = 0.1342

x = 0.0334 - 0.2684*x

1.2684*x = 0.0334

x = 0.0263 M

So, at equilibrium,

[H2] = 0.249 - 2x

= 0.249 - 2*0.0263

= 0.1964 M

[H2] = x = 0.0263 M

[I2] = x = 0.0263 M