Question

A 1.00 L container holds 0.015 mol of H2 (g) , 0.015 mol of I2 (g), and 0.015 mol of HI (g) at 721 K. What are the concentrations(pressures) of H2 (g), I2 (g), and HI (g) after the system achieved a state of equilibrium? The value of Kc is 50.0 for reaction: H2 (g) + I2 (g)  2HI (g)

Answer 1

Given that [H2] = 0.015 mol/ 1L = 0.015 M

[I2] = 0.015 mol/ 1L = 0.015 M

[HI] = 0.015 mol/ 1L = 0.015 M

H2 (g) + I2 (g) <----------> 2HI (g) Kc = 50

Initial   0.015 M    0.015 M    0.015 M

At equilibrium 0.015 -x 0.015 -x 0.015 + 2x

  

Kc = [HI]²/ [H2] [I2]

50 = (0.015 + 2x)² / (0.015 -x ) ( 0.015 -x )

50 (0.015-x)² =  (0.015 + 2x)²

(7.07)² (0.015-x)² =  (0.015 + 2x)²

(7.07) (0.015-x)=  (0.015 + 2x)

0.10605 - 7.07 x = 0.015 + 2x

2x + 7.07 x = 0.10605 -  0.015

9.07 x = 0.09105

x = 0.01 M

Therefore,

Equilibrium concentrations are

[H2] = 0.015 - x = 0.015 - 0.01 = 0.005

[I2] =  0.015 - x = 0.015 - 0.01 = 0.005

[HI] =  0.015 +2x = 0.015 + 2 x 0.01 = 0.035 M