Question

A.) Initially, 1.0 mol of H₂ and 2.0 mol of I₂ are placed into a 10.0 L reaction flask. When equilibrium was established, the [HI] was 0.080 mol/L. Find the equilibrium concentrations of H₂ and I₂ and evaluate K_c for the following reaction:

H₂ (g)+ I₂ (g) <--> 2HI (g)

B.) The equilibrium constant for the reaction above is 45.0 at 490 C. What is the equilibrium concentration of each species if 1.0 mol of HI were put in a 2.0 L flask at 490 C?

Answer 1

concentration of H2 = 1 / 10 = 0.1 M

concentration of I2 = 2 / 10 = 0.2 M

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

0.1    0.2                             0

0.1-x       0.2- x                            2x

concnentration of [HI] = 0.080 M

2x = 0.08

x = 0.04

equilibrium concentrations of

[H2] = 0.060 M

[I2] = 0.16 M

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

     = 0.08^2 / 0.06 x 0.16

Kc = 0.67

B)

concentration of HI = 1 / 2 = 0.5 M

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

0               0                            0.5

x                x                           0.5 - x

Kc = 0.5 - x / x^2

45 = 0.5 - x / x^2

x = 0.095

equilibrium concentration of each species :

[H2] = 0.095 M

[I2] = 0.095 M

[HI] = 0.405 M