Chemistry?? Ice table??

Question

2. Calculating Equilibrium Constants from Initial and Equilibrium Concentrations

A closed system initially containing 1.000 x 10−3 M of H2 and 2.000 x 10−3 M of I2.

At 448°C is allowed to reach equilibrium. Analysis of the equilibrium mixture shows that the concentration of HI is 1.87 x 10−3 M.

Calculate Kc at 448°C for the reaction taking place. H2 (g) + I2 (g) --- >2 HI(g) 

 

Answer 1

Step 1:

(2)

The chemical equation is .

Initial concentration of H₂ is 1.00 * 10^-3 M.

Initial concentration of I₂ is 2.00 * 10^-3 M.

Concentration of HI at equilibrium is 1.87 * 10^-3 M.

1 moles of H₂ and 1 mole of I₂ decomposes to 2 moles of HI.

Let 2x be the change in concentration of HI,

then change in concentration of H₂ be   and

change in concentration of I₂ be .     (negative sign indicates decomposition)

Step 2:

Construct an ICE table.

Equilibrium Amount = Initial amount + Change in amount.

(i)

Find the change in amount of HI.

Equilibrium amount of HI is 1.87 * 10^-3 M.

Initial amount of HI is 0 M.

Equilibrium Amount = Initial amount + Change in amount.

1.87 * 10^-3 = 0 + 2x

2x = 1.87 * 10^-3

x = 0.935 * 10^-3.

(ii)

Find the equilibrium amount of H₂.

Initially the amount of H₂ is 1.00 * 10^-3 M.

Equilibrium Amount = Initial amount + Change in amount.

Equilibrium Amount =

Equilibrium Amount =

Equilibrium Amount =

Equilibrium Amount of H₂ is M.

(iii)

Find the equilibrium amount of I₂.

Initially the amount of I₂ is 2.00 * 10^-3 M.

Equilibrium Amount = Initial amount + Change in amount.

Equilibrium Amount =

Equilibrium Amount =

Equilibrium Amount =

Equilibrium Amount of I₂ is M.

Comments

Step 3:

Find the equilibrium constant of the reaction.

The chemical equation is .

Equilibrium constant .

Equilibrium constant of the reaction is 50.51.

Solution:

Equilibrium constant of the reaction is 50.51.