Step 1:

(a)

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 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 of I₂ is $\"\"$ M.

\ 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.