Question

The reaction of iron and water vapor results in an equilibrium 3Fe(s) + 4H2O (g) ⇋ Fe3O4(s) + 4H2(g) Kc = 5.0 at 800°C. What is the concentration of water present (in M) at equilibrium if the reaction is initiated with 7.5 g of H2 and excess Fe3O4 in a 15.0 liter container?

Answer 1

Answer: The concentration of water at equilibrium for the given chemical reaction is 0.1 M

Explanation:

To calculate the molarity of solution, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{Volume of solution (in L)}}$

Given mass of hydrogen gas = 7.5 g  

Molar mass of hydrogen gas = 2 g/mol

Volume of solution = 15.0 L

Putting values in above equation, we get:

$\text{Molarity of hydrogen gas}=\frac{7.5}{2g/mol\times 15.0}\\\\\text{Molarity of hydrogen gas}=0.25M$

For the given chemical equation:

$3Fe(s)+4H_2O(g)\rightleftharpoons Fe_3O_4(s)+4H_2(g)$

As, the reaction is started by hydrogen gas. So, the reaction will proceed backwards and the equilibrium constant will be the reciprocal of equilibrium constant for the given reaction.

Now, the reaction becomes:

                      $Fe_3O_4(s)+4H_2(g)\rightleftharpoons 3Fe(s)+4H_2O(g)$

Initial:                               0.25        

At eqllm:                         0.25-4x                     4x

We are given:

$K_c=5.0$

So, $K_c'=\frac{1}{5.0}=0.2$

The expression of $K_c'$ for above equation follows:

$K_c'=\frac{[H_2O]^4}{[H_2]^4}$

The concentration of pure solids and liquids are taken as 1  in the equilibrium constant expression. So, iron and $Fe_3O_4$ will not appear in the expression.

Putting values in above equation, we get:

$0.2=\frac{(4x)^4}{(0.25-4x)^4}\\\\x=-0.126,0.025$

Neglecting the value of x = -0.126 because concentration cannot be negative.

So, equilibrium concentration of water = $4x=(4\times 0.025)=0.1M$

Hence, the concentration of water at equilibrium for the given chemical reaction is 0.1 M