Question

Iron(III) oxide and hydrogen react to form iron and water, like this: (s)(g)(s)(g) At a certain temperature, a chemist finds that a reaction vessel containing a mixture of iron(III) oxide, hydrogen, iron, and water at equilibrium has the following composition: compound amount Calculate the value of the equilibrium constant for this reaction. Round your answer to significant digits.

Answer 1

The question is incomplete, here is the complete question:

Iron(III) oxide and hydrogen react to form iron and water, like this:

$Fe_2O_3(s)+3H_2(g)\rightarrow 2Fe(s)+3H_2O(g)$

At a certain temperature, a chemist finds that a 5.4 L reaction vessel containing a mixture of iron(III) oxide, hydrogen, iron, and water at equilibrium has the following composition:

Compound        Amount

   $Fe_2O_3$         3.54 g

      $H_2$             3.63 g

      Fe             2.37 g

     $H_2O$           2.13 g

Calculate the value of the equilibrium constant for this reaction. Round your answer to 2 significant digits

Answer: The value of equilibrium constant for the given reaction is $2.8\times 10^{-4}$

Explanation:

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

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

• For hydrogen gas:

Given mass of hydrogen gas = 3.63 g

Molar mass of hydrogen gas = 2 g/mol

Volume of solution = 5.4 L

Putting values in above equation, we get:

$\text{Molarity of hydrogen gas}=\frac{3.63}{2\times 5.4}\\\\\text{Molarity of hydrogen gas}=0.336M$

• For water:

Given mass of water = 2.13 g

Molar mass of water = 18 g/mol

Volume of solution = 5.4 L

Putting values in above equation, we get:

$\text{Molarity of water}=\frac{2.13}{18\times 5.4}\\\\\text{Molarity of water}=0.0219M$

The given chemical equation follows:

$Fe_2O_3(s)+3H_2(g)\rightarrow 2Fe(s)+3H_2O(g)$

The expression of $K_c$ for above equation follows:

$K_c=\frac{[H_2O]^3}{[H_2]^3}$

The concentration of pure solids and pure liquids are taken as 1 in equilibrium constant expression. So, the concentration of iron and iron (III) oxide is not present in equilibrium constant expression.

Putting values in above equation, we get:

$K_c=\frac{(0.0219)^3}{(0.336)^3}\\\\K_c=2.77\times 10^{-4}$

Hence, the value of equilibrium constant for the given reaction is $2.8\times 10^{-4}$