Question

One mole of H2O and one mole of CO are taken in 10 L vessel and heated to
725 K. At equilibrium 40% of water (by mass) reacts with CO according to the equation, H2O (g) + CO (g) = H, (g) + Co (g) Calculate the equilibrium constant for the reaction.​

Answer 1

Solution :

Initially,

• Moles of H₂O = 1

• Moles of CO = 1

$\sf{H_2O + CO_2 \leftrightharpoons H_2 + CO_2 }\\ \\ \\ \sf{1 } \: \: \: \: \: \: \: \: \: \: \: \: \sf{1} \: \: \: \: \: \: \: \: \: \: \: \: \sf{0} \: \: \: \: \: \: \: \: \: \: \: \: \sf{0}$

Now, at equilibrium 40% of water reacts with CO in order to form H₂ and CO₂

$\sf{H_2O + CO_2 \leftrightharpoons H_2 + CO_2 }\\ \\ \\ \sf{1 - 0.4} \: \: \: \sf{1-0.4} \: \: \: \: \sf{0} \: \: \: \: \: \sf{0}$

Moles of reactants and products at equilibrium:

$\sf{H_2O + CO_2 \leftrightharpoons H_2 + CO_2 }\\ \\ \\ \sf{0.6} \: \: \: \: \: \: \: \: \: \: \: \: \sf{0.6} \: \: \: \: \: \: \: \: \: \: \sf{0.4} \: \: \: \: \: \: \: \: \: \: \sf{0.4}$

Now that we have the moles of reactants and products let us find out the equilibrium constant.

$\implies\sf{K_c = \dfrac{[H_2] . [CO_2]}{[H_2O].[CO]}}$

Substitute the obtained values in the above formula,

$\implies\sf{K_c = \dfrac{0.4 \times 0.4 }{0.6 \times 0.6}}$

$\implies\sf{K_c = \dfrac{0.16 }{0.36}}$

$\implies\boxed{\sf{\bf{K_c = 0.4}}}$

The equilibrium constant for the reaction is 0.4