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  • One mole of H_2O 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, H_2O (g) + CO (g) H_2 (g) + CO_2 (g)

7.14     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,

                H_{2}O_{(g)}+CO_{(g)}\rightleftharpoons H_{2}_{(g)}+CO_{2}_{(g)}

               Calculate the equilibrium constant for the reaction.  

Answers (1)

best_answer

The given reaction is-

                               H_{2}O_{(g)}+CO_{(g)}\rightleftharpoons H_{2}_{(g)}+CO_{2}_{(g)}
initial conc                   1/10            1/10              0               0
At equilibrium             0.6/10           0.6/10          0.04          0.04

Now, the equilibrium constant for the reaction can be calculated as;

K_c = \frac{[H_2][CO_2]}{[H_{2}O][CO]}
         = \frac{.04 \times.04}{(.06)^2}

           = 0.44 (approx)

Posted by

manish

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