Ncert Exercises & Solutions

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,

H2O (g) + CO (g) H2 (g) + CO2 (g)

Calculate the equilibrium constant for the reaction.

The given reaction is:

H_{2} O_{(g)} + {CO}_{(g)} \leftrightarrow H_{2 (g)} + {CO}_{2 (g)}

Initial conc . \frac{1}{10} M \frac{1}{10} M 0 0

At equilibrium \frac{1 - 0.4}{10} M \frac{1 - 0.4}{10} M \frac{0.4}{10} M \frac{0.4}{10} M

= 0.06 M = 0.06 M = 0.04 M = 0.04 M

Therefore, the equilibrium constant for the reaction,

Therefore, the equilibrium constant for the reaction,

K_{c} = \frac{[H_{2}] [{CO}_{2}]}{[H_{2} O] [CO]}

= \frac{0.04 \times 0.04}{0.06 \times 0.06}

= 0.444 (approximately)

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