A mixture of 1 mol of H_2O and 1 mol of CO | vedclass

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(D) The balanced chemical equation is:$CO_{(g)} + H_2O_{(g)} \rightleftharpoons CO_{2(g)} + H_{2(g)}$Initial moles:$CO = 1 \ mol$,$H_2O = 1 \ mol$,$CO

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$44$

Vedclass · Answer

(D) The balanced chemical equation is:$CO_{(g)} + H_2O_{(g)} ightleftharpoons CO_{2(g)} + H_{2(g)}$Initial moles:$CO = 1 \ mol$,$H_2O = 1 \ mol$,$CO_2 = 0 \ mol$,$H_2 = 0 \ mol$At equilibrium:$CO = (1 - x) \ mol$,$H_2O = (1 - x) \ mol$,$CO_2 = x \ mol$,$H_2 = x \ mol$Given that $40\%$ of water reacts,$x = 0.4 \ mol$.Therefore,at equilibrium:$[CO] = \frac{1 - 0.4}{10} = 0.06 \ M$$[H_2O] = \frac{1 - 0.4}{10} = 0.06 \ M$$[CO_2] = \frac{0.4}{10} = 0.04 \ M$$[H_2] = \frac{0.4}{10} = 0.04 \ M$The equilibrium constant $K_C$ is given by:$K_C = \frac{[CO_2][H_2]}{[CO][H_2O]} = \frac{0.04 	imes 0.04}{0.06 	imes 0.06} = \frac{0.0016}{0.0036} = \frac{16}{36} = \frac{4}{9} \approx 0.444$Thus,$K_C 	imes 10^2 = 0.444 	imes 100 = 44.4 \approx 44$.

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