One mole H_2O_{(g)} and one mole CO_{(g)} are | vedclass

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(A) The reaction is: $H_2O_{(g)} + CO_{(g)} \rightleftharpoons H_{2(g)} + CO_{2(g)}$Initial moles: $H_2O = 1.0 \ mol$,$CO = 1.0 \ mol$,$H_2 = 0 \ mol$

Vedclass · Accepted Answer

$0.444$

Vedclass · Answer

(A) The reaction is: $H_2O_{(g)} + CO_{(g)} ightleftharpoons H_{2(g)} + CO_{2(g)}$Initial moles: $H_2O = 1.0 \ mol$,$CO = 1.0 \ mol$,$H_2 = 0 \ mol$,$CO_2 = 0 \ mol$.Given that $40 \%$ of water reacts,the moles of $H_2O$ reacted $= 0.4 	imes 1.0 = 0.4 \ mol$.According to the stoichiometry of the reaction,$0.4 \ mol$ of $CO$ will also react,and $0.4 \ mol$ of $H_2$ and $0.4 \ mol$ of $CO_2$ will be formed.Moles at equilibrium:$n(H_2O) = 1.0 - 0.4 = 0.6 \ mol$$n(CO) = 1.0 - 0.4 = 0.6 \ mol$$n(H_2) = 0.4 \ mol$$n(CO_2) = 0.4 \ mol$Since the volume of the flask is $1 \ L$,the concentrations are equal to the number of moles.$[H_2O] = 0.6 \ M, [CO] = 0.6 \ M, [H_2] = 0.4 \ M, [CO_2] = 0.4 \ M$$K_c = \frac{[H_2][CO_2]}{[H_2O][CO]} = \frac{0.4 	imes 0.4}{0.6 	imes 0.6} = \frac{0.16}{0.36} = 0.444$

One mole $H_2O_{(g)}$ and one mole $CO_{(g)}$ are taken in a $1 \ L$ flask and heated to $725 \ K$. At equilibrium,$40 \%$ of water reacted with $CO_{(g)}$ as follows:
$H_2O_{(g)} + CO_{(g)} \rightleftharpoons H_{2(g)} + CO_{2(g)}$
Its $K_c$ value is:

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One mole $H_2O_{(g)}$ and one mole $CO_{(g)}$ are taken in a $1 \ L$ flask and heated to $725 \ K$. At equilibrium,$40 \%$ of water reacted with $CO_{(g)}$ as follows:$H_2O_{(g)} + CO_{(g)} \rightleftharpoons H_{2(g)} + CO_{2(g)}$Its $K_c$ value is: