
Given,\[F{e_2}{O_3}\left( s \right)\] may be converted to \[Fe\] by the reaction, \[F{e_2}{O_3} + 3{H_2}\left( g \right) \rightleftharpoons 2Fe\left( s \right) + 3{H_2}O\left( g \right)\] ; for which \[{K_p} = 8\] at temperature\[720C\] . What \[\% \] of \[{H_2}\] remains unreacted after the reaction has come to equilibrium?
A. $ \approx 22\% $
B. $ \approx 34\% $
C. $ \approx 66\% $
D. $ \approx 78\% $
Answer
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Hint:\[Kp\] is referred to as the equilibrium constant of a reaction which is equal to the ratio of the pressure of the products and reactants. For any reaction the percentage of the reactants decreases as the reaction progresses towards equilibrium.
Complete step by step answer:The reaction given is the reduction of iron oxide to metallic iron. The reaction is written as:
\[F{e_2}{O_3} + 3{H_2}\left( g \right) \rightleftharpoons 2Fe\left( s \right) + 3{H_2}O\left( g \right)\]
The equilibrium constant Kp of the above reaction is given as \[8\] . Let the initial pressure of hydrogen at time \[t = 0\] be \[1atm\] . This change of the pressure of hydrogen and water is shown below:
The pressure of the solids in the given reaction i.e. \[F{e_2}{O_3}\] and \[Fe\] is constant and does not change at equilibrium so they are neglected. The equilibrium constant is expressed as \[\dfrac{{P_{{H_2}O}^3}}{{P_{{H_2}}^3}} = {K_p}\]
$\dfrac{{{{(3 P)}^3}}}{{{{(1 - 3P)}^3}}} = 8$
$\dfrac{{3 P}}{{1 - 3P}} = 2$
=>\[9P = 2\]
$P = 0.22$
The percentage of hydrogen that has unreacted after the hydrogen comes to equilibrium is
$ = \dfrac{{1 - 3P}}{1} \times 100 = \dfrac{{1 - 3 \times 0.22}}{1} \times 100 = 34\% $ .
Hence option B is the correct answer.
Note:Like \[{K_p}\] , \[{K_c}\] is the equilibrium constant of the reaction expressed as a ratio of molar concentration of products and reactants. \[{K_p}\] and \[{K_c}\] are related to each other by a mathematical relationship.\[{K_p} = {K_c}{\left( {RT} \right)^{\Delta n}}\] , where \[R\] is gas constant, \[T\] is temperature, and \[\Delta n\] is the change in number of gas molecules. When the change in number of gas molecules is \[0\], then the value of \[{K_p}\] is equal to \[{K_c}\]. If the value of \[K\]is large then the equilibrium concentration of products is large. If the value of \[K\] is small then the equilibrium concentration of reactants is large.
Complete step by step answer:The reaction given is the reduction of iron oxide to metallic iron. The reaction is written as:
\[F{e_2}{O_3} + 3{H_2}\left( g \right) \rightleftharpoons 2Fe\left( s \right) + 3{H_2}O\left( g \right)\]
The equilibrium constant Kp of the above reaction is given as \[8\] . Let the initial pressure of hydrogen at time \[t = 0\] be \[1atm\] . This change of the pressure of hydrogen and water is shown below:
| Time | Pressure of \[{H_2}\left( g \right)\] | Pressure of \[{H_2}O\left( g \right)\] |
| At \[t = 0\] | \[1\] | \[0\] |
| At \[t = {t_{equilibrium}}\] | \[1 - 3P\] | \[3P\] |
The pressure of the solids in the given reaction i.e. \[F{e_2}{O_3}\] and \[Fe\] is constant and does not change at equilibrium so they are neglected. The equilibrium constant is expressed as \[\dfrac{{P_{{H_2}O}^3}}{{P_{{H_2}}^3}} = {K_p}\]
$\dfrac{{{{(3 P)}^3}}}{{{{(1 - 3P)}^3}}} = 8$
$\dfrac{{3 P}}{{1 - 3P}} = 2$
=>\[9P = 2\]
$P = 0.22$
The percentage of hydrogen that has unreacted after the hydrogen comes to equilibrium is
$ = \dfrac{{1 - 3P}}{1} \times 100 = \dfrac{{1 - 3 \times 0.22}}{1} \times 100 = 34\% $ .
Hence option B is the correct answer.
Note:Like \[{K_p}\] , \[{K_c}\] is the equilibrium constant of the reaction expressed as a ratio of molar concentration of products and reactants. \[{K_p}\] and \[{K_c}\] are related to each other by a mathematical relationship.\[{K_p} = {K_c}{\left( {RT} \right)^{\Delta n}}\] , where \[R\] is gas constant, \[T\] is temperature, and \[\Delta n\] is the change in number of gas molecules. When the change in number of gas molecules is \[0\], then the value of \[{K_p}\] is equal to \[{K_c}\]. If the value of \[K\]is large then the equilibrium concentration of products is large. If the value of \[K\] is small then the equilibrium concentration of reactants is large.
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