
What is the equilibrium expression for the following reaction?
$FeO(s) + {H_2}(g) \rightleftharpoons Fe(s) + {H_2}O(g)?$
Answer
509.1k+ views
Hint: We have to know that the forward and reverse reactions have the same rate when a reaction is in equilibrium. And as the forward and backward reactions continue, the concentrations of the reaction components remain stable at equilibrium. A chemical reaction's equilibrium constant is the value of its reaction quotient at chemical equilibrium, a condition reached by a dynamic chemical mechanism after a period of time has passed in which its structure shows no discernible propensity to alter.
Complete answer:
We have to remember that the ratio of concentrations at equilibrium for a reaction at a given temperature is described by equilibrium constants. In general, the letters $K$ or ${K_c}$ are used. Both concentrations are expressed in terms of molar concentration, as shown by the subscript $c$. For a generic balanced equation as given below,
\[aA{\text{ }} + bB{\text{ }} \rightleftharpoons {\text{ }}cC{\text{ }} + {\text{ }}dD\]
In this equation, a moles of reactant A reacts with b moles of reactant B to give c moles of product C and d moles of product D. The equilibrium constant is
${K_c} = \dfrac{{{{[C]}^c}{{[D]}^d}}}{{{{[A]}^a}{{[B]}^b}}}$
Hence, for the given reaction, $FeO(s) + {H_2}(g) \rightleftharpoons Fe(s) + {H_2}O(g)$, The equilibrium constant is likely to be ${K_c} = \dfrac{{[Fe][{H_2}O]}}{{[FeO][{H_2}]}}$. Since ferrous oxide and iron are solids, they are unable to convey concentration and hence are absent from the equilibrium expression. The equilibrium is solely determined by the quantities of water and dihydrogen, or partial pressures.
Therefore, the equilibrium expression becomes ${K_c} = \dfrac{{[{H_2}O]}}{{[{H_2}]}}$.
Note:
We must be notes that an equilibrium constant, also known as ${K_{eq}}$, ${K_c}$, or ${K_p}$, can be used to describe any chemical equilibrium. The ${K_{eq}}$, ${K_c}$, or ${K_p}$ expressions are formed by dividing component amounts by reactant numbers, with each quantity (concentration or pressure) elevated to the power of its coefficient in the balanced chemical equation. Equilibrium constant is often written without the use of units.
Complete answer:
We have to remember that the ratio of concentrations at equilibrium for a reaction at a given temperature is described by equilibrium constants. In general, the letters $K$ or ${K_c}$ are used. Both concentrations are expressed in terms of molar concentration, as shown by the subscript $c$. For a generic balanced equation as given below,
\[aA{\text{ }} + bB{\text{ }} \rightleftharpoons {\text{ }}cC{\text{ }} + {\text{ }}dD\]
In this equation, a moles of reactant A reacts with b moles of reactant B to give c moles of product C and d moles of product D. The equilibrium constant is
${K_c} = \dfrac{{{{[C]}^c}{{[D]}^d}}}{{{{[A]}^a}{{[B]}^b}}}$
Hence, for the given reaction, $FeO(s) + {H_2}(g) \rightleftharpoons Fe(s) + {H_2}O(g)$, The equilibrium constant is likely to be ${K_c} = \dfrac{{[Fe][{H_2}O]}}{{[FeO][{H_2}]}}$. Since ferrous oxide and iron are solids, they are unable to convey concentration and hence are absent from the equilibrium expression. The equilibrium is solely determined by the quantities of water and dihydrogen, or partial pressures.
Therefore, the equilibrium expression becomes ${K_c} = \dfrac{{[{H_2}O]}}{{[{H_2}]}}$.
Note:
We must be notes that an equilibrium constant, also known as ${K_{eq}}$, ${K_c}$, or ${K_p}$, can be used to describe any chemical equilibrium. The ${K_{eq}}$, ${K_c}$, or ${K_p}$ expressions are formed by dividing component amounts by reactant numbers, with each quantity (concentration or pressure) elevated to the power of its coefficient in the balanced chemical equation. Equilibrium constant is often written without the use of units.
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