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For the reaction CO(g)+H2O(g)CO2( g)+H2( g)\mathrm{CO}(\mathrm{g})+\mathrm{H}_2 \mathrm{O}(\mathrm{g}) \rightleftharpoons \mathrm{CO}_2(\mathrm{~g})+\mathrm{H}_2(\mathrm{~g}) at a certain temperature, Keq K_{\text {eq }} is 1 . At cquilibrium would you cxpect to find

(a) only CO\mathrm{CO} and H2\mathrm{H}_2
(b) mostly CO2\mathrm{CO}_2 and H2\mathrm{H}_2
(c) about equal concentrations of CO\mathrm{CO} and H2O\mathrm{H}_2 \mathrm{O}, compared to CO2\mathrm{CO}_2 and H2\mathrm{H}_2
(d) mostly CO\mathrm{CO} and H2O\mathrm{H}_2 \mathrm{O} (e) only CO\mathrm{CO} and H2O\mathrm{H}_2 \mathrm{O} Explain your answer briefly.

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In this task, we need to explain what we expect to find at equilibrium when the KeqK_{eq}=1.

For the reaction:

aA+bBcC+dD\text{aA}+\text{bB} \rightleftharpoons \text{cC} +\text{dD}

the Keq_{eq} is:

Keq=[C]c[D]d[Bb][A]aK_{eq}=\dfrac{[\text{C}]^c[\text{D}]^d}{[\text{B}^b][\text{A}]^a}

The equilibrium constant is equal to the product of the concentration of the products raised to the potency of its stoichiometric factor divided by the product of the concentration of the reactants raised to the potency of its stoichiometric factor.

If a KeqK_{eq} is greater than 1, the amount of products at equilibrium is greater than the amount of reactants, so the forward reaction is favored. If a KeqK_{eq} is less than 1, the amount of reactants at equilibrium is greater than the amount of products, so the reverse reaction is favored.

If a KeqK_{eq} is equal to 1, it means that the concentrations of the reactants and products are about the same.

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