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# (a)At $500 mathrm{~K}$, the equilibrium constant for the reaction $mathrm{H}_{2(mathrm{~g})}+mathrm{I}_{2(g)} ightleftharpoons 2 mathrm{HI}_{(g)}$ is 24.8. If $frac... ## Question ###### (a)At$500 mathrm{~K}$, the equilibrium constant for the reaction$mathrm{H}_{2(mathrm{~g})}+mathrm{I}_{2(g)} ightleftharpoons 2 mathrm{HI}_{(g)}$is 24.8. If$frac{1}{2} mathrm{~mol} / mathrm{L}$of$mathrm{HI}$ispresent at equilibrium, what are the concentrations of$mathrm{H}_{2}$and$mathrm{I}_{2}$, assuming that we started by taking HI and reached the equilibrium at$500 mathrm{~K} ?$(a)$0.068 mathrm{~mol} mathrm{~L}$(b)$1.020 mathrm{~mol} mathrm{~L}^{-1}$(c)$0.10 mathrm{~mol} mathrm{~

(a) At $500 mathrm{~K}$, the equilibrium constant for the reaction $mathrm{H}_{2(mathrm{~g})}+mathrm{I}_{2(g)} ightleftharpoons 2 mathrm{HI}_{(g)}$ is 24.8. If $frac{1}{2} mathrm{~mol} / mathrm{L}$ of $mathrm{HI}$ is present at equilibrium, what are the concentrations of $mathrm{H}_{2}$ and $mathrm{I}_{2}$, assuming that we started by taking HI and reached the equilibrium at $500 mathrm{~K} ?$ (a) $0.068 mathrm{~mol} mathrm{~L}$ (b) $1.020 mathrm{~mol} mathrm{~L}^{-1}$ (c) $0.10 mathrm{~mol} mathrm{~L}^{-1}$ (d) $1.20 mathrm{~mol} mathrm{~L}^{-1}$ #### Similar Solved Questions

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