Class 11 Chemistry 6-1.Equilibrium (Chemical Equilibrium)Law of equilibrium and Equilibrium constant

Medium

For the reaction $CO_{(g)} + 2H_{2_{(g)}} \rightleftharpoons CH_3OH_{(g)}$,the equilibrium constant $K_c$ is $0.5$. If the concentrations of $CO$ and $H_2$ at equilibrium are $0.18 \ M$ and $0.22 \ M$ respectively,what is the concentration of $CH_3OH$?

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(D) The equilibrium constant expression for the reaction is $K_c = \frac{[CH_3OH]}{[CO][H_2]^2}$.
Given $K_c = 0.5$,$[CO] = 0.18 \ M$,and $[H_2] = 0.22 \ M$.
Substituting the values: $0.5 = \frac{[CH_3OH]}{(0.18)(0.22)^2}$.
$[CH_3OH] = 0.5 \times 0.18 \times 0.0484$.
$[CH_3OH] = 0.004356 \ M$ or $4.356 \times 10^{-3} \ M$.

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6-1.Equilibrium (Chemical Equilibrium)

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Law of equilibrium and Equilibrium constant

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For the reaction $CO_{(g)} + 2H_{2_{(g)}} \rightleftharpoons CH_3OH_{(g)}$,the equilibrium constant $K_c$ is $0.5$. If the concentrations of $CO$ and $H_2$ at equilibrium are $0.18 \ M$ and $0.22 \ M$ respectively,what is the concentration of $CH_3OH$?

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The equilibrium constant for the reaction $SO_{2(g)} + \frac{1}{2} O_{2(g)} \rightleftharpoons SO_{3(g)}$ is $5 \times 10^{-2} \ atm^{-1/2}$. The equilibrium constant of the reaction $2 SO_{3(g)} \rightleftharpoons 2 SO_{2(g)} + O_{2(g)}$ would be

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