For the reaction H_{2(g)} + I_{2(g)} rightlef | vedclass

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(A) The reaction is $H_{2(g)} + I_{2(g)} \rightleftharpoons 2HI_{(g)}$.Initial moles: $H_2 = 0.4$,$I_2 = 0.4$,$HI = 0$.At equilibrium,$n(HI) = 0.5 \ m

Vedclass · Accepted Answer

$11.11$

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(A) The reaction is $H_{2(g)} + I_{2(g)} \rightleftharpoons 2HI_{(g)}$.Initial moles: $H_2 = 0.4$,$I_2 = 0.4$,$HI = 0$.At equilibrium,$n(HI) = 0.5 \ mol$.From stoichiometry,$2 \ mol$ of $HI$ are formed from $1 \ mol$ of $H_2$ and $1 \ mol$ of $I_2$.So,$0.5 \ mol$ of $HI$ is formed from $0.25 \ mol$ of $H_2$ and $0.25 \ mol$ of $I_2$.Equilibrium moles: $n(H_2) = 0.4 - 0.25 = 0.15 \ mol$,$n(I_2) = 0.4 - 0.25 = 0.15 \ mol$,$n(HI) = 0.5 \ mol$.Concentrations in $2 \ L$ vessel: $[H_2] = 0.15/2 = 0.075 \ M$,$[I_2] = 0.075 \ M$,$[HI] = 0.5/2 = 0.25 \ M$.$K_c = \frac{[HI]^2}{[H_2][I_2]} = \frac{(0.25)^2}{(0.075)(0.075)} = \frac{0.0625}{0.005625} = 11.11$.Since $\Delta n_g = 2 - (1+1) = 0$,$K_p = K_c(RT)^{\Delta n_g} = K_c = 11.11$.

For the reaction $H_{2(g)} + I_{2(g)} \rightleftharpoons 2HI_{(g)}$,$0.4 \ mol$ of $H_2$ and $I_2$ each are taken in a $2 \ L$ vessel. If $0.5 \ mol$ of $HI$ is formed at equilibrium,calculate the equilibrium constant $K_c$ and $K_p$.

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