If 2 mol of H_2 and I_2 are taken initially | vedclass

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(B) The reaction is $H_{2(g)} + I_{2(g)} \rightleftharpoons 2HI_{(g)}$.Initial moles: $H_2 = 2, I_2 = 2, HI = 0$.At equilibrium,$[HI] = 2 \ mol/L$. Si

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$4$

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(B) The reaction is $H_{2(g)} + I_{2(g)} ightleftharpoons 2HI_{(g)}$.Initial moles: $H_2 = 2, I_2 = 2, HI = 0$.At equilibrium,$[HI] = 2 \ mol/L$. Since the volume is $1 \ L$,the moles of $HI$ at equilibrium is $2$.Let $x$ be the degree of dissociation. The equilibrium moles are: $H_2 = (2-x), I_2 = (2-x), HI = 2x$.Given $2x = 2$,so $x = 1$.Equilibrium concentrations are: $[H_2] = 2-1 = 1 \ M$,$[I_2] = 2-1 = 1 \ M$,$[HI] = 2 \ M$.The equilibrium constant $K_c = \frac{[HI]^2}{[H_2][I_2]} = \frac{2^2}{1 	imes 1} = 4$.For the reaction,$\Delta n_g = (2) - (1+1) = 0$.Since $\Delta n_g = 0$,$K_p = K_c(RT)^{\Delta n_g} = K_c = 4$.

If $2 \ mol$ of $H_2$ and $I_2$ are taken initially in a $1 \ L$ vessel,and the equilibrium concentration of $HI$ is $2 \ mol/L$,find the $K_p$ for the reaction $H_{2(g)} + I_{2(g)} \rightleftharpoons 2HI_{(g)}$.

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