Class 11Chemistry6-1.Equilibrium (Chemical Equilibrium)Mix Examples- 6-1.Equilibrium (Chemical Equilibrium)
DifficultIn a closed vessel at $448^{\circ} C$,$0.5 \ mol$ of $H_2$ and $0.5 \ mol$ of $I_2$ react to form hydrogen iodide.
Reaction: $H_{2(g)} + I_{2(g)} \rightleftharpoons 2HI_{(g)}$,$K_c = 50$.
$(i)$ Calculate the moles of $I_2$ that remain unreacted at equilibrium.
$(ii)$ Calculate $K_p$.
Reaction: $H_{2(g)} + I_{2(g)} \rightleftharpoons 2HI_{(g)}$,$K_c = 50$.
$(i)$ Calculate the moles of $I_2$ that remain unreacted at equilibrium.
$(ii)$ Calculate $K_p$.
(N/A) The reaction is $H_{2(g)} + I_{2(g)} \rightleftharpoons 2HI_{(g)}$.
Let $x$ be the moles of $H_2$ and $I_2$ reacted at equilibrium.
Initial moles: $[H_2] = 0.5$,$[I_2] = 0.5$,$[HI] = 0$.
Equilibrium moles: $[H_2] = 0.5 - x$,$[I_2] = 0.5 - x$,$[HI] = 2x$.
$K_c = \frac{[HI]^2}{[H_2][I_2]} = \frac{(2x)^2}{(0.5-x)(0.5-x)} = 50$.
Taking the square root: $\frac{2x}{0.5-x} = \sqrt{50} = 7.071$.
$2x = 3.5355 - 7.071x \implies 9.071x = 3.5355 \implies x = 0.39$.
Moles of $I_2$ remaining = $0.5 - 0.39 = 0.11 \ mol$.
Since $\Delta n_g = 2 - (1+1) = 0$,$K_p = K_c(RT)^{\Delta n_g} = K_c(RT)^0 = K_c$.
Therefore,$K_p = 50$.
Let $x$ be the moles of $H_2$ and $I_2$ reacted at equilibrium.
Initial moles: $[H_2] = 0.5$,$[I_2] = 0.5$,$[HI] = 0$.
Equilibrium moles: $[H_2] = 0.5 - x$,$[I_2] = 0.5 - x$,$[HI] = 2x$.
$K_c = \frac{[HI]^2}{[H_2][I_2]} = \frac{(2x)^2}{(0.5-x)(0.5-x)} = 50$.
Taking the square root: $\frac{2x}{0.5-x} = \sqrt{50} = 7.071$.
$2x = 3.5355 - 7.071x \implies 9.071x = 3.5355 \implies x = 0.39$.
Moles of $I_2$ remaining = $0.5 - 0.39 = 0.11 \ mol$.
Since $\Delta n_g = 2 - (1+1) = 0$,$K_p = K_c(RT)^{\Delta n_g} = K_c(RT)^0 = K_c$.
Therefore,$K_p = 50$.
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