Class 11Chemistry1 Equilibrium: Chemical EquilibriumMix Examples- 6-1.Equilibrium (Chemical Equilibrium)
DifficultAt $717 \ K$,$3.2 \ mol$ of $HI$ is heated in a closed tube. $20\%$ of $HI$ decomposes at equilibrium according to the reaction $2HI_{(g)} \rightleftharpoons H_{2_{(g)}} + I_{2_{(g)}}$. Find $K_c$ and the moles of $HI$,$H_2$,and $I_2$ at equilibrium.
(N/A) The reaction is $2HI_{(g)} \rightleftharpoons H_{2_{(g)}} + I_{2_{(g)}}$.
Initial moles: $HI = 3.2 \ mol$,$H_2 = 0 \ mol$,$I_2 = 0 \ mol$.
At equilibrium,$20\%$ of $HI$ decomposes,so the amount reacted is $3.2 \times 0.20 = 0.64 \ mol$.
Remaining $HI = 3.2 - 0.64 = 2.56 \ mol$.
According to the stoichiometry,$2 \ mol$ of $HI$ produces $1 \ mol$ of $H_2$ and $1 \ mol$ of $I_2$.
Thus,$H_2 = 0.64 / 2 = 0.32 \ mol$ and $I_2 = 0.64 / 2 = 0.32 \ mol$.
$K_c = [H_2][I_2] / [HI]^2 = (0.32 / V) \times (0.32 / V) / (2.56 / V)^2 = (0.32 \times 0.32) / (2.56)^2 = 0.1024 / 6.5536 = 0.015625$.
Initial moles: $HI = 3.2 \ mol$,$H_2 = 0 \ mol$,$I_2 = 0 \ mol$.
At equilibrium,$20\%$ of $HI$ decomposes,so the amount reacted is $3.2 \times 0.20 = 0.64 \ mol$.
Remaining $HI = 3.2 - 0.64 = 2.56 \ mol$.
According to the stoichiometry,$2 \ mol$ of $HI$ produces $1 \ mol$ of $H_2$ and $1 \ mol$ of $I_2$.
Thus,$H_2 = 0.64 / 2 = 0.32 \ mol$ and $I_2 = 0.64 / 2 = 0.32 \ mol$.
$K_c = [H_2][I_2] / [HI]^2 = (0.32 / V) \times (0.32 / V) / (2.56 / V)^2 = (0.32 \times 0.32) / (2.56)^2 = 0.1024 / 6.5536 = 0.015625$.
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