Class 11Chemistry6-1.Equilibrium (Chemical Equilibrium)Mix Examples- 6-1.Equilibrium (Chemical Equilibrium)
MediumIn a $10 \ L$ vessel,$SO_3, SO_2$,and $O_2$ gases are present at a definite temperature with $K_c = 100$. For the reaction $2SO_{2(g)} + O_{2(g)} \rightleftharpoons 2SO_{3(g)}$,if $SO_3$ and $SO_2$ are present in the same number of moles at equilibrium,find the moles of $O_2$. If $SO_3$ is double the moles of $SO_2$,what is the number of moles of $O_2$?
(N/A) The equilibrium constant expression is $K_c = \frac{[SO_3]^2}{[SO_2]^2 [O_2]} = 100$.
Given volume $V = 10 \ L$.
Case $1$: If moles of $SO_3 = \text{moles of } SO_2$,then $[SO_3] = [SO_2]$.
Substituting into the expression: $100 = \frac{1}{[O_2]} \implies [O_2] = 0.01 \ M$.
Moles of $O_2 = [O_2] \times V = 0.01 \times 10 = 0.1 \ mol$.
Case $2$: If moles of $SO_3 = 2 \times \text{moles of } SO_2$,then $[SO_3] = 2[SO_2]$.
Substituting into the expression: $100 = \frac{(2[SO_2])^2}{[SO_2]^2 [O_2]} = \frac{4}{[O_2]} \implies [O_2] = 0.04 \ M$.
Moles of $O_2 = [O_2] \times V = 0.04 \times 10 = 0.4 \ mol$.
Given volume $V = 10 \ L$.
Case $1$: If moles of $SO_3 = \text{moles of } SO_2$,then $[SO_3] = [SO_2]$.
Substituting into the expression: $100 = \frac{1}{[O_2]} \implies [O_2] = 0.01 \ M$.
Moles of $O_2 = [O_2] \times V = 0.01 \times 10 = 0.1 \ mol$.
Case $2$: If moles of $SO_3 = 2 \times \text{moles of } SO_2$,then $[SO_3] = 2[SO_2]$.
Substituting into the expression: $100 = \frac{(2[SO_2])^2}{[SO_2]^2 [O_2]} = \frac{4}{[O_2]} \implies [O_2] = 0.04 \ M$.
Moles of $O_2 = [O_2] \times V = 0.04 \times 10 = 0.4 \ mol$.
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