For the equilibrium , . At a given time, , , and . In which direction will the reaction take place? — Chemical Equilibrium Chemistry Question
Question
For the equilibrium $2CO_2(g) \rightleftharpoons 2CO(g) + O_2(g)$, $K_c = 6.4 \times 10^{-7}$. At a given time, $[CO_2] = 0.178\text{ M}$, $[CO] = 0.021\text{ M}$, and $[O_2] = 5.7 \times 10^{-5}\text{ M}$. In which direction will the reaction take place?
Answer: B
💡 Solution & Explanation
$Q_c = ([CO]^2[O_2])/[CO_2]^2 = (0.021)^2(5.7 \times 10^{-5}) / (0.178)^2 \approx 7.93 \times 10^{-7}$. Since $Q_c$ ($7.93 \times 10^{-7}$) is strictly greater than $K_c$ ($6.4 \times 10^{-7}$), the reaction moves in the backward direction to consume products.
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