For the equilibrium , . At a given time, the concentrations are , , and . In which direction will th — 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, the concentrations are $[CO_2] = 1.78 \times 10^{-1}\text{ M}$, $[CO] = 2.1 \times 10^{-2}\text{ M}$, and $[O_2] = 5.7 \times 10^{-5}\text{ M}$. In which direction will the reaction take place to reach equilibrium?
Answer: B
💡 Solution & Explanation
$Q_c = \frac{[CO]^2[O_2]}{[CO_2]^2} = \frac{(2.1 \times 10^{-2})^2(5.7 \times 10^{-5})}{(1.78 \times 10^{-1})^2} = \frac{(4.41 \times 10^{-4})(5.7 \times 10^{-5})}{3.168 \times 10^{-2}} \approx 7.93 \times 10^{-7}$. 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 and form reactants.
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