For the reaction , the equilibrium constant at . In a flask, the instantaneous molar concentrations — Chemical Equilibrium Chemistry Question
Question
For the reaction $H_2(g) + I_2(g) \rightleftharpoons 2HI(g)$, the equilibrium constant $K_c = 57.0$ at $700\text{ K}$. In a flask, the instantaneous molar concentrations are $[H_2] = 0.10\text{ M}$, $[I_2] = 0.20\text{ M}$, and $[HI] = 0.40\text{ M}$. Calculate the exact numerical value of the reaction quotient $Q_c$.
Answer: 8
💡 Solution & Explanation
The reaction quotient $Q_c = \frac{[HI]^2}{[H_2][I_2]}$. Plugging in the instantaneous concentrations: $Q_c = \frac{(0.40)^2}{(0.10)(0.20)} = \frac{0.16}{0.02} = 8.0$. Since $Q_c (8.0) < K_c (57.0)$, the reaction will proceed in the forward direction.
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