Exactly of an ideal gas is taken in a vessel of volume . It dissociates according to the reaction at — Chemical Equilibrium Chemistry Question
Question
Exactly $1\text{ mole}$ of an ideal gas $A$ is taken in a vessel of volume $1\text{ L}$. It dissociates according to the reaction $A(g) \rightleftharpoons B(g) + C(g)$ at $27^\circ\text{C}$. The elementary forward and backward reaction rate constants are $1.5 \times 10^{-2}\text{ s}^{-1}$ and $3 \times 10^{-2}\text{ M}^{-1}\text{s}^{-1}$ respectively. Calculate the numerical value of $100 \times K_c$ for this reaction.
Answer: 50
💡 Solution & Explanation
At equilibrium, the equilibrium constant $K_c$ is the ratio of the forward rate constant to the backward rate constant. $K_c = \frac{k_f}{k_b} = \frac{1.5 \times 10^{-2}}{3 \times 10^{-2}} = 0.5$. The question asks for $100 \times K_c$, which is $100 \times 0.5 = 50$.
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