of and of are mixed and allowed to attain equilibrium at . At equilibrium, the formed is . The equil — Chemical Equilibrium Chemistry Question
Question
$15\text{ moles}$ of $H_2$ and $5.2\text{ moles}$ of $I_2$ are mixed and allowed to attain equilibrium at $500^\circ\text{C}$. At equilibrium, the $HI$ formed is $10\text{ moles}$. The equilibrium constant $K_c$ for $H_2 + I_2 \rightleftharpoons 2HI$ is:
Answer: A
💡 Solution & Explanation
$HI = 2x = 10 \Rightarrow x = 5$. Eq moles: $H_2 = 15 - 5 = 10$, $I_2 = 5.2 - 5 = 0.2$. Volume cancels out since $\Delta n_g = 0$. $K_c = [HI]^2 / ([H_2][I_2]) = (10^2) / (10 \times 0.2) = 100 / 2 = 50$.
💬Ask on WhatsApp →
Still have doubts about this question?
Send it to our AI chemistry tutor on WhatsApp — gets answered in minutes