For the dissociation reaction , let be the degree of dissociation and be the total pressure. If , th — Chemical Equilibrium Chemistry Question
Question
For the dissociation reaction $PCl_5(g) \rightleftharpoons PCl_3(g) + Cl_2(g)$, let $\alpha$ be the degree of dissociation and $P$ be the total pressure. If $\alpha \ll 1$, the expression relating $\alpha$, $K_p$, and $P$ is:
Answer: A
💡 Solution & Explanation
Total moles $= 1+\alpha$. $P_{PCl_3} = P_{Cl_2} = (\alpha/(1+\alpha))P$, $P_{PCl_5} = ((1-\alpha)/(1+\alpha))P$. $K_p = (\alpha^2 / (1-\alpha^2)) P$. If $\alpha \ll 1$, $1-\alpha^2 \approx 1$, so $K_p \approx \alpha^2 P$, yielding $\alpha = (K_p/P)^{1/2}$.
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