JEE Mains · Chemistry · STD 11 - 6.1. Equilibrium - 1 (chemical Equilibrium)
At temperature T, compound \(\mathrm{AB}_{2(\mathrm{~g})}\) dissociates as \(\mathrm{AB}_{2(\mathrm{~g})} \rightleftharpoons \mathrm{AB}_{(\mathrm{g})}+\frac{1}{2} \mathrm{~B}_{2(\mathrm{~g})}\) having degree of dissociation \(x\) (small compared to unity). The correct expression for \(x\) in terms of \(\mathrm{K}_{\mathrm{p}}\) and p is _______.
- A \(\sqrt[4]{\frac{2 K_p}{p}}\)
- B \(\sqrt[3]{\frac{2 K_{\mathrm{p}}}{\mathrm{p}}}\)
- C \(\sqrt[3]{\frac{2 \mathrm{~K}_{\mathrm{p}}^2}{\mathrm{p}}}\)
- D \(\sqrt{K_p}\)
Answer & Solution
Correct Answer
(C) \(\sqrt[3]{\frac{2 \mathrm{~K}_{\mathrm{p}}^2}{\mathrm{p}}}\)
Step-by-step Solution
Detailed explanation
\(\begin{array}{lcc} & \mathrm{AB}_2(\mathrm{g}) & \rightleftharpoons & \mathrm{AB}(\mathrm{g}) & + & \frac{1}{2} \mathrm{~B}_2(\mathrm{g}) \\ t=0 & p_0 & \\ t=t_{\text {eq }} & p_0(1-x) & & p_0 x & & \frac{p_0 x}{2} \end{array}\)…
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