JEE Mains · Chemistry · STD 11 - 6.1. Equilibrium - 1 (chemical Equilibrium)
\(\mathrm{A}(\mathrm{s}) \rightleftharpoons \mathrm{M}(\mathrm{s})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{~g})\) The equilibrium constant for the reaction is \(K_{p}=4\), At equilibrium, the partial pressure of \(\mathrm{O}_{2}\) is \(....\,atm.\) (Round off to the nearest integer).
- A \(16\)
- B \(20\)
- C \(25\)
- D \(31\)
Answer & Solution
Correct Answer
(A) \(16\)
Step-by-step Solution
Detailed explanation
\(\mathrm{k}_{\mathrm{p}}=\mathrm{Po}_{2}^{\frac{1}{2}}=4\) \(\therefore \mathrm{Po}_{2}=16\, \text { bar }=16\, \text { atm }\)
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