JEE Mains · Chemistry · STD 11 - 6.1. Equilibrium - 1 (chemical Equilibrium)
\(4.0\,moles\) of argon and \(5.0\, moles\) of \(PCI _{5}\) are introduced into an evacuated flask of \(100\, litre\) capacity at \(610\, K\). The system is allowed to equilibrate. At equilibrium, the total pressure of mixture was found to be \(6.0 \,atm\). The \(K _{ p }\) for the reaction is ...... [Given : \(R=0.082 \,L \,atm\, K ^{-1}\, mol ^{-1}\) ]
- A \(2.25\)
- B \(6.24\)
- C \(12.13\)
- D \(15.24\)
Answer & Solution
Correct Answer
(A) \(2.25\)
Step-by-step Solution
Detailed explanation
\(PCl _{5}=5\) \(mole\) \(Ar =4\) \(mole\) \(P _{\text {Total }}=\frac{9 \times 0.82 \times 610}{100}=4.5 \,atm\) \(P _{ PCI _{5}}=\frac{5 \times 4.5}{9}=2.5 ; P _{ Ar }=\frac{4 \times 4.5}{9}=2\) \(PCl _{5} \rightleftharpoons PCl _{3}+ Cl _{2}\) \(2.5- P \quad P \quad P\)…
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