JEE Mains · Chemistry · STD 11 - 6.1. Equilibrium - 1 (chemical Equilibrium)
For reaction : \(SO _2( g )+\frac{1}{2} O _2( g ) \rightleftharpoons SO _3( g )\) \(K _{ P }=2 \times 10^{12}\) at \(27^{\circ}\,C\) and 1 atm pressure. The \(K _{ c }\) for the same reaction is \(.........\times 10^{13}\). (Nearest integer) (Given \(R =0.082\,L\,atm\,K ^{-1}\,mol ^{-1}\) )
- A \(2\)
- B \(3\)
- C \(4\)
- D \(1\)
Answer & Solution
Correct Answer
(D) \(1\)
Step-by-step Solution
Detailed explanation
\(SO _{2( g )}+\frac{1}{2} O _{2( g )} \rightleftharpoons SO _{3( g )}\) \(K _{ P }=2 \times 10^{12} \text { at } 300\,K\) \(K _{ P }= K _{ C } \times( RT )^{\Delta n _{ g }}\) \(2 \times 10^{12}= K _{ C } \times(0.082 \times 300)^{-1 / 2}\) \(K _{ C }=9.92 \times 10^{12}\)…
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