JEE Mains · Chemistry · STD 11 - 6.1. Equilibrium - 1 (chemical Equilibrium)
For the reaction \(SO_2(g) + \frac{1}{2} O_2(g) \rightleftharpoons SO_3(g),\) if \(K_P = K_C ( RT)^x\) where the symbols have usual meaning then the value of \(x\) is (assuming ideality) :
- A \(-1\)
- B \(-0.5\)
- C \(0.5\)
- D \(1\)
Answer & Solution
Correct Answer
(B) \(-0.5\)
Step-by-step Solution
Detailed explanation
\(S O_{2}(g)+\frac{1}{2} O_{2}(g) \rightleftharpoons S O_{3}(g)\) \(K_{p}=K_{C}(R T)^{x}\) where \(x=\Delta n_{g}=\) number of gaseous moles in product - number of gaseous moles in reactant \(=1-\left(1+\frac{1}{2}\right)\) \(=1-\frac{3}{2}=-\frac{1}{2}\)
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