WBJEE · Chemistry · Chemical Equilibrium
For the reaction \(S O_2+\frac{1}{2} O_2=S O_3\), if we write \(K_p=K_c(R T)^x\), then \(\mathrm{x}\) becomes
- A \(-1\)
- B \(-\frac{1}{2}\)
- C \(\frac{1}{2}\)
- D \(1\)
Answer & Solution
Correct Answer
(B) \(-\frac{1}{2}\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & \text { } \mathrm{K}_{\mathrm{p}}=\mathrm{K}_{\mathrm{C}}(\mathrm{RT})^x \\ & x=\left(\sum \mathrm{n}_{(\mathrm{g})}\right)_{\mathrm{P}}-\left(\sum \mathrm{n}_{(\mathrm{g})}\right)_{\mathrm{R}} \\ & =1-\frac{3}{2}=-\frac{1}{2}\end{aligned}\)
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