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JEE Mains · Chemistry · STD 11 - 6.1. Equilibrium - 1 (chemical Equilibrium)

अभिक्रिया : \(\mathrm{SO}_2(\mathrm{~g})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g}) \rightleftharpoons \mathrm{SO}_3(\mathrm{~g})\) के लिए \(27^{\circ} \mathrm{C}\) तथा \(1 \mathrm{~atm}\) दाब पर \(\mathrm{K}_{\mathrm{P}}=2 \times 10^{12}\) है। इसी अभिक्रिया के लिए \(\mathrm{K}_{\mathrm{c}}\) है \(\quad \times 10^{13}\). (निकटतम पूर्णांक में) (दिया है \(\mathrm{R}=0.082 \mathrm{~L} \mathrm{~atm} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\) )

  1. A \(2\)
  2. B \(3\)
  3. C \(4\)
  4. D \(1\)
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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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