JEE Mains · Chemistry · STD 11 - 6.1. Equilibrium - 1 (chemical Equilibrium)
The values of \(K_p/K_c\) for the following reactions at \(300\, K\) are respectively (At \(300\, K, RT = 24.62\, dm^3\, atm \,mol^{-1}\)) \({N_2}(g) + {O_2}(g) \rightleftharpoons 2NO(g)\) \({N_2}{O_4}(g) \rightleftharpoons 2N{O_2}(g)\) \({N_2}(g) + 3{H_2}(g) \rightleftharpoons 2N{H_3}(g)\)
- A \(1, 24.62\, dm^3\, atm\, mol^{-1}\), \(606.0\, dm^6\, atm^2 \,mol\,^{-2}\)
- B \(1, 24.62\, dm^3\, atm\, mol^{-1}\), \(1.65 \times 10^{-3}\, dm^{-6}\, atm^{-2}\, mol\,^{2}\)
- C \(1,4.1 \times 10^{-2}\, dm^{-3}\, atm^{-1}\, mol\), \(606.0\, dm^6\, atm^2 \,mol\,^{-2}\)
- D \(24.62\, dm^3\, atm\, mol^{-1}\), \(606.0\, dm^6\, atm^2\,atm^2\, mol\,^{-2}\) , \(1.65 \times 10^{-3}\,dm^{-6}\,atm^{-2}\,mol^2\)
Answer & Solution
Correct Answer
(B) \(1, 24.62\, dm^3\, atm\, mol^{-1}\), \(1.65 \times 10^{-3}\, dm^{-6}\, atm^{-2}\, mol\,^{2}\)
Step-by-step Solution
Detailed explanation
\({K_p} = {K_{c\,}}{\left( {RT} \right)^{\Delta ng}}\)
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