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JEE Mains · Chemistry · STD 11 - 6.1. Equilibrium - 1 (chemical Equilibrium)
\(\left( 1 \right)\,{N_2}\left( g \right) + 3{H_2}\left( g \right) \rightleftharpoons 2N{H_3}\left( g \right)\,,\,{K_1}\) \(\left( 2 \right)\,{N_2}\left( g \right) + {O_2}\left( g \right) \rightleftharpoons 2NO\left( g \right)\,,\,{K_2}\) \(\left( 3 \right)\,{H_2}\left( g \right) + \frac{1}{2}{O_2}\left( g \right) \rightleftharpoons {H_2}O\left( g \right)\,,\,{K_3}\) The equation for the equilibrium constart of the reaction \(2N{H_3}\left( g \right) + \frac{5}{2}{O_2}\left( g \right) \rightleftharpoons 2NO\left( g \right) + 3{H_2}O\left( g \right)\) \((K_4)\) in terms of \(K_1 , K_2\) , and \(K_3\) is
- A \(\frac{{{K_1}{K_2}}}{{{K_3}}}\)
- B \(\frac{{{K_1}{K_3^2}}}{{{K_2}}}\)
- C \({K_1}{K_2}{K_3}\)
- D \(\frac{{{K_2}{K_3^3}}}{{{K_1}}}\)
Answer & Solution
Correct Answer
(D) \(\frac{{{K_2}{K_3^3}}}{{{K_1}}}\)
Step-by-step Solution
Detailed explanation
To calculate the value of \(K_4\) in the given equation we should apply eqn. \((2)\) + eqn. \((3)\times3 \) - eqn. \((1)\) hence \({K_4} = \frac{{{K_2}{K_3^3}}}{{{K_1}}}\)
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