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

The value of \(K _{ C }\) is \(64\) at \(800\, K\) for the reaction \(N _{2}( g )+3 H _{2}( g ) \rightleftharpoons 2 NH _{3}( g )\) The value of \(K _{ C }\) for the following reaction is : \(NH_{3}(g) \rightleftharpoons \frac{1}{2} N _{2}(g)+\frac{3}{2} H_{2}(g)\)

  1. A \(\frac{1}{4}\)
  2. B \(\frac{1}{8}\)
  3. C \(8\)
  4. D \(\frac{1}{64}\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(\frac{1}{8}\)

Step-by-step Solution

Detailed explanation

\(N _{2}+3 H _{2} \rightleftharpoons 2 NH _{3} \rightarrow K _{ C }=64\) \(2 NH _{3} \rightleftharpoons N _{2}+3 H _{2} \rightarrow K _{ C }=\frac{1}{64}\)…