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 : \(NH_{3}(g) \rightleftharpoons \frac{1}{2} N _{2}(g)+\frac{3}{2} H_{2}(g)\)
- A \(\frac{1}{4}\)
- B \(\frac{1}{8}\)
- C \(8\)
- D \(\frac{1}{64}\)
Answer & Solution
Correct Answer
(B) \(\frac{1}{8}\)
Step-by-step Solution
Detailed explanation
\(N _{2}+3 H _{2} \rightleftharpoons 2 NH _{3} \rightarrow K _{ C }=64\) \(2 NH _{3} \rightleftharpoons N _{2}+3 H _{2} \rightarrow K _{ C }=\frac{1}{64}\)…
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