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

For the reaction : \(Fe _{2} N ( s )+\frac{3}{2} H _{2}( g ) \rightleftharpoons 2 Fe ( s )+ NH _{3}( g )\)

  1. A \(K _{ C }= K _{ P }( RT )\)
  2. B \(K _{ C }= K _{ P }( RT )^{-1 / 2}\)
  3. C \(K _{ C }= K _{ P }( RT )^{-3 / 2}\)
  4. D \(K _{ C }= K _{ P }( RT )^{1 / 2}\)
Verified Solution

Answer & Solution

Correct Answer

(D) \(K _{ C }= K _{ P }( RT )^{1 / 2}\)

Step-by-step Solution

Detailed explanation

\(Fe _{2} N ( s )+\frac{3}{2} H _{2}( g ) \rightleftharpoons 2 Fe ( s )+ NH _{3}( g )\) \(\Delta n _{9}=1-\frac{3}{2}=\frac{-1}{2}\) \(\frac{ K _{ p }}{ K _{ c }}=( RT )^{\Delta n_g}=( RT )^{-1 / 2}\) \(K _{ c }=\frac{ K _{ p }}{( RT )^{-1 / 2}}= K _{ p ^{\prime}}( RT )^{1 / 2}\)
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