JEE Main & Advanced Chemistry Equilibrium / साम्यावस्था Question Bank Topic Test - Chemical Equilibrium (6-5-21)

The reaction between \[{{N}_{2}}\] and \[{{H}_{2}}\] to form ammonia has \[{{K}_{c}}=6\times {{10}^{-2}}\] at the temperature 500°C. The numerical value of \[{{K}_{p}}\] for this reaction is

A) \[1.5\times {{10}^{-5}}\]

B) \[1.5\times {{10}^{5}}\]\[\]

C) \[1.5\times {{10}^{-6}}\]

D) \[1.5\times {{10}^{6}}\]

Correct Answer: A

Solution :

[a] \[{{K}_{p}}={{K}_{c}}{{(RT)}^{\Delta n}}\]; \[\Delta n=2-4=-2\]

\[{{K}_{p}}=6\times {{10}^{-2}}\times {{(0.0812\times 773)}^{-2}}\]

\[{{K}_{p}}=\frac{6\times {{10}^{-2}}}{{{(0.0812\times 773)}^{2}}}=1.5\times {{10}^{-5}}\].