Equilibrium Question 882

Question: 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

Options:

A) $ 1.5\times {10^{-5}} $

B) $ 1.5\times 10^{5} $

$ $

C) $ 1.5\times {10^{-6}} $

D) $ 1.5\times 10^{6} $

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Answer:

Correct Answer: A

Solution:

$ 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}} $ .



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