Equilibrium Question 569
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 [UPSEAT 1999]
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} $
Show Answer
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}} $ .
