Equilibrium Question 722
Question: At $ 527{}^\circ C $ , the reaction given below has $ K_{c}=4 $
$ NH_3(g)\rightarrow \frac{1}{2}N_2(g)+\frac{3}{2}H_2(g) $ What is the $ K_{p} $ for the reaction’ $ N_2(g)+3H_2(g)\rightarrow 2NH_3(g) $
Options:
A) $ 16\times {{(800R)}^{2}} $
B) $ {{( \frac{800R}{4} )}^{-2}} $
C) $ {{( \frac{1}{4\times 800R} )}^{2}} $
D) None of these
Show Answer
Answer:
Correct Answer: C
Solution:
If, $ K_{c}=\frac{{{[N_2]}^{1/2}}{{[H_2]}^{3/2}}}{{{[NH_3]}^{1}}}=4 $ Then, $ \frac{{{[NH_3]}^{2}}}{{{[ [N_2][H_2] ]}^{3}}}={{(4)}^{-2}}=\frac{1}{16} $ also $ K_{p}=K_{c}\cdot {{(RT)}^{\Delta n_{g}}} $
$ =\frac{1}{16}\times {{(800R)}^{-2}}={{( \frac{1}{4\times 800R} )}^{2}} $
BETA
