Equilibrium Question 773

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}} $



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