Equilibrium Question 624
Question: 28 g of $ N_2 $ and 6 g of $ H_2 $ were kept at $ 400^{o}C $ in 1 litre vessel, the equilibrium mixture contained $ 27.54g $ of $ NH_3 $ . The approximate value of $ K_{c} $ for the above reaction can be (in $ mol{e^{-2}}litre^{2} $ ) [CBSE PMT 1990]
Options:
A) 75
B) 50
C) 25
D) 100
Show Answer
Answer:
Correct Answer: A
Solution:
$ N_2+3H_2 $ ⇌ $ 2NH_3 $ Initial conc. 1 3 0 at equilibrium 1-0.81 3-2.43 1.62 0.19 0.57 No. of moles of $ N_2=\frac{28}{28}=1 $ mole No. of moles of $ H_2=\frac{6}{2}=3 $ mole No. of moles of $ NH_3=\frac{27.54}{17}=1.62 $ mole $ K_{c}=\frac{{{[NH_3]}^{2}}}{[N_2]{{[H_2]}^{3}}}=\frac{{{[1.62]}^{2}}}{[0.19]{{[0.57]}^{3}}} $ =75
