Equilibrium Question 647
Question: Two moles of $ NH_3 $ when put into a previously evacuated vessel (one litre), partially dissociate into $ N_2 $ and $ H_2 $ . If at equilibrium one mole of NH3 is present, the equilibrium constant is [MP PET 2003]
Options:
A) 3/4 $ mol^{2},litr{e^{-2}} $
B) 27/64 $ mol^{2},litr{e^{-2}} $
C) 27/32 $ mol^{2},litr{e^{-2}} $
D) 27/16 $ mol^{2},litr{e^{-2}} $
Show Answer
Answer:
Correct Answer: D
Solution:
$\begin{aligned} 2NH_3(g) &\xrightarrow{2-2x} N_2(g)x + 3H_2(g)3x \\ \text{Conc. of } NH_3 \text{ at equilibrium} &= 2-2x \\ \text{Given that Conc. of } NH_3 \text{ at equilibrium} &= 1 \\ \therefore 2-2x &= 1 \Rightarrow x = 0.5 \\ \therefore \text{ Conc. of } H_2 \text{ at equilibrium} &= 3x = 3 \times 0.5 = 1.5 \\ \text{Conc. of } H_2 \text{ at equilibrium} &= x = 0.5 \\ \therefore \text{ Equilibrium constant } (K_c) &= \frac{[N_2][H_2]^3}{[NH_3]^2} \\ \Rightarrow K_c &= \frac{0.5 \times (1.5)^3}{(1)^2} = \frac{27}{16} \end{aligned}$
