Equilibrium Question 252

Question: Ammonia under a pressure of 15 atm at 27°C is heated to 347°C in a closed vessel in the presence of a catalyst. Under the conditions, $ NH_3 $ is partially decomposed according to the equation, $ 2NH_3 $ ⇌ $ N_2+3H_2 $ .The vessel is such that the volume remains effectively constant whereas pressure increases to 50 atm. Calculate the percentage of $ NH_3 $ actually decomposed. [IIT 1981; MNR 1991; UPSEAT 2001]

Options:

A) 65%

B) 61.3%

C) 62.5%

D) 64%

Show Answer

Answer:

Correct Answer: B

Solution:

$ 2NH_3 $ ⇌ $ N_2+ $ $ 3H_2 $

Initial mole $ a $ 0 0
Mole at equilibrium $ (a-2x) $ $ x $ $ 3x $

Initial pressure of $ NH_3 $ of a mole = 15 atm at $ 27^{o}C $

The pressure of ‘a’ mole of $ NH_3=p $ atm at $ 347^{o}C $

$ \therefore $ $\frac{15}{300}=\frac{p}{620} $

$ \therefore $ $p=31 $ atm

At constant volume and at $ 347^{o}C $ , mole $ \propto $ pressure $ a\propto 31 $ (before equilibrium)

$ \therefore $ $a+2x,\propto 50 $ (after equilibrium)
$ \therefore $ $\frac{a+2x}{a}=\frac{50}{31} $

$ \therefore $ $x=\frac{19}{62}a $

$ \therefore $ % of $ NH_3 $ decomposed $ =\frac{2x}{a}\times 100 $

$ =\frac{2\times 19a}{62\times a}\times 100=61.33% $



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