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% $
