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, NH3 is partially decomposed according to the equation, 2NH3⇋N2+3H2 .The vessel is such that the volume remains effectively constant whereas pressure increases to 50atm. Calculate the percentage of NH3 actually decomposed.
Options:
A .  65%
B .  61.3%
C .  62.5%
D .  64%
Answer: Option B : B 2NH3⇋N2+3H2 Initialmolea00 Mole at equilibrium (a−2x)x3x Initial pressure of NH3 of a mole = 15 atm 27∘C The pressure of 'a' mole of NH3=patmat347∘C ∴15300=p620 ∴ p=31 atm At constant volume and at 347∘C,mole α pressure ∴a+2xa=5031 ∴x=1962 ∴%ofNH3 decomposed = 2xa×100 =2×19a62×a×100=61.33%
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