Question

Ammonia is heated at 15 atm for ${27}^{o}C$ to ${347}^{o}C$ assuming volume constant. The new pressure becomes 50 atm at equilibrium. The % of ammonia actually decompose is :

• A. 17.3
• B. 61.3
• C. 15.3
• D. 14.3

Answer 1

The correct option is B 61.3

pressure of ammonia at ${27}^{o}C$ = 15 atm
pressure of ammonia at ${347}^{o}C$ = p atm
p/620 = 15/300 p= 31 atm
Let a moles of ammonia be present. Total pressure at equilibrium = 50 atm
$N{H}_3\to N_2+3H_2$
at equilibrium (a - 2x) + x + 3x = a + 2x
Total moles: initial no. of moles/moles at equilibrium = initial pressure/ equilibrium pressure
a/(a+2x) = 31/50
x = 19/62 a
ammount of ammonia dicomposed= 2x = 2X19/62a = 19/31a
% of ammonia decomposed = $19\times a/31\times a$ $\times 100$
= 61.3