Question

Ammonia gas at 15 atm is introduced in a rigid vessel at 300 K. At equilibrium the total pressure of the vessel is found to be 40.11 atm at 300⁰C. The degree of dissociation of NH3 will be:

• A. 0.6
• B. 0.4
• C. Unpredictable
• D. None

Answer 1

Answer: (B)

0.4

Answer 2

P1 = 15 atm ; T1 = 300 K.

Equilibrium temperature is 300⁰C that is 573 K.

So first of all we have to calculate pressure of NH3 at 573 K.

$\frac { \mathrm { P } _ { 1 } } { \mathrm {~T} _ { 1 } } = \frac { \mathrm { P } _ { 2 } } { \mathrm {~T} _ { 2 } } = \frac { 15 } { 300 } = \frac { \mathrm { P } _ { 2 } } { 573 }$

P2 = 28.65 atm at 300ºC.

NH3 (g) $\rightleftharpoons$ $\frac { 1 } { 2 }$N2(g) + $\frac { 3 } { 2 }$ H2(g).

t = 0 28.65 atm 0 0

t = teq. [28.65–x] atm $\frac { 3 } { 2 }$ x

But according to question.

Ptotal = 28.65 – x + + $\frac { 3 } { 2 }$ x

or 28.65 + x = 40.11.

x = 11.46.

Degree of dissociation of NH3 $= \frac { 11.46 } { 28.65 } = 0.4$.