JEE MainChemistryModeratePhysical Chemistry

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 300ºC. The degree of dissociation of NH3 will be

Show answer and solution

P_{1} = 15\ atm\;T_{1} = 300K

Equilibrium temperature is 300ºC that is 573 K.
So first of all we have to calculate pressure of NH3 at 573 K.

\frac{P_{1}}{T_{1}} = \frac{P_{2}}{T_{2}} = \frac{15}{300} = \frac{P_{2}}{573}

P_{2} = 28.65atm\ at\ 300^{\circ}C

{NH}_{3}(g) \rightleftharpoons \frac{1}{2}N_{2}(g) + \frac{3}{2}H_{2}(g)

t=0 28.65 atm 0 0

t = t_{eq}

\lbrack 28.65 - x\rbrack

\frac{x}{2}\ atm\

\frac{3}{2}x

But according to question.

P_{total\ } = 28.65 - x + \frac{x}{2} + \frac{3}{2}x

28.65 + x = 40.11

x = 11.46

Degree of dissociation of

{NH}_{3} = \frac{11.46}{28.65} = 0.4

Source: Grass-G / CH-22 Chemistry Paper 19 Dec. Eng.docx

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