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 `300^@C`.The degree of dissociation of `NH_3` will be :

Options

A. `0.6`
B. `0.4`
C. Unpredictable
D. None of these

Answer

B

Explanation

`P_(1)=15 "atm" , T_(1)=300 K`.
Equilibrium temperature is `300^(@)C` that is `573 K`.
`(P_(1))/(T_(1))=(P_(2))/(T_(2))=(15)/(300)=(P_(2))/(573)`
`P_(2)=28.65 "atm at" 300^(@)`.
`NH_(3)(g)hArr(1)/(2)N_(2)(g)+(3)/(2)H_(2)(g)`.
`{:(t=0,28.65"atm",0,0),(t=t_(eq.),[28.65-x],(x)/(2)"atm,(3)/(2)x):}`
But according to question.
`P_(total)=28.65-x+(x)/(2)+(3)/(2)x`
or `28.65+x=40.11`
`x=11.46`.
Degree of dissociation of `NH_(3)=(11.46)/(28.65)=0.4`.