NO and `Br_(2)` at initial pressures of `98.4` and `41.3` torr respectively were allowed react at `300 K`. At equilibrium the total pressure was `110.5` torr. Calculate the value of equilibrium constant, `K_(p)` and the standard free energy change at `300 K` for the reaction:
`2NO(g)+Br_(2)(g) hArr 2NOBr(g)`

Answer

A, B, C, D

Explanation

`{:(NO,rarr,Br_(2),hArr,2NOBr),(98.4,,41.3,,),((98.4-2x),,(41.3-x),,2x):}`
`P_(t)` (at equilibrium)`=(98.4-2x)+(41.3-x)+2x`
`=110.5` (given)
`rArr x=29.2`
`{:(P_(NO),=98.4-2x=40,),(P_(Br_(2)),=41.3-x=12.1,),(,P_(NOBr)=2x=58.4,):}] rArr K_(P)=(P_(NOBr)^(2))/(P_(NO)^(2)P_(Br))`
`=0.1762 "torr"^(-1)`
`K_(p)`= should be in atm units for calculation of `DeltaG^(ɵ)`.
`K_(p)=0.1762xx760=133.9 "atm"^(-1)`
`DeltaG^(ɵ)=-2.303 RT log K_(p)=-12216.26 J "mol"^(-1)`
`=-12.22 kJ "mol"^(-1)`