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)`

NO and `Br_(2)` at initial pressures of `98.4` and `41.3` torr respect

1.	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» Correct Answer - A::B::C::D `{:(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)`

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