Question

The free energy of formation of NO is `78 kJ mol^(-1)` at the temperature of an authomobile engine `(1000 K)`. What is the equilibrium constant for this reaction at `1000 K`?
`1/2 N_(2)(g)+1/2 O_(2)(g) hArr NO(g)`
A. `8.4xx10^(-5)`
B. `7.1xx10^(-9)`
C. `4.2xx10^(-10)`
D. `1.7xx10^(-19)`

Answer 1

Correct Answer - A
`DeltaG=78 kJ "mol"^(-1)`
`T=1000 K`
`DeltaG^(ɵ)=-nRT "In" K_(p)`
`:. In K_(p)=(-DeltaG^(ɵ))/(nRT)=(-78)/(8.314xx1000)`
or `K_(p)= "antilog" ((-78)/(8.314xx1000))=8.4xx10^(-5)`