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The free energy of formation of NO is 78...

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)`

Text Solution

Verified by Experts

The correct Answer is:
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)`
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