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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
`K="antilog"[(-DeltaG^(@))/(2.303RT)]`
`="antilog"[(-78xx1000)/(2.303xx8.314xx1000)]`
`=8.4xx10^(-5)`
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