Answer : The value of
is, 
Explanation :
The formula used for
is:
............(1)
where,
= Gibbs free energy for the reaction
= standard Gibbs free energy
R = gas constant = 8.314 J/mole.K
T = temperature = 
Q = reaction quotient
First we have to calculate the
.
Formula used :

Now put all the given values in this formula, we get:


Now we have to calculate the value of 'Q'.
The given balanced chemical reaction is,

The expression for reaction quotient will be :

In this expression, only gaseous or aqueous states are includes and pure liquid or solid states are omitted.
Now put all the given values in this expression, we get


Now we have to calculate the value of
by using relation (1).

Now put all the given values in this formula, we get:


Therefore, the value of
is, 