Answer : The value of
is, 0.5 and
= positive.
Explanation :
The unimolecular reaction is:

In unimolecular reaction, the starting material is 2 times to the product.
.........(1)
As we know that:
...........(2)
Now substitute equation 1 in 2, we get:


Now we have to calculate the value of
at 298 K.

where,
= standard Gibbs free energy = ?
R = gas constant = 8.314 J/mol.K
T = temperature = 298 K
= equilibrium constant = 0.5
Now put all the given values on the above formula, we get:


Thus, the value of
at 298 K is, 1717.32 J/mol
As we know that:
= +ve, reaction is non spontaneous
= -ve, reaction is spontaneous
= 0, reaction is in equilibrium
Thus, the
= +ve. So, the reaction is non spontaneous.