Explanation:
As the given reaction is as follows.

The given data is as follows.
Initially,
= 0.109 atm,
= 0.109 atm,
= 0.109 atm
And, at equilibrium
= 0.047 atm
Therefore, change in
will be calculated as follows.
0.109 - 0.047
= 0.062 atm
Hence,
= (0.109 + 0.062) atm = 0.171 atm
=
atm
= 0.233 atm
Now, calculate the value of
as follows.

= 
= 5.06
Also, we known that 
Hence, calculate the value of
as follows.

= 
= 
= -4017.05 J/mol
or, = -4.017 kJ/mol (as 1 kJ = 1000 J)
Thus, we can conclude that the value of
for the given reaction is -4.017 kJ/mol.