For the reaction a(g) + 2 b(g) ↔ c(g) the initial partial pressures of gases a, b, and c are all 0.109 atm. once equilibrium has been established, it is found that pc = 0.047 atm. what is δg° for this reaction (in kj/mol) at 25°c?

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.

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