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EastWind [94]
3 years ago
5

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?
Chemistry
1 answer:
strojnjashka [21]3 years ago
3 0

Explanation:

As the given reaction is as follows.

             A(g) + 2B(g) \leftrightarrow C(g)

The given data is as follows.

Initially,    P_{A} = 0.109 atm,     P_{B} = 0.109 atm,

                P_{C} = 0.109 atm

And, at equilibrium

        P_{C} = 0.047 atm

Therefore, change in P_{C} will be calculated as follows.

                    0.109 - 0.047

                 = 0.062 atm

Hence, P_{A} = (0.109 + 0.062) atm = 0.171 atm

P_{B} = [0.109 + (2 \times 0.062)] atm

                        = 0.233 atm

Now, calculate the value of K_{p} as follows.

         K_{p} = \frac{P_{C}}{(P_{A})(P_{B})^{2}}

                    = \frac{0.047}{(0.171) \times (0.233)^{2}}

                    = 5.06

Also, we known that \Delta G^{o} = -RT lnK_{p}

Hence, calculate the value of \Delta G^{o} as follows.

        \Delta G^{o} = -RT lnK_{p}

                      = -8.314 J/mol K \times 298 K \times ln 5.06

                       = -8.314 J/mol K \times 298 K \times 1.62

                      = -4017.05 J/mol

or,                   = -4.017 kJ/mol              (as 1 kJ = 1000 J)

Thus, we can conclude that the value of \Delta G^{o} for the given reaction is -4.017 kJ/mol.

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