Question

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?

Answer 1

Step-by-step 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 * 0.062)]` atm

= 0.233 atm

Now, calculate the value of
`K_(p)` as follows.

`K_(p) = (P_(C))/((P_(A))(P_(B))^(2))`

=
`(0.047)/((0.171) * (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 * 298 K * ln 5.06`

=
`-8.314 J/mol K * 298 K * 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.