Respuesta :
Explanation:
As the given reaction is as follows.
[tex]A(g) + 2B(g) \leftrightarrow C(g)[/tex]
The given data is as follows.
Initially, [tex]P_{A}[/tex] = 0.109 atm, [tex]P_{B}[/tex] = 0.109 atm,
[tex]P_{C}[/tex] = 0.109 atm
And, at equilibrium
[tex]P_{C}[/tex] = 0.047 atm
Therefore, change in [tex]P_{C}[/tex] will be calculated as follows.
0.109 - 0.047
= 0.062 atm
Hence, [tex]P_{A}[/tex] = (0.109 + 0.062) atm = 0.171 atm
[tex]P_{B}[/tex] = [tex][0.109 + (2 \times 0.062)][/tex] atm
= 0.233 atm
Now, calculate the value of [tex]K_{p}[/tex] as follows.
[tex]K_{p} = \frac{P_{C}}{(P_{A})(P_{B})^{2}}[/tex]
= [tex]\frac{0.047}{(0.171) \times (0.233)^{2}}[/tex]
= 5.06
Also, we known that [tex]\Delta G^{o} = -RT lnK_{p}[/tex]
Hence, calculate the value of [tex]\Delta G^{o}[/tex] as follows.
[tex]\Delta G^{o} = -RT lnK_{p}[/tex]
= [tex]-8.314 J/mol K \times 298 K \times ln 5.06[/tex]
= [tex]-8.314 J/mol K \times 298 K \times 1.62[/tex]
= -4017.05 J/mol
or, = -4.017 kJ/mol (as 1 kJ = 1000 J)
Thus, we can conclude that the value of [tex]\Delta G^{o}[/tex] for the given reaction is -4.017 kJ/mol.
