Respuesta :
Explanation:
The given data is as follows.
[tex]\Delta_{r} G[/tex] = -16.7 kJ/mol = [tex]-16.7 \times 10^{3}[/tex], T = 298 K
R = 8.314 J/mol K, [tex]K_{eq}[/tex] = ?
Relation between [tex]\Delta_{r} G[/tex] and [tex]K_{eq}[/tex] is as follows.
[tex]\Delta_{r} G[/tex] = [tex]-RT ln K_{eq}[/tex]
Hence, putting the values into the above equation as follows.
[tex]\Delta_{r} G[/tex] = [tex]-RT ln K_{eq}[/tex]
[tex]-16.7 \times 10^{3} J/mol[/tex] = [tex]-8.314 J/mol K \times 298 K ln K_{eq}[/tex]
[tex]ln K_{eq}[/tex] = [tex]\frac{-16.7 \times 10^{3} J/mol}{-8.314 J/mol K \times 298 K}[/tex]
= 6.740
[tex]K_{eq}[/tex] = antilog (6.740)
= 846
Thus, we can conclude that [tex]K_{eq}[/tex] for given values is 846.
