If under a given set of conditions the reaction A → B occurs with ΔG = -14 kJ/mol, and the reaction C→ B occurs with ΔG =- 16 kJ/mol, then: a. Conversion of A to C is exergonic. b. A and C can never be at equilibrium, even under different reaction conditions. c. Oconversion of A to C is entropically driven. d. Conversion of C to A is freely reversible.

a. Conversion of A to C is exergonic.

Explanation:

In the problem:  C→ B occurs with ΔG = + 16 kJ/mol

It is possible to sum ΔG of reactions to obtain ΔG of another related reaction:

A → B ΔG = -14kJ/mol

C → B ΔG = +16kJ/mol

A → C ΔG = -14kJ/mol - (+16kJ/mol) = -30kJ/mol

As ΔG < 0

The reaction is exergonic

Under these condition, the reaction occurs. But under another conditions, the reaction will be at equilibrium (ΔG = 0)

A reaction is entropically driven if ΔG < 0 and ΔS is high. But we don't have information of ΔS.

As ΔG <0, the reaction is not spontaneous in the reverse direction

Right option is:

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