Question

Suppose Delta G^o is 25.0 KJ/mol for a hypothetical reaction in which A converts into B. Which of the following statements describes an equlibrium mixture of A and B?

a) B has a higher concentration than A.
b) Only A is present.
c) A and B have equal concentrations.
d) Only B is present.
e) A has a higher concentration than B.

Could you please explain your reasoning behind the answer as well?

Answer 1

Ans. Standard Gibb’s free energy, dG⁰ = -RT ln k          - equation 1

                        Where,

                        dG⁰ = Standard Gibb's free energy

                        R = Universal gas constant = 0.0083146 kJ mol^-1 K^-1

T = Temperature in kelvin

k = Equilibrium constant (Keq)

# Given: dG⁰ = + 25.0 kJ/mol                   ;                       A -------> B

# To explain what the (+) sign of dG⁰ value means- note the following point-

# Case 1: When Keq is less than 1.

For simplicity of clarification, we assign some values of Keq (= k) less than 1.0

So,       Keq = 0.5 (less than 1.0)

Putting the values in equation 1-

            dG⁰ = - RT ln k = - RT ln (0.5) = - RT (- 0.693) = + 0.693RT

ln of any value less than 1.0 is always a NEGATIVE value.

# Note that when Keq is less than 1.0, the resultant dG⁰ value is POSITIVE, i.e. the step of reaction is endothermic.

The term “k” in “ln k” is the equilibrium constant.

That is, k = [B] / [A]            - to get the value of k smaller than 1.0, and in turn to get a positive dG⁰ value, there must be [A] > [B]. Say, if [A] is 2 and [B] is 1 molar, then k =0.5.

# When [A] > [B], it means that the equilibrium concentration of reactants is more than that of product. It means two things – I. the reaction is NOT going in forward direction; II. Or, the reaction proceeds in backward direction.

# Case 2: When Keq is greater than 1.

For simplicity of clarification, we assign some values of Keq greater than 1.0

So,       Keq = 2.0 (greater than 1.0)

Putting the values in equation 1-

            dG⁰ = - RT ln k = - RT ln (2) = - RT (+ 0.693) = - 0.693RT

Note that when Keq is greater than 1.0, the resultant dG⁰ value is NEGATIVE, i.e. the step of reaction is exothermic.

# To make dG⁰ a negative value, there must be [B] > [A]

A [B] > [A] means that the equilibrium concertation of product is more than that of reactant. I happens when the reaction goes to forwards direction.   

## Also, a reaction with positive dG⁰ is endergonic and requires energy input for forward direction. It is non-spontaneous because of being endergonic.

A reaction with negative dG⁰ is exergonic and releases energy for forward direction. It is spontaneous because of being exergonic.

Case 3: When dG⁰ = 0, the [B] / [A] ratio is equal to 1.0 (note: ln 1 = 0). Also, k = 1. The system is said to be at equilibrium.

Conclusion: Correct option E. A has higher concentration than B, that is [A] > [B]

To get a positive dG⁰ value, there must be [A] > [B].