Question

For a reaction carried out at 25°C with an equilibrium constant of 1 × 10⁻³, in order to increase the equilibrium constant by a factor of 10:

a. How much must ΔG° change?

b. How much must ΔH° change if ΔS° = 0 kcal mol^-1 K^-1?

c. How much must ΔS° change if ΔH° = 0 kcal mol^-1?

Short answer

a) ΔG^o = ^_RT ln K_eq

= - 1.36 kcal/mol

b)ΔG^o = ΔH^o - T ΔS^o

ΔH^o= -1.36 kcal/mol

c)ΔG^o = ΔH^o - T ΔS^o

ΔS^o= 4.56 × 10^-3 kcal/(mol deg)

Step-by-step solution

Some definition

What is ΔS° value =

S value is always positive. But ΔS° denotes change in entropy. Entropy means the degree of randomness and disorder of the molecules. It can be negative or positive. If the products are less than the reactants then disorder decreases so ΔS° value becomes negative. If the reactant and products are in same number then the ΔS° value becomes zero. If the products are more in number than the reactant then ΔS° value will be positive.

What is ΔH° value =

A positive ΔH°value value represents addition of energy from the reaction then the reaction is endothermic reaction. A negative value represents removal of energy from the reaction then the reaction is exothermic reaction.

What is equilibrium constant=

It provides relationship between the products and reactants when a chemical reaction reaches equilibrium.

Final answer

Given:

T = 25°C = 298 K

K_eq = 1 × 10⁻³

a)We know,

ΔG^o = ^_RT ln K_eq

= - 1.986 × 10^-3 × 298 K × ln (10)

= - 1.986 × 10^-3 × 298 K × 2.3

= - 1.36 kcal/mol

b) Now, we will have to calculate Δ H^o, from the given data by using the formula

ΔG^o = ΔH^o - T ΔS^o

-1.36 = ΔH^o – 0

ΔH^o= -1.36 kcal/mol

c) Now, we will have to calculate Δ S^o, from the given data by using the formula

ΔG^o = ΔH^o - T ΔS^o

-1.36 = 0 – 298 ΔS^o

ΔS^o= 4.56 × 10^-3 kcal/(mol deg)