Question

a. What is the equilibrium constant for a reaction that is carried out at 25°C (298 K) with ΔH° = 20 kcal/mol and ΔS° = 5.0 x 10^-2 kcal mol^-1 K^-1?

b. What is the equilibrium constant for the same reaction carried out at 125°C?

Short answer

Here we have to found out the equilibrium constant:

a)K_eq=1.8×10^-4

b)K_eq= 1.0

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

a)Now, we will have to calculateΔ G^o, from the given data by using the formula

Δ G^o = ΔH^o - T ΔS^o

= 20Kcal/mol - (298 K0.05 Kcal mol^-1K^-1)

= (20-14.9) = 5.1 kcal/mol

The equilibrium constant K_eq can be calculated as follows,

Δ G^o = - RT lnK_eq

5.1= -1.986×10³×298×lnK_eq

-8.6=lnK_eq

K_eq=1.8×10^-4

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

Δ G^o = ΔH^o - T ΔS^o

= 20Kcal/mol - (398 K0.05 Kcal mol^-1K^-1)

= (20-20) = 0 kcal/mol

The equilibrium constant K_eq can be calculated as follows,

Δ G^o = - RT lnK_eq

0= -1.986×10³×398×lnK_eq

0=lnK_eq

K_eq=1