Question

a. For a reaction with ∆H° = -12 kcal/mol and ∆S° = 0.01kcal mol^-1 K^-1, calculate the ∆G° and the equilibrium constant at: 1. 30 °C and 2. 150 °C.

b. How does ∆G° change as T increases?

c. How does Keq change as T increases?

Short answer

a)K_eq=1.8x 10⁸

b) increasing the temperature leads ∆G° to become more negative in this reaction.

c) the temperature rises, K_eq decreases.

Step-by-step solution

Definition of ∆G°-

G denotes the change in free energy of a system as it transitions from one state to another, such as all reactants to all products.

The maximum useful energy released (or absorbed) in transitioning from the initial to the end state is represented by this value.

∆G° becomes negative as a result of the production of products with stronger bonds and greater freedom of motion, resulting in a favourable reaction.

Calculation of Keq at 30˚C-

Given-

∆H° = -12 kcal/mol

∆S° = 0.01kcal mol^-1 K^-1

T=30 °C= (30+273) K

∆ G^°= ∆H° -T∆S°

Recall that (T = ˚C+273)

∆ G^°=-12-(273+300) (0.01)

∆ G^°=-12- 3=-15Kcal/mol

∆ G^°=RT lnK_eq

∆ G^°= -(1.986x 10^-3) (303)lnK_eq

-15=-0.60 lnK_eq

lnK_eq=25

K_eq =7.2 x 10¹⁰

Calculation of Keq at 150˚C-

∆H° = -12 kcal/mol

∆S° = 0.01kcal mol^-1 K^-1

T=150 °C= (150+273) K

∆ G^°= ∆H° -T∆S°

∆ G^°=-12-(273+150) (0.001)

∆ G^°=-12-4=-16kcal/mol

∆ G^°=-(1.986x 10^-3) (423) ln K_eq

-16=-0.84ln K_eq

ln K_eq=19

K_eq=1.8x 10⁸

b) Reason why ∆G° change as T increases-

Calculations demonstrate that increasing the temperature leads∆G° to become more negative in this reaction.

c) Reason why Keq change as T increases –

The calculation for this reaction reveals that as the temperature rises, K_eq decreases. As a result of k_eq = -∆G°/RT