Chemical equilibrium |

H₂ (g) + ½ S₂ (g) = H₂S (g), KP₁

H₂ (g) + S₂ (g) = 2H₂S (g), KP₂

Show that, KP₂ = (KP₁)²

SolutionH₂ (g) + S₂ (g) = 2H₂S (g), KP₂

Show that, KP₂ = (KP₁)²

For the first reaction,

ΔG⁰₁ = - RT ln KP₁

For the second reaction,

ΔG⁰₂ = - RT ln KP₂

Since, ΔG⁰₂ = 2ΔG⁰₁

Therefore, it follows that - RT ln KP₂ = - 2RT ln KP₁

∴ Kp₂ = (Kp₁)²

ProblemΔG⁰₁ = - RT ln KP₁

For the second reaction,

ΔG⁰₂ = - RT ln KP₂

Since, ΔG⁰₂ = 2ΔG⁰₁

Therefore, it follows that - RT ln KP₂ = - 2RT ln KP₁

∴ Kp₂ = (Kp₁)²

- Use Gibbs - Helmholtz equation to derive the Vant Hoff reaction isochore. What condition do you expect a linear relationship between log k and 1/T?

The Gibbs - Helmholtz equation is,

ΔG⁰ = ΔH⁰ + T[d(ΔG⁰)/dT]

Zero superscript is indicating the stranded values.

or, - (ΔH⁰/T² ) = -(ΔG⁰/T² ) + 1/T[d(ΔG⁰)/dT]p

or, - (ΔH⁰/T² ) = [d/dT(ΔG⁰/T)]p

Again Van't Hoff isotherm is,

- RT lnKp = ΔG⁰

or, - R lnKp = ΔG⁰/T

Differentiating with respect to T at constant P,

- R [dlnKP/dT]p = [d/dT(ΔG⁰/T)]p

Comparing the above two equation we have,

ΔG⁰ = ΔH⁰ + T[d(ΔG⁰)/dT]

_{P}Zero superscript is indicating the stranded values.

or, - (ΔH⁰/T² ) = -(ΔG⁰/T² ) + 1/T[d(ΔG⁰)/dT]p

or, - (ΔH⁰/T² ) = [d/dT(ΔG⁰/T)]p

Again Van't Hoff isotherm is,

- RT lnKp = ΔG⁰

or, - R lnKp = ΔG⁰/T

Differentiating with respect to T at constant P,

- R [dlnKP/dT]p = [d/dT(ΔG⁰/T)]p

Comparing the above two equation we have,

dlnKp/dT = (ΔH⁰/T² |

- This is the Van't Hoff equation isochore.

- Greater the value of ΔH⁰, the faster the equilibrium constant(Kp)changes with temperature(T), ΔH⁰ should remain constant for the linear plot of log Kp vs 1/T.

- How does the equilibrium constant for a reaction, 2A + 3B ⇆ 4C + Heat, change when (i) pressure is increasing (ii) temperature is decreasing (iii) a catalyst added?

- Equilibrium constant remains the same when P is increased. The reaction is exothermic, hence ΔH = (-) ve so the equilibrium constant is increased with decreases of temperature.
- Equilibrium constant remains the same though a catalyst is added.

ΔG⁰ = - RT lnKa

but ΔG⁰ is not changed due to the addition of catalyst as the latter does not participate in the reaction. - The initial state and the final state of a chemical reaction remains the same whether the catalyst is added or not. Hence the
(Ka) remains unchanged.*equilibrium constant*

Problem

- For a reaction,2A + B ⇆ 2C, ΔG⁰(500 K) = 2 KJ mol

^{-1}. Find the Kp at 500 K for the reaction A + ½B ⇆ C.

ΔG⁰ (500 K) for the reaction,

A + ½B ⇆ C

= (2 KJ mol

= 1 KJ mol

The relation is, ΔG⁰ = - RT lnKp

or, 1 = - (8.31 × 10-3) × (500) lnKp

or, lnKp = 1/(8.31 × 0.5)

= 0.2406

∴ Kp = 1.27

ProblemA + ½B ⇆ C

= (2 KJ mol

^{-1})/2= 1 KJ mol

^{-1}The relation is, ΔG⁰ = - RT lnKp

or, 1 = - (8.31 × 10-3) × (500) lnKp

or, lnKp = 1/(8.31 × 0.5)

= 0.2406

∴ Kp = 1.27

- Justify or criticize the following: "The equilibrium yield of products can not change if the equilibrium constant is kept fixed."

- This statement is not correct.

- Equilibrium yield of the product is changed if pressure is changed (Δγ ≠ 0), if an inert gas is added at constant P (Δγ ≠ 0), and any of the reacting components are added a depleted.

- For example, N₂ + 3 H₂ ⇆ 2NH₃

- The equilibrium yield of NH₃ is increased if P is increased though equilibrium constant kept fixed.

- Justify or criticize the following: heat of reaction is the same whether a catalyst is used or not.

- H is a state function hence ΔH (heat of a reaction) does not change if the initial state and final state of a process are the same. A catalyst cannot change the initial and final state of a chemical reaction, hence ΔH remains the same whether a catalyst is used or not.

- Therefore the statement is correct.