For a reaction taking place in a container in equilibrium with its surroundings, the effect of temperature on its equilibrium constant K in terms of change in entropy is described by

Engineering

Chemistry

Equilibrium Constant and Mass Action Law

Calculation of Various Parameters at Equilibrium

Le Chateliers Principle

Gibbs Free Energy Calculations and Third Law

Question

For a reaction taking place in a container in equilibrium with its surroundings, the effect of temperature on its equilibrium constant K in terms of change in entropy is described by

With increase in temperature, the value of K for exothermic reaction decreases because favourable change in entropy of the surroundings decreases.

With increase in temperature, the value of K for endothermic reaction increases because the entropy change of the system is negative.

With increase in temperature, the value of K for exothermic reaction decreases because the entropy change of the system is positive.

With increase in temperature, the value of K for endothermic reaction increases because unfavourable change in entropy of the surroundings decreases

With increase in temperature, the value of K for exothermic reaction decreases because favourable change in entropy of the surrounding decreases

To understand how temperature affects the equilibrium constant K in terms of entropy change, we need to recall the relationship between Gibbs free energy (ΔG), enthalpy (ΔH), entropy (ΔS), and the equilibrium constant. The key equation is:

$Δ G = Δ H - T Δ S$

At equilibrium, ΔG = 0, so:

$0 = Δ H - T Δ S$

Rearranging gives:

$Δ S = \frac{Δ}{H}$

Also, the relationship between ΔG and K is:

$Δ G = - R T \ln K$

Combining these, we get:

$\ln K = - \frac{Δ}{H} + \frac{Δ}{S}$

This shows that lnK (and thus K) depends on temperature. The sign of ΔH (enthalpy change) determines how K changes with T:

For an exothermic reaction (ΔH < 0), increasing T makes the term -ΔH/(RT) less negative, so K decreases.
For an endothermic reaction (ΔH > 0), increasing T makes -ΔH/(RT) less negative (more positive), so K increases.

Now, considering entropy: the total entropy change (ΔS_total) includes both the system (ΔS_sys) and surroundings (ΔS_surr). At equilibrium, ΔS_total = 0. The entropy change of the surroundings is related to the heat transferred: ΔS_surr = -ΔH/T.

For an exothermic reaction (ΔH < 0), ΔS_surr is positive (favourable). When T increases, the magnitude of ΔS_surr decreases (because |ΔH/T| is smaller), making the favourable change in entropy of the surroundings less favourable. This contributes to the decrease in K.

For an endothermic reaction (ΔH > 0), ΔS_surr is negative (unfavourable). When T increases, the magnitude of ΔS_surr decreases (|ΔH/T| is smaller), making the unfavourable change less unfavourable. This contributes to the increase in K.

Therefore, the correct statement is: "With increase in temperature, the value of K for exothermic reaction decreases because favourable change in entropy of the surroundings decreases." This matches the first option.

Detailed Solution

Related Topics and Formulae

Detailed Solution

Related Topics and Formulae

Student who searched for similar questions

Questions 1

Questions 2

Questions 3

Questions 4

Questions 5