For a reaction taking place in a container in equilibrium with its surroundings, the effect of tempe

Q: Why is the heat of the surroundings equal to the negative enthalpy of the system?

By the first law of thermodynamics (conservation of energy), any heat lost by the system is exactly the heat gained by the surroundings. At constant pressure, the heat exchanged by the system is defined as its enthalpy change, $\Delta H_{\text{sys}}$. Therefore, $q_{\text{surr}} = -\Delta H_{\text{sys}}$.

Q: Does the entropy of the system change with temperature too?

Yes, it does. However, the options specifically ask about the effect of temperature on $K$ *in terms of the change in entropy of the surroundings*. The dominant factor explaining the shift in $K$ with $T$ is usually the $1/T$ dependence of $\Delta S_{\text{surr}}$.

Q: Why does a decrease in favourable entropy change lead to a decrease in K?

The equilibrium constant $K$ is directly related to the total entropy change of the universe ($\Delta S_{\text{total}} = R \ln K$). If the positive (favourable) contribution from the surroundings decreases, the total entropy change decreases, leading to a smaller $K$.

Q: What exactly does 'unfavourable change' mean in thermodynamics?

An unfavourable change refers to a decrease in entropy ($\Delta S < 0$). The second law of thermodynamics states that the total entropy of the universe must increase for a spontaneous process. A negative entropy change in any part of the universe opposes the process.

Q: Can we just solve this using Le Chatelier's principle instead of entropy?

Absolutely! Le Chatelier's principle gives the exact same result (e.g., increasing $T$ shifts an exothermic reaction backward, decreasing $K$). However, this specific question is testing your understanding of the *thermodynamic reasoning* behind Le Chatelier's principle using the concept of entropy.

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

Options:

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

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

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

(D) With increase in temperature, the value of K for endothermic reaction increases because the entropy change of the system negative

Sigma Percentile · Accepted Answer

### The Thermodynamic Dance of System and Surroundings Imagine a chemical reaction taking place inside a closed container. This container is our **system**, and everything outside it is the **surroundings**. When a reaction occurs, it doesn't happen in isolation; it constantly exchanges heat with its surroundings. This heat exchange is the key to understanding how temperature affects the equilibrium constant, $K$. To unlock this, we need to look at the entropy change of the surroundings, denoted as $\Delta S_{ ext{surr}}$. ### The Master Equation How do we calculate the entropy change of the surroundings? The fundamental definition of entropy change is the reversible heat exchanged divided by the temperature. $$\Delta S_{ ext{surr}} = \frac{q_{ ext{surr}}}{T}$$ By the first law of thermodynamics, any heat gained by the surroundings must have been lost by the system, and vice versa. Therefore, $q_{ ext{surr}} = -q_{ ext{sys}}$. At constant pressure, the heat exchanged by the system is simply its enthalpy change, $\Delta H_{ ext{sys}}$. This gives us our master equation: $$\Delta S_{ ext{surr}} = \frac{-\Delta H_{ ext{sys}}}{T}$$ This elegant equation tells us exactly how the surroundings react to the system's thermal behavior. ### Case 1: The Exothermic Reaction Let's first consider an exothermic reaction. In this case, the system releases heat, meaning $\Delta H_{ ext{sys}} < 0$. If we plug a negative value into our master equation, the two negative signs cancel out, resulting in a positive $\Delta S_{ ext{surr}}$. $$\Delta S_{ ext{surr}} = \frac{-(-ve)}{T} > 0$$ An increase in entropy is a **favourable change**. The surroundings are happy to receive this heat. But what happens if we increase the temperature, $T$? Looking at the equation, $T$ is in the denominator. As $T$ increases, the overall value of the fraction decreases. This means the positive, favourable entropy change of the surroundings becomes *less* positive. Because the process becomes less favourable overall, the forward reaction is hindered, and the equilibrium constant $K$ decreases. This perfectly aligns with **Option (C)**. ### Case 2: The Endothermic Reaction Now, let's flip the scenario and look at an endothermic reaction. Here, the system absorbs heat from the surroundings, so $\Delta H_{ ext{sys}} > 0$. Plugging this positive value into our equation yields a negative $\Delta S_{ ext{surr}}$. $$\Delta S_{ ext{surr}} = \frac{-(+ve)}{T} < 0$$ A decrease in entropy is an **unfavourable change**. The surroundings are resisting giving up their heat. What happens when we increase the temperature here? Again, $T$ is in the denominator. As $T$ increases, the magnitude of the fraction decreases. This means our negative $\Delta S_{ ext{surr}}$ becomes *less* negative. The unfavourable change in the surroundings is shrinking! Because the penalty is reduced, the reaction becomes more favourable, driving the equilibrium forward and increasing $K$. This confirms that **Option (B)** is also correct. ### The Grand Conclusion By simply analyzing the equation $\Delta S_{ ext{surr}} = -\Delta H_{ ext{sys}}/T$, we can logically deduce Le Chatelier's principle from a purely thermodynamic standpoint. Both options (B) and (C) beautifully describe this interplay between heat, temperature, and entropy.

Animated Solution for Chemistry - Chemical Thermodynamics: 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

Select Answer:

Visualized Solution

\text{System and Surroundings}

\Delta S_{\text{surr}} \text{ Formula}

\text{Exothermic Reaction}

\text{Effect of } T \uparrow \text{ on Exothermic}

\text{Endothermic Reaction}

\text{Effect of } T \uparrow \text{ on Endothermic}

\text{Final Conclusion}

\text{The Way Forward}

The Sigma Insight: Entropy and Free Energy

The Thermodynamic Dance of System and Surroundings

The Master Equation

Case 1

Case 2

The Grand Conclusion

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