Problem 13 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

A reaction at 1 bar is non-spontaneous at low temperature but becomes spontaneous at high temperature. Identify the correct statement about the reaction among the following :(a) \(\Delta \mathrm{H}\) is negative while \(\Delta \mathrm{S}\) is positive (b) Both \(\Delta \mathrm{H}\) and \(\Delta \mathrm{S}\) are negative (c) \(\Delta \mathrm{H}\) is positive while \(\Delta \mathrm{S}\) is negative (d) Both \(\Delta \mathrm{H}\) and \(\Delta \mathrm{S}\) are positive.

Step-by-Step Solution

Answer

Both \(\Delta H\) and \(\Delta S\) are positive (option d).

Step 1: Understanding Gibbs Free Energy

The spontaneity of a reaction is determined using the Gibbs free energy change, \( \Delta G = \Delta H - T \Delta S \), where \( \Delta G \) is Gibbs free energy change, \( \Delta H \) is enthalpy change, \( T \) is temperature, and \( \Delta S \) is entropy change. A negative \( \Delta G \) implies a spontaneous reaction.

Step 2: Analyzing Temperature Effects

For a reaction to change from non-spontaneous at low temperature to spontaneous at high temperature, \( \Delta G \) must become negative as temperature increases. This implies that \( T \Delta S \) becomes large enough to outweigh \( \Delta H \).

Step 3: Considering Enthalpy and Entropy

Given that increasing temperature makes reaction spontaneous, \( \Delta S \) should be positive. If \( \Delta H \) were negative, the reaction would be spontaneous at low temperatures as well. Therefore, \( \Delta H \) must be positive.

Step 4: Select the Correct Option

From the analysis, for the reaction to be non-spontaneous at low temperatures and spontaneous at high temperatures, \( \Delta H \) is positive and \( \Delta S \) is positive. Thus, the correct statement is (d) Both \( \Delta \mathrm{H} \) and \( \Delta \mathrm{S} \) are positive.

Key Concepts

Spontaneous Reactions

Understanding if a reaction is spontaneous or not is vital in predicting how it will proceed. Spontaneity refers to a process that, given the right initial conditions, can proceed without needing additional energy input. This is determined by Gibbs Free Energy, represented by the equation:

• \( \Delta G = \Delta H - T \Delta S \)

A reaction is spontaneous when \( \Delta G \) is negative. If it is positive, the reaction is non-spontaneous and requires energy to proceed.
One interesting case is when a reaction shifts from non-spontaneous to spontaneous at different temperatures. This typically happens because the effect of the temperature on entropy (\( T \Delta S \)) starts to outweigh the impact of enthalpy (\( \Delta H \)). In the given example, a reaction that is non-spontaneous at low temperatures but becomes spontaneous at high temperatures likely has both positive \( \Delta H \) and positive \( \Delta S \).

Enthalpy Change

Enthalpy change, denoted as \( \Delta H \), represents the heat absorbed or released during a chemical reaction. It is an essential factor in determining the energy changes and stability of a reaction.

• A positive \( \Delta H \) (endothermic reaction) means the system absorbs heat, making products less stable at lower temperatures.
• A negative \( \Delta H \) (exothermic reaction) indicates that the system releases heat, often driving the reaction forward at lower temperatures.

Enthalpy change on its own does not define spontaneity. It must be considered alongside entropy and temperature. In the context of the problem, a positive \( \Delta H \) suggests that energy input (heat) is needed initially, making the reaction non-spontaneous at low temperatures. However, if entropy change is also positive, increased temperature can tip the scales toward spontaneity.

Entropy Change

Entropy, represented by \( \Delta S \), is a measure of disorder or randomness in a system. It reflects how energy is dispersed within a process.

• A positive \( \Delta S \) suggests an increase in disorder, favoring the reaction at higher temperatures.
• A negative \( \Delta S \) indicates a decrease in disorder, which doesn't benefit from added temperature.

Entropy change plays a critical role in determining reaction spontaneity at varying temperatures. When \( \Delta S \) is positive, as in the exercise, the product formation increases the disorder. This enhances the temperature's role in driving the spontaneity. With a reaction that shifts to being spontaneous at higher temperatures, \( \Delta S \) being positive is essential. Since \( T \Delta S \) becomes significant enough to overcome a positive \( \Delta H \), it points toward increasing entropy as a main contributor to the spontaneity.