Problem 55 - Chemistry The Central Science In Si Units

Question

For a certain chemical reaction, \(\Delta H^{\circ}=-40.0 \mathrm{~kJ}\) and \(\Delta S^{\circ}=-150.0 \mathrm{~J} / \mathrm{K} .(\mathbf{a})\) Does the reaction lead to an increase or decrease in the randomness or disorder of the system? (b) Does the reaction lead to an increase or decrease in the randomness or disorder of the surroundings? (c) Calculate \(\Delta G^{\circ}\) for the reaction at \(298 \mathrm{~K}\). (d) Is the reaction spontaneous at \(298 \mathrm{~K}\) under standard conditions?

Step-by-Step Solution

Answer

(a) Decrease in system disorder. (b) Increase in surroundings' disorder. (c) \( \Delta G^{\circ} = 4.7 \, \mathrm{kJ} \). (d) Reaction is not spontaneous.

Step 1: Determine Change in System's Disorder

To evaluate the change in the disorder of the system, we examine the sign of \( \Delta S^{\circ} \). Here, \( \Delta S^{\circ} = -150.0 \, \text{J/K} \), meaning the system's disorder decreases.

Step 2: Assess Surroundings’ Disorder

The change in the surroundings’ disorder is determined by the sign of \( \Delta H^{\circ} \). Here, \( \Delta H^{\circ} = -40.0 \, \text{kJ} \), which implies that the heat is released to the surroundings, thus increasing the surroundings' disorder.

Step 3: Calculate \( \Delta G^{\circ} \) for the Reaction

Use the equation \( \Delta G^{\circ} = \Delta H^{\circ} - T \Delta S^{\circ} \). Convert \( \Delta S^{\circ} \) to kJ by dividing by 1000: \( \Delta S^{\circ} = -0.150 \, \text{kJ/K} \). Substituting the values: \[ \Delta G^{\circ} = -40.0 \, \text{kJ} - 298 \, \text{K} \times (-0.150 \, \text{kJ/K}) = -40.0 \, \text{kJ} + 44.7 \, \text{kJ} =4.7 \, \text{kJ} \].

Step 4: Determine the Spontaneity of the Reaction

A reaction is spontaneous if \( \Delta G^{\circ} \) is negative. Here, \( \Delta G^{\circ} = 4.7 \, \text{kJ} \), which is positive, indicating the reaction is not spontaneous at 298 K under standard conditions.

Key Concepts

Standard Entropy Change

Entropy is a fundamental concept in chemistry, helping to describe the degree of disorder or randomness in a system. The standard entropy change, denoted as \( \Delta S^{\circ} \), tells us how the disorder of a system changes during a reaction. In the given exercise, we have \( \Delta S^{\circ} = -150.0 \, \text{J/K} \). The negative sign indicates a decrease in entropy, meaning the reaction results in a more ordered system. When molecules in a reaction become more structured or align in an ordered fashion, \( \Delta S^{\circ} \) becomes negative.

However, it is also important to consider the entropy change of the surroundings. Despite the system's decrease in entropy, the overall process, especially exothermic reactions which release heat (as observed with a negative \( \Delta H^{\circ} \)), can increase the disorder, or entropy, in the surroundings. Thus, even if entropy decreases within the system, the total change involving both system and surroundings can reveal much about how energy dispersion affects the universe as a whole.

Standard Enthalpy Change

Enthalpy change, \( \Delta H^{\circ} \), reflects the heat exchange between a system and its surroundings at constant pressure. In the problem, \( \Delta H^{\circ} = -40.0 \, \text{kJ} \), suggesting an exothermic reaction. An exothermic process releases heat, thus increasing the entropy of the surroundings, even as the system itself may become more ordered from the perspective of standard entropy change.

The release of energy to the surroundings during an exothermic reaction has consequences beyond simple heat flow. It introduces energy into the surroundings, increasing molecular motion and effectively enhancing the randomness of the particles in the surroundings. This is why an exothermic reaction, like the one described, can run counter to the system's decrease in entropy, further playing into the great balancing act that thermodynamics describes.

Spontaneity of Reactions

Spontaneity in chemical reactions is determined by the sign of the Gibbs Free Energy change, \( \Delta G^{\circ} \). Calculated via \( \Delta G^{\circ} = \Delta H^{\circ} - T \Delta S^{\circ} \), it considers both the enthalpy and entropy components of a reaction. In our problem, substituting the given values, \( \Delta G^{\circ} = 4.7 \, \text{kJ} \), is positive.

A positive \( \Delta G^{\circ} \) indicates non-spontaneity under standard conditions. Essentially, it means the reaction as described is unlikely to proceed on its own at 298 K. The interplay between energy release and storage (enthalpy and entropy) informs us whether the universe 'wants' the reaction to occur without external influence. Spontaneity is not just about energy favorability but also how energy spreads or becomes more chaotic on a whole, reaffirming the delicate balance in thermodynamics.