Problem 55 - Chemistry: The Central Science in SI Units, Global Edition

Question

For a certain chemical reaction, \(\Delta H^{\circ}=-40.0 \mathrm{k} \mathrm{J}\) and \(\Delta S^{\circ}=-150.0 \mathrm{~J} / \mathrm{K}\). (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) The reaction leads to a decrease in the randomness or disorder of the system since \(\Delta S^{\circ} = -150.0 \, J/K\), which is negative. (b) The reaction leads to an increase in the randomness or disorder of the surroundings because \(\Delta H^{\circ} = -40.0 \, kJ\), which is negative, meaning heat is released by the system to the surroundings. (c) Using the formula \(\Delta G^{\circ} = \Delta H^{\circ} - T\Delta S^{\circ}\), we find \(\Delta G^{\circ} = 4,700 \, J\). (d) The reaction is not spontaneous at \(298 \, K\) under standard conditions since \(\Delta G^{\circ} > 0\).

Step 1: (a) Effect on the randomness of the system

We are given \(\Delta S^{\circ} = -150.0 \, J/K\). Since \(\Delta S^{\circ}\) represents the change in entropy, a negative value means that the entropy decreases, and there is a decrease in the randomness or disorder of the system.

Step 2: (b) Effect on the randomness of the surroundings

We are given \(\Delta H^{\circ} = -40.0 \, kJ\). \(\Delta H^{\circ}\) represents the change in enthalpy, which is equal to the heat absorbed or released by the system at constant pressure. A negative value means that heat is released by the system to the surroundings. When heat is transferred, the randomness of the surroundings increases.

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

To calculate \(\Delta G^{\circ}\), we must use the following formula: \[\Delta G^{\circ} = \Delta H^{\circ} - T\Delta S^{\circ}\] where \(T\) is the temperature in Kelvin. We are given \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\), and we know the temperature is \(298 K\). So, we can plug in the values and calculate \(\Delta G^{\circ}\): \[\Delta G^{\circ} = (-40.0 \times 10^3 \, J) - (298 \, K)(-150.0 \, J/K) \]

Step 4: Calculate the value for \(\Delta G^{\circ}\)

Now we can compute the result for \(\Delta G^{\circ}\): \[\Delta G^{\circ} = (-40.0 \times 10^3 \, J) + (298 \times 150.0 J) \] \[\Delta G^{\circ} = -40.0 \times 10^3 \, J + 44,700 \, J \] \[\Delta G^{\circ} = -40,000 \, J + 44,700 \, J = 4,700 \, J\]

Step 5: (d) Spontaneity of the reaction

A reaction is spontaneous at a given temperature under standard conditions if \(\Delta G^{\circ} < 0\). In this case, we calculated \(\Delta G^{\circ} = 4,700 \, J\), which is greater than zero. Therefore, the reaction is not spontaneous at \(298 \, K\) under standard conditions.

Key Concepts

Entropy Change

Entropy is a measure of the randomness or disorder within a system. In chemical reactions, the change in entropy, denoted as \(\Delta S\), indicates whether the system becomes more disordered or more ordered. If \(\Delta S\) is positive, the system's disorder increases; if negative, the system's disorder decreases.

For the given reaction, we have \(\Delta S^\circ = -150.0 \, \mathrm{J}/\mathrm{K}\). This negative entropy change implies that the system becomes more organized, reducing its randomness or disorder as the reaction proceeds.

Several factors can affect entropy change:

• Phase transitions: Gases have higher entropy than liquids or solids.
• Complexity of molecules: More complex molecules can increase entropy.
• Temperature: Higher temperatures generally lead to higher entropy.

Enthalpy Change

Enthalpy measures the total heat content of a system, affecting how energy is absorbed or released during a chemical reaction. \(\Delta H\), the change in enthalpy, is crucial in determining the direction of heat transfer.

The negative \(\Delta H^\circ = -40.0 \, \mathrm{kJ}\) for this reaction means that the system releases heat to the surroundings. Hence, it is an exothermic reaction, as heat is given off. Exothermic reactions usually feel warm as they occur and can affect the system's and surroundings' states and behaviors.

Key aspects of enthalpy change:

• Heat released in exothermic vs. heat absorbed in endothermic reactions.
• Role in determining the balance between system and surroundings.
• Calculations at constant pressure give insight into energy transfers.

Spontaneity of Reaction

The spontaneity of a chemical reaction is fundamentally determined by the Gibbs Free Energy change (\(\Delta G\)). The equation \(\Delta G = \Delta H - T \Delta S\) integrates both enthalpy and entropy to predict whether a reaction will proceed on its own.

A negative \(\Delta G\) indicates a spontaneous reaction, while a positive value suggests non-spontaneity under the given conditions. For this particular reaction at \(298 \, \mathrm{K}\), the calculated \(\Delta G^\circ = 4,700 \, \mathrm{J}\), which is positive.

This indicates the reaction is not spontaneous under standard conditions. The interplay between \(\Delta H\) and \(T\Delta S\) is crucial:

• High temperatures can make entropy favorable, affecting spontaneity.
• Enthalpy must often be minimal for spontaneity if entropy is unfavorable.
• Reactions can be spontaneous at certain temperatures, even with positive \(\Delta G\) values at other temps.