Problem 55 - Chemistry The Central Science

Question

For a certain chemical reaction, \(\Delta H^{\circ}=-35.4 \mathrm{kJ}\) and \(\Delta S^{\circ}=-85.5 \mathrm{J} / \mathrm{K}\) . (a) Is the reaction exothermic or endothermic? (b) Does the reaction lead to an increase or decrease in the randomness or disorder of the system?(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

The reaction is exothermic, as \(\Delta H^{\circ} = -35.4 \mathrm{kJ}\) is negative. There is a decrease in randomness, as \(\Delta S^{\circ} = -85.5 \mathrm{J} / \mathrm{K}\) is negative. The \(\Delta G^{\circ}\) at 298 K is -9921 J, and the reaction is spontaneous under standard conditions at this temperature since \(\Delta G^{\circ}\) is negative.

Step 1: Determine if the reaction is exothermic or endothermic

We are given \(\Delta H^{\circ} = -35.4 \mathrm{kJ}\), which is a negative value. Therefore, the reaction is exothermic.

Step 2: Determine if the reaction increases or decreases the randomness

We are given \(\Delta S^{\circ} = -85.5 \mathrm{J} / \mathrm{K}\), which is a negative value. Therefore, there is a decrease in randomness (or disorder) in the reaction.

Step 3: Calculate \(\Delta G^{\circ}\) at 298 K

We are given the temperature as \(T=298\,\mathrm{K}\). To calculate \(\Delta G^{\circ}\), we will use the equation: $$ \Delta G^{\circ} = \Delta H^{\circ} - T\Delta S^{\circ} $$ First, we convert \(\Delta H^{\circ}\) to J/mol: $$ \Delta H^{\circ} = -35.4\,\mathrm{kJ} * 1000\,\mathrm{J}/\mathrm{kJ} = -35400\,\mathrm{J} $$ Now, we plug in the values and get: $$ \Delta G^{\circ} = -35400\,\mathrm{J} - (298\,\mathrm{K} * -85.5\,\mathrm{J}/\mathrm{K}) = -35400\,\mathrm{J} + 25479\,\mathrm{J} = -9921\,\mathrm{J} $$

Step 4: Determine if the reaction is spontaneous at 298 K

Since the value of \(\Delta G^{\circ} = -9921\,\mathrm{J}\) is negative, the reaction is spontaneous at 298 K under standard conditions. In summary, the reaction is exothermic, decreases randomness, has a \(\Delta G^{\circ}\) of -9921 J, and is spontaneous at 298 K under standard conditions.

Key Concepts

Exothermic Reaction

In chemistry, reactions are often classified based on how they exchange energy with their surroundings. An exothermic reaction is one that releases energy, typically in the form of heat, to its surroundings. This occurs when the energy required to break the bonds in the reactants is less than the energy released when the new bonds are formed in the products.

So, what does this mean for the values we see? In terms of enthalpy, which is represented as \( \Delta H \), an exothermic reaction features a negative \( \Delta H \) value. This tells us that energy is leaving the system. In the example from our exercise, \( \Delta H^{\circ} = -35.4\,\mathrm{kJ} \), indicating that 35.4 kJ of energy per mole is released. When handling this kind of reaction, you might observe heat being emitted or even a temperature rise in the surrounding environment.

It's important to understand that exothermic reactions are generally favored energetically because they tend to lead toward a more stable state. Think of burning wood, where the heat emitted is a clear sign of an exothermic process.

Entropy

Entropy is a fascinating concept in thermodynamics that deals with the disorder or randomness in a system. Represented by \( \Delta S \), a change in entropy can tell us much about the molecular dance that occurs during a reaction.

In our exercise, we encounter a negative \( \Delta S^{\circ} = -85.5 \,\mathrm{J}/\mathrm{K} \), indicating a decrease in entropy. But what does this signify? When entropy decreases, the system becomes more ordered. This is counterintuitive because most processes in nature lead to an increase in entropy, or more disorder. Imagine a room getting messy—it's the natural way things tend to go.

However, certain reactions and conditions require increased order. For example, when gas molecules condense into a liquid, they become more ordered, and entropy decreases. Understanding entropy helps us predict the direction of a reaction, especially when paired with other factors like enthalpy and temperature.

Spontaneity

Spontaneity in chemistry refers to whether a reaction proceeds on its own without external input. This is determined by the Gibbs Free Energy change \( \Delta G \), which combines enthalpy and entropy changes to predict reaction spontaneity.

\( \Delta G \) is calculated using the formula: \( \Delta G^{\circ} = \Delta H^{\circ} - T\Delta S^{\circ} \). Here, \( T \) is the temperature in Kelvin. If \( \Delta G \) is negative, the reaction is spontaneous under those conditions, meaning it can occur without outside help.

In our worked exercise example, \( \Delta G^{\circ} = -9921\,\mathrm{J} \), which is less than zero. This indicates that the reaction is spontaneous at 298 K. It's crucial to note that spontaneity doesn't mean immediate or fast—it just implies that the process is energetically favorable.

Conceptually, a negative \( \Delta G \) shows that the reaction will move toward forming products, driven naturally by a decrease in free energy, aligning with the system's tendency to reach more stable configurations.