Question

A certain reaction has \(\Delta H^{\circ}=-19.5 \mathrm{~kJ}\) and \(\Delta S^{\circ}=\) \(+42.7 \mathrm{~J} / \mathrm{K} .\) (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?

Short answer

(a) The reaction is exothermic as \(\Delta H^{\circ}<0\). (b) The reaction leads to an increase in the randomness or disorder of the system as \(\Delta S^{\circ}>0\). (c) For the reaction at \(298 \mathrm{~K}\), we calculate \(\Delta G^{\circ} \approx -7665.4 \mathrm{~J}\). (d) The reaction is spontaneous at \(298 \mathrm{~K}\) under standard conditions as \(\Delta G^{\circ}<0\).

Step-by-step solution

Understand the terms and determine exothermic/endothermic and change in disorder

The enthalpy change of a reaction, \(\Delta H^{\circ}\), indicates if the reaction is exothermic (releases heat) or endothermic (absorbs heat). If \(\Delta H^{\circ}<0\) then the reaction is exothermic, and if \(\Delta H^{\circ}>0\) then the reaction is endothermic. The entropy change, \(\Delta S^{\circ}\), gives an idea of the change in randomness or disorder during the reaction. If \(\Delta S^{\circ}>0\), the reaction results in an increase of disorder and if \(\Delta S^{\circ}<0\), the reaction results in a decrease in disorder. Given, \(\Delta H^{\circ}=-19.5 \mathrm{~kJ}\) and \(\Delta S^{\circ}=\) \(+42.7 \mathrm{~J} / \mathrm{K} .\) Since \(\Delta H^{\circ}<0\), the reaction is exothermic. As \(\Delta S^{\circ}>0\), the reaction leads to an increase in the randomness or disorder of the system.

Calculate \(\Delta G^{\circ}\) using the Gibbs-Helmholtz equation

Using the Gibbs-Helmholtz equation, we can find \(\Delta G^{\circ}\) for the reaction at \(298 \mathrm{~K}.\) The equation is given by: \(\Delta G^{\circ} = \Delta H^{\circ} - T \Delta S^{\circ}\) Given temperature T is \(298 \mathrm{~K}\). However, we need to ensure the units are consistent, so convert \(\Delta H^{\circ}\) from kJ to J. \(\Delta H^{\circ} = -19.5 \mathrm{~kJ} \times 1000 = -19500 \mathrm{~J}\) Now substitute the values and calculate \(\Delta G^{\circ}\). \(\Delta G^{\circ} = -19500 \mathrm{~J} - (298 \mathrm{~K} \times +42.7 \mathrm{~J} / \mathrm{K}) \approx -7665.4 \mathrm{~J}\)

Determine if the reaction is spontaneous

To determine if the reaction is spontaneous at \(298 \mathrm{~K},\) we need to look at the sign of \(\Delta G^{\circ}\). If \(\Delta G^{\circ}<0\) the reaction is spontaneous. If \(\Delta G^{\circ}>0\) the reaction is non-spontaneous. As we calculated, \(\Delta G^{\circ} \approx -7665.4 \mathrm{~J}\) Since \(\Delta G^{\circ}<0\), the reaction is spontaneous at \(298 \mathrm{~K}\) under standard conditions.

Key concepts

Enthalpy

Enthalpy, symbolized as \( \Delta H \), is a thermodynamic quantity related to the heat content of a system. It's a key concept in understanding how energy is absorbed or released during a chemical reaction. When a reaction involves heat transfer, the enthalpy change can tell us whether the process is exothermic or endothermic.

• An exothermic reaction releases heat, leading to a temperature increase in the surrounding environment. This is seen when \( \Delta H < 0 \).
• An endothermic reaction absorbs heat, resulting in a temperature drop, represented by \( \Delta H > 0 \).

In the provided exercise, the enthalpy change is \( -19.5 \ \mathrm{kJ} \), indicating an exothermic reaction. This tells us that the reaction releases energy, making the surroundings warmer.

Entropy

Entropy, denoted as \( \Delta S \), is a measure of randomness or disorder within a system. It helps us understand the directionality of processes such as mixing, spreading, and natural events.

• When \( \Delta S > 0 \), it indicates an increase in disorder, suggesting the products of a reaction are less ordered than the reactants.
• Conversely, \( \Delta S < 0 \) suggests a decrease in disorder, meaning the products are more ordered.

In the given exercise, \( \Delta S \) is \( +42.7 \ \mathrm{J/K} \), showing an increase in disorder. This implies that the system becomes more chaotic as the reaction proceeds.

Gibbs Free Energy

Gibbs Free Energy, represented as \( \Delta G \), is crucial for determining the spontaneity of a reaction. It combines enthalpy and entropy to predict whether a process can occur under constant temperature and pressure without outside intervention.
The formula used is:
\( \Delta G = \Delta H - T \Delta S \)
where \( T \) is the temperature in Kelvin.

• When \( \Delta G < 0 \), the reaction is spontaneous, or it can proceed without additional energy.
• If \( \Delta G > 0 \), the reaction is non-spontaneous and requires external energy.
• A \( \Delta G = 0 \) indicates a system at equilibrium.

Using the exercise values for enthalpy and entropy, and a temperature of \( 298 \ \mathrm{K} \), \( \Delta G \) was calculated to be approximately \( -7665.4 \ \mathrm{J} \), confirming the reaction is spontaneous.

Exothermic Reaction

An exothermic reaction is characterized by the release of energy, usually in the form of heat, into its surroundings. As mentioned earlier, such a reaction has a negative enthalpy change (\( \Delta H < 0 \)). This energy release results in a warming effect, often felt in everyday instances.
Some common examples of exothermic reactions include:

• Combustion, such as burning wood or fuel.
• Respiration in living organisms, where glucose is broken down.
• Mixing acids with bases, which often releases heat.

In the discussed exercise, with \( \Delta H = -19.5 \ \mathrm{kJ} \), the reaction is clearly exothermic, indicating that it releases heat.

Spontaneity

The concept of spontaneity in thermodynamics refers to whether a process can proceed on its own under given conditions. An important factor in determining spontaneity is the Gibbs Free Energy change (\( \Delta G \)).

• A negative \( \Delta G \) means the process is spontaneous at the given temperature, indicating it can occur without any input of external energy.
• A positive \( \Delta G \) denotes a non-spontaneous process, thus needing energy input to occur.

The calculated \( \Delta G \) in our exercise was approximately \( -7665.4 \ \mathrm{J} \), showcasing that the reaction is indeed spontaneous at \( 298 \ \mathrm{K} \). This means the conditions naturally favor the forward progression of the reaction, contributing to its complete conversion of reactants to products.