Question

Given the values of \(\Delta H\) and \(\Delta S\), which of the following changes will be spontaneous at constant \(T\) and \(P ?\) a. \(\Delta H=+25 \mathrm{~kJ}, \Delta S=+5.0 \mathrm{~J} / \mathrm{K}, T=300 . \mathrm{K}\) b. \(\Delta H=+25 \mathrm{~kJ}, \Delta S=+100 . \mathrm{J} / \mathrm{K}, T=300 . \mathrm{K}\) c. \(\Delta H=-10 . \mathrm{kJ}, \Delta S=+5.0 \mathrm{~J} / \mathrm{K}, T=298 \mathrm{~K}\) d. \(\Delta H=-10 . \mathrm{kJ}, \Delta S=-40 . \mathrm{J} / \mathrm{K}, T=200 . \mathrm{K}\)

Short answer

Options b, c, and d will undergo spontaneous changes at constant \(T\) and \(P\).

Step-by-step solution

Gather the given values for each option

For each option, we are given the values for \(\Delta H\), \(\Delta S\), and \(T\). We will simply plug these values into the formula for \(\Delta G\) to determine whether the reaction will be spontaneous.

Calculate \(\Delta G\) for option a

We have \(\Delta H = +25 \textrm{ kJ}\), \(\Delta S = +5.0 \textrm{ J/K}\), and \(T = 300 \textrm{ K}\). Converting \(\Delta H\) to J: \(\Delta H = 25 \times 1000 = 25,000 \textrm{ J}\). Now we will calculate \(\Delta G = \Delta H - T \Delta S\): \(\Delta G = 25,000 \textrm{ J} - (300 \textrm{ K})(5.0 \textrm{ J/K}) = 25,000 \textrm{ J} - 1500 \textrm{ J} = 23,500 \textrm{ J}\). Since \(\Delta G > 0\), option a is not spontaneous.

Calculate \(\Delta G\) for option b

For option b, we have \(\Delta H = +25 \textrm{ kJ}\), \(\Delta S = +100 \textrm{ J/K}\), and \(T = 300 \textrm{ K}\). Using the same conversion as before, we have \(\Delta H = 25,000 \textrm{ J}\). Calculating \(\Delta G = \Delta H - T \Delta S\): \(\Delta G = 25,000 \textrm{ J} - (300 \textrm{ K})(100 \textrm{ J/K}) = 25,000 \textrm{ J} - 30,000 \textrm{ J} = -5,000 \textrm{ J}\). Since \(\Delta G < 0\), option b is spontaneous.

Calculate \(\Delta G\) for option c

In option c, we have \(\Delta H = -10 \textrm{ kJ}\), \(\Delta S = +5.0 \textrm{ J/K}\), and \(T = 298 \textrm{ K}\). Converting \(\Delta H\) to J: \(\Delta H = -10 \times 1000 = -10,000 \textrm{ J}\). Calculating \(\Delta G = \Delta H - T \Delta S\): \(\Delta G = -10,000 \textrm{ J} - (298 \textrm{ K})(5.0 \textrm{ J/K}) = -10,000 \textrm{ J} - 1,490 \textrm{ J} = -11,490 \textrm{ J}\). Since \(\Delta G < 0\), option c is spontaneous.

Calculate \(\Delta G\) for option d

For option d, we have \(\Delta H = -10 \textrm{ kJ}\), \(\Delta S = -40 \textrm{ J/K}\), and \(T = 200 \textrm{ K}\). Using the same conversion as before, we have \(\Delta H = -10,000 \textrm{ J}\). Calculating \(\Delta G = \Delta H - T \Delta S\): \(\Delta G = -10,000 \textrm{ J} - (200 \textrm{ K})(-40 \textrm{ J/K}) = -10,000 \textrm{ J} + 8,000 \textrm{ J} = -2,000 \textrm{ J}\). Since \(\Delta G < 0\), option d is spontaneous.

Conclusion

Options b, c, and d will undergo spontaneous changes at constant \(T\) and \(P\).

Key concepts

Enthalpy Change

Enthalpy change, represented by \(\Delta H\), is a measure of the total heat content in a system. It is an essential concept in thermodynamics as it indicates whether heat is absorbed or released during a chemical reaction. When the value of \(\Delta H\) is negative, the reaction is exothermic, meaning it releases heat to the surroundings. Conversely, a positive \(\Delta H\) indicates an endothermic reaction, where heat is absorbed from the surroundings.

Understanding the enthalpy change helps in predicting the behavior of reactions. In the context of determining spontaneity, enthalpy change is crucial but needs to be assessed alongside entropy changes and temperature to provide a full picture.

Entropy Change

Entropy change \(\Delta S\) reflects the degree of disorder or randomness in a system. A positive \(\Delta S\) suggests that a system is becoming more disordered, while a negative \(\Delta S\) indicates a decrease in disorder. In thermodynamics, systems tend to move towards higher entropy, aligning with the Second Law of Thermodynamics.

When assessing reaction spontaneity, the entropy change is a critical factor. A reaction with a significant increase in entropy (positive \(\Delta S\)) could become spontaneous if the increase offsets any positive enthalpy change. However, just like enthalpy, it is only one part of the equation. It must be considered with temperature and enthalpy change when evaluating reaction spontaneity.

Spontaneity of Reactions

The spontaneity of a chemical reaction is defined by its ability to occur without external intervention. Calculating the Gibbs Free Energy change \(\Delta G\), which integrates both the entropy change \(\Delta S\) and enthalpy change \(\Delta H\), provides insights into this spontaneity. The formula \(\Delta G = \Delta H - T\Delta S\) helps determine it.

A negative \(\Delta G\) indicates that a reaction is spontaneous, signaling that the process can occur without any additional energy input. If \(\Delta G\) is positive, the reaction is non-spontaneous, requiring energy to proceed. This calculation allows chemists to predict and manipulate chemical processes effectively, ensuring reactions are efficient and favorable under certain conditions.

Thermodynamics

Thermodynamics is the branch of physical science that deals with the relations between heat and other forms of energy. In chemistry, it is crucial for understanding how energy changes in a system during a reaction.

Thermodynamic principles, including the First and Second Laws, provide a foundation for understanding concepts such as enthalpy, entropy, and Gibbs Free Energy. These principles help explain why certain reactions are favorable and how alterations to a system's temperature or pressure can affect a reaction's direction.

• The First Law, Conservation of Energy, states that energy cannot be created or destroyed, only transformed.
• The Second Law introduces the concept of entropy, indicating that processes naturally progress towards disorder.

Temperature Dependence of Reactions

Temperature plays a pivotal role in chemical reactions and their spontaneity. It directly influences the Gibbs Free Energy equation \(\Delta G = \Delta H - T\Delta S\).

An increase in temperature can potentially make a non-spontaneous reaction (with positive \(\Delta G\)) spontaneous if the change in entropy \(\Delta S\) is positive and large enough to offset the enthalpy term. Conversely, a decrease in temperature may halt a previously spontaneous reaction, especially if \(\Delta S\) is negative, thereby raising \(\Delta G\) to a positive value.

Thus, understanding the temperature dependence is essential for controlling and optimizing reactions, as it determines energy requirements and feasibility in various practical applications.