Problem 35 - Chemistry An Atoms First Approach

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. \(\Delta H=-10 . \mathrm{kJ}, \Delta S=-40 . \mathrm{J} / \mathrm{K}, T=200 . \mathrm{K}\)

Step-by-Step Solution

Answer

The spontaneous changes at constant \(T\) and \(P\) are options (c) and (d).

Step 1: Calculating Gibbs free energy change for option (a)

Calculate the Gibbs free energy change using the given values: \(\Delta G = \Delta H - T \Delta S = +25 \times 10^3 \mathrm{kJ} - 300 \mathrm{K} \times (5.0 \times 10^{-3} \mathrm{kJ/K}) = 25000 \mathrm{kJ} - 1.5 \mathrm{kJ} = 24898.5 \mathrm{kJ}\)

Step 2: Checking for spontaneity in option (a)

Since \(\Delta G\) is positive, the process is not spontaneous.

Step 3: Calculating Gibbs free energy change for option (b)

Calculate the Gibbs free energy change using the given values: \(\Delta G = \Delta H - T \Delta S = +25 \times 10^3 \mathrm{kJ} - 300 \mathrm{K} \times (100 \times 10^{-3} \mathrm{kJ/K}) = 25000 \mathrm{kJ} - 30 \mathrm{kJ} = 24970 \mathrm{kJ}\)

Step 4: Checking for spontaneity in option (b)

Since \(\Delta G\) is positive, the process is not spontaneous.

Step 5: Calculating Gibbs free energy change for option (c)

Calculate the Gibbs free energy change using the given values: \(\Delta G = \Delta H - T \Delta S = (-10 \times 10^3) \mathrm{kJ} - 298 \mathrm{K} \times (5.0 \times 10^{-3} \mathrm{kJ/K}) = -10000 \mathrm{kJ} - 1.49 \mathrm{kJ} = -10001.49 \mathrm{kJ}\)

Step 6: Checking for spontaneity in option (c)

Since \(\Delta G\) is negative, the process is spontaneous.

Step 7: Calculating Gibbs free energy change for option (d)

Calculate the Gibbs free energy change using the given values: \(\Delta G = \Delta H - T \Delta S = (-10 \times 10^3) \mathrm{kJ} - 200 \mathrm{K} \times (-40 \times 10^{-3} \mathrm{kJ/K}) = -10000 \mathrm{kJ} + 8 \mathrm{kJ} = -9992 \mathrm{kJ}\)

Step 8: Checking for spontaneity in option (d)

Since \(\Delta G\) is negative, the process is spontaneous. From the analysis, the changes that will be spontaneous under the specified conditions are options (c) and (d).

Key Concepts

Spontaneity

In the realm of thermodynamics, spontaneity refers to the inherent tendency of a process to occur without any external intervention. A key indicator of spontaneity is the \(\Delta G\), or Gibbs Free Energy change, of a process. For a reaction or process to be spontaneous under certain conditions of constant temperature and pressure, \(\Delta G\) must be negative.

The Gibbs Free Energy formula is essential here:

• \(\Delta G = \Delta H - T \Delta S\)

Where:

• \(\Delta H\) is the change in enthalpy
• \(T\) is the absolute temperature in Kelvin
• \(\Delta S\) is the change in entropy

If \(\Delta G < 0\), the process is spontaneous, indicating it can occur naturally. Conversely, \(\Delta G > 0\) denotes a non-spontaneous process.

For example, in options (c) and (d) from the original exercise, calculations showed negative \(\Delta G\) values, suggesting that these processes are spontaneous.

Thermodynamics

Thermodynamics is a branch of physics that deals with the relationships between heat, work, temperature, and energy. Within this scope, the Gibbs Free Energy is a central concept that helps predict the direction and feasibility of processes. Consider thermodynamics as a framework that allows us to understand why some reactions happen spontaneously while others require energy.

It’s important to note the roles of enthalpy (\(\Delta H\)) and entropy (\(\Delta S\)) in determining the spontaneity of a reaction. Both contribute to the total Gibbs Free Energy change, influencing whether a process is feasible. The influence of temperature, as a multiplier of entropy, plays a crucial role in deciding the outcome.

For option (a) in the original exercise, even though the process had a \(\Delta S\) indicating an increase in disorder, the resulting \(\Delta G\) was still positive, showing non-spontaneity due to the overpowering effect of \(\Delta H\).

Enthalpy and Entropy

Enthalpy (\(\Delta H\)) and entropy (\(\Delta S\)) are both critical in the field of thermodynamics.

Enthalpy represents the total heat content of a system. Its change during a reaction often indicates heat release or absorption. A negative \(\Delta H\) suggests an exothermic process, which frequently indicates favorability for spontaneity.

On the other hand, Entropy measures the disorder or randomness in a system. An increase in entropy, denoted by \(\Delta S\) being positive, typically favors spontaneity, as systems naturally tend towards greater disorder.

In the original exercise, we see that options (c) and (d) had a negative \(\Delta H\), which aligned with observed spontaneity. In (c), both \(\Delta S\) and \(\Delta H\) contributed to a negative \(\Delta G\), aligning with spontaneous behavior. In contrast, option (d) balanced out a negative entropy change with a significant exothermic enthalpy change, resulting in a spontaneous process.