Problem 101 - Chemistry Matter and Change

Question

For the change \(\mathrm{H}_{2} \mathrm{O}(1) \rightarrow \mathrm{H}_{2} \mathrm{O}(\mathrm{g}), \Delta G_{\mathrm{sssstem}}^{\circ}\) is 8.557 \(\mathrm{kJ}\) and \(\Delta H_{\text { system is } 44.01 \mathrm{kJ} . \text { What is } \Delta S_{\text { system }}^{\circ}}\) for the change?

Step-by-Step Solution

Answer

\( \Delta S_{\text{system}}^{\circ} = 118.9 \, \text{J/K} \)

Step 1: Identify the Gibbs Free Energy Equation

The relationship between Gibbs free energy, enthalpy, and entropy is given by the equation \( \Delta G^{\circ} = \Delta H^{\circ} - T \Delta S^{\circ} \). We are given \( \Delta G^{\circ} = 8.557 \, \text{kJ} \) and \( \Delta H^{\circ} = 44.01 \, \text{kJ} \). Our task is to find \( \Delta S^{\circ} \).

Step 2: Rearrange Equation to Solve for ΔS

Rearrange the equation to solve for the entropy change: \( \Delta S^{\circ} = \frac{\Delta H^{\circ} - \Delta G^{\circ}}{T} \). Before substituting the values, we first need to know the temperature. Standard conditions imply the reaction is carried out at 298 K.

Step 3: Convert Units (If Necessary)

Ensure all units are consistent. Both \( \Delta G \) and \( \Delta H \) are already in kJ, so we don't need to convert to J. However, when calculating \( \Delta S \), the unit of entropy will be in kJ/K unless converted to J/K.

Step 4: Substitute and Calculate

Substitute the given values into the rearranged equation: \( \Delta S^{\circ} = \frac{44.01 \, \text{kJ} - 8.557 \, \text{kJ}}{298 \, \text{K}} \). Calculate \( \Delta S^{\circ} \) to find: \( \Delta S^{\circ} = \frac{35.453 \, \text{kJ}}{298 \, \text{K}} \approx 0.1189 \, \text{kJ/K} \).

Step 5: Convert to Standard Units for Entropy

Convert the entropy from kJ/K to J/K: \( 0.1189 \, \text{kJ/K} = 118.9 \, \text{J/K} \). Therefore, \( \Delta S^{\circ} \) is \( 118.9 \, \text{J/K} \).

Key Concepts

Enthalpy

In the realm of thermodynamics, enthalpy is an important concept. It is a measure of the total energy of a system, which includes both its internal energy and the energy required to make space for it by displacing its surroundings. Enthalpy is denoted by the symbol \( H \).
For chemical reactions, the change in enthalpy, \( \Delta H \), tells us whether the reaction is absorbing or releasing heat under constant pressure:

• Exothermic reactions release heat, leading to a negative \( \Delta H \).
• Endothermic reactions absorb heat, resulting in a positive \( \Delta H \).

In our example reaction, the water is changing from liquid to gas, which requires energy. This is represented by a positive \( \Delta H^{\circ} = 44.01 \text{kJ} \), indicating it's an endothermic process.

Entropy

Entropy is a measure of disorder or randomness in a system, denoted by the symbol \( S \). In most natural processes, entropy tends to increase, signifying a move towards greater disorder.
The change in entropy, \( \Delta S \), indicates how the disorder changes during a process:

• If \( \Delta S \) is positive, the disorder of the system increases.
• If \( \Delta S \) is negative, the disorder decreases.

For the aforementioned chemical change, \( \Delta S^{\circ} \) is calculated to be approximately \( 118.9 \, \text{J/K} \). This positive value aligns with the transition from liquid to gas, as gases have higher entropy compared to liquids due to their increased molecular freedom.

Thermodynamics

Thermodynamics is the study of energy, heat, work, and how they interrelate. It is crucial for understanding processes in physical systems at equilibrium or during changes. At the heart of thermodynamics are its laws, which govern the principles of energy conservation and entropy.
The Gibbs Free Energy equation, \( \Delta G^{\circ} = \Delta H^{\circ} - T \Delta S^{\circ} \), is a fundamental part of thermodynamics, combining these key elements:

• \( \Delta G^{\circ} \) tells us about the spontaneity of a process. A negative \( \Delta G^{\circ} \) indicates a spontaneous or feasible process under constant temperature and pressure.
• Using this equation allows us to determine unknown values like \( \Delta S^{\circ} \) when \( \Delta G^{\circ} \) and \( \Delta H^{\circ} \) are known, as demonstrated in the example above.

Standard Conditions

In chemistry, reactions are often analyzed under standard conditions to maintain consistency. These conditions include certain predefined values for temperature and pressure at which reactions are typically considered:

• Standard temperature is 298 K (25°C).
• Standard pressure is 1 atmosphere (atm).

Utilizing standard conditions allows us to simplify calculations and easily compare thermodynamic data for various reactions. In this context, using these conditions ensures that our calculation for \( \Delta S^{\circ} \) accurately reflects the change under controlled, typical circumstances.