Problem 34 - Chemistry An Atoms First Approach

Question

Calculate \(\Delta S_{\text {sur }}\) for the following reactions at \(25^{\circ} \mathrm{C}\) and 1 atm. $$\text { a. } \mathrm{C}_{3} \mathrm{H}_{8}(g)+5 \mathrm{O}_{2}(g) \longrightarrow 3 \mathrm{CO}_{2}(g)+4 \mathrm{H}_{2} \mathrm{O}(l)$$$$\begin{aligned} &\Delta H^{\circ}=-2221 \mathrm{kJ}\\\ &\text { b. } 2 \mathrm{NO}_{2}(g) \longrightarrow 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) \quad \Delta H^{\circ}=112 \mathrm{kJ} \end{aligned}$$

Step-by-Step Solution

Answer

The change in entropy of the surroundings for reaction (a) is 7453 J/mol·K and for reaction (b) is -376 J/mol·K.

Step 1: Convert the given temperature to kelvin

To convert 25°C to kelvin, add 273.15. \[25^{\circ}\text{C} + 273.15 = 298.15\, \text{K}\] Temperature, T = 298.15 K

Step 2: Calculate \(\Delta S_{\text{sur}}\) for reaction (a)

Using the given \(\Delta H^{\circ} = -2221\, \text{kJ}\) for reaction (a) and temperature T = 298.15 K, we can plug these values into the formula for \(\Delta S_{\text{sur}}\): \[\Delta S_{\text{sur}} = -\frac{-2221\, \text{kJ}}{298.15\, \text{K}}\] Convert \(\Delta H^{\circ}\) to joules: \(-2221\, \text{kJ} = -2221000\, \text{J}\) \[\Delta S_{\text{sur}} = -\frac{-2221000\, \text{J}}{298.15\, \text{K}}\]

Step 3: Calculate \(\Delta S_{\text{sur}}\) for reaction (b)

Using the given \(\Delta H^{\circ} = 112\, \text{kJ}\) for reaction (b) and temperature T = 298.15 K, we can plug these values into the formula for \(\Delta S_{\text{sur}}\): \[\Delta S_{\text{sur}} = -\frac{112\, \text{kJ}}{298.15\, \text{K}}\] Convert \(\Delta H^{\circ}\) to joules: \(112\, \text{kJ} = 112000\, \text{J}\) \[\Delta S_{\text{sur}} = -\frac{112000\, \text{J}}{298.15\, \text{K}}\]

Step 4: Express the results in J/mol·K

Now, we can calculate \(\Delta S_{\text{sur}}\) for each reaction: For reaction (a): \[\Delta S_{\text{sur}} = \frac{2221000\, \text{J}}{298.15\, \text{K}} = 7453\, \text{J/mol·K}\] For reaction (b): \[\Delta S_{\text{sur}} = -\frac{112000\, \text{J}}{298.15\, \text{K}} = -376\, \text{J/mol·K}\] Therefore, \(\Delta S_{\text{sur}}\) for reaction (a) is 7453 J/mol·K and for reaction (b) is -376 J/mol·K.

Key Concepts

Thermodynamics

Thermodynamics is a branch of physics dealing with heat, work, temperature, and the related properties of matter. The field explores how different forms of energy can be transformed into one another and how they affect matter. It's fundamental in explaining why physical and chemical phenomenon occur in a certain way.

At the heart of thermodynamics are four laws which can be applied to all sorts of thermodynamic systems, albeit closed or open. One of the key principles related to the given exercise is the second law of thermodynamics, which states that the entropy of an isolated system always increases or remains constant. In simple terms, entropy is a measure of the disorder or randomness in a system, and this concept helps us understand the irreversible nature of natural processes. When we calculate the change in entropy (abla S) during a chemical reaction, we are assessing how the disorder of the system and its surroundings has been altered by that reaction.

Chemical Reactions

Chemical reactions involve the transformation of one set of chemical substances to another through the breaking and forming of chemical bonds. These reactions are influenced by various thermodynamic properties, such as the internal energy, enthalpy (Delta H), entropy (Delta S), and temperature (T), all of which determine the spontaneity and extent of the reaction.

Different types of chemical reactions, such as combustion in reaction (a) from the exercise or the decomposition in reaction (b), involve changes in enthalpy and entropy. Exothermic reactions, like combustion, release heat and typically have a negative Delta H value, which reflects the release of energy. In contrast, endothermic reactions, such as the decomposition of NO2, absorb heat, resulting in a positive Delta H. These changes, combined with the temperature of the surroundings, allow us to calculate the change in the entropy of the surroundings, further revealing insights into the second law of thermodynamics.

Gibbs Free Energy

Gibbs free energy (G) integrates enthalpy, entropy, and temperature to predict the direction of a chemical reaction and whether it’s spontaneous. The Gibbs free energy change (Delta G) for a process is defined as: Delta G = Delta H - TDelta S, where Delta H is the change in enthalpy, T is the absolute temperature in Kelvin (K), and Delta S is the change in entropy of the system. A spontaneous reaction occurs when Delta G is negative, indicating that the process can proceed without the input of additional energy.

When we apply this to chemical reactions, a negative Delta G suggests a reaction is likely to occur, while a positive value suggests it is not favored under the given conditions. For example, the calculation in the exercise helps us understand the entropy change of the surroundings (Delta S_sur), which is directly related to the concept of Gibbs free energy. By measuring these changes, scientists can better predict the behavior of chemical reactions in various conditions.