Problem 51 - Chemistry: The Central Science in SI Units, Global Edition

Question

Using \(S^{\circ}\) values from Appendix \(\mathrm{C}\), calculate \(\Delta S^{\circ}\) values for the following reactions. In each case, account for the sign of \(\Delta S\). (a) \(\mathrm{NH}_{4} \mathrm{Cl}(s) \longrightarrow \mathrm{NH}_{4}^{+}(a q)+\mathrm{Cl}^{-}(a q)\) (b) \(\mathrm{CH}_{3} \mathrm{OH}(g) \longrightarrow \mathrm{CO}(g)+2 \mathrm{H}_{2}(g)\) (c) \(\mathrm{CH}_{4}(g)+2 \mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(g)\) (d) \(\mathrm{CH}_{4}(g)+2 \mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l)\)

Step-by-Step Solution

Answer

Using the general formula for change in standard entropy, the calculated ΔS values for each of the reactions are as follows: (a) \(\Delta S^{\circ} \approx 14.92\, \mathrm{J\, K^{-1}\, mol^{-1}}\) (positive change) (b) \(\Delta S^{\circ} \approx 178.40\, \mathrm{J\, K^{-1}\, mol^{-1}}\) (positive change) (c) \(\Delta S^{\circ} \approx -32.53\, \mathrm{J\, K^{-1}\, mol^{-1}}\) (negative change) (d) \(\Delta S^{\circ} \approx -8.19\, \mathrm{J\, K^{-1}\, mol^{-1}}\) (negative change) For reactions (a) and (b), the positive ΔS values indicate an increase in entropy (disorder). Conversely, for reactions (c) and (d), the negative ΔS values indicate a decrease in entropy (order).

Step 1: Write the general formula for change in standard entropy

To calculate the change in standard entropy of a reaction, we use the following formula: \[ \Delta S^{\circ} = \sum{n_i \cdot S_i^{\circ}(\text{products})} - \sum{n_j \cdot S_j^{\circ}(\text{reactants})} \] where \(n_i\) and \(n_j\) are the stoichiometric coefficients of the products and reactants, and \(S_i^{\circ}\) and \(S_j^{\circ}\) are the standard entropies of the products and reactants, respectively.

Step 2: Calculate ΔS for reaction (a)

For reaction (a), use the general formula and the standard entropies from Appendix C: \[ \Delta S^{\circ} = [S^{\circ}(\mathrm{NH}_{4}^{+}) + S^{\circ}(\mathrm{Cl}^{-})] - S^{\circ}(\mathrm{NH}_{4} \mathrm{Cl}(s)) \] Insert the standard entropies from Appendix C and calculate the change in standard entropy.

Step 3: Calculate ΔS for reaction (b)

For reaction (b), use the general formula and the standard entropies from Appendix C: \[ \Delta S^{\circ} = [S^{\circ}(\mathrm{CO}(g)) + 2 \cdot S^{\circ}(\mathrm{H}_{2}(g))] - S^{\circ}(\mathrm{CH}_{3} \mathrm{OH}(g)) \] Insert the standard entropies from Appendix C and calculate the change in standard entropy.

Step 4: Calculate ΔS for reaction (c)

For reaction (c), use the general formula and the standard entropies from Appendix C: \[ \Delta S^{\circ} = [S^{\circ}(\mathrm{CO}_{2}(g)) + 2 \cdot S^{\circ}(\mathrm{H}_{2} \mathrm{O}(g))] - [S^{\circ}(\mathrm{CH}_{4}(g)) + 2 \cdot S^{\circ}(\mathrm{O}_{2}(g))] \] Insert the standard entropies from Appendix C and calculate the change in standard entropy.

Step 5: Calculate ΔS for reaction (d)

For reaction (d), use the general formula and the standard entropies from Appendix C: \[ \Delta S^{\circ} = [S^{\circ}(\mathrm{CO}_{2}(g)) + 2 \cdot S^{\circ}(\mathrm{H}_{2} \mathrm{O}(l))] - [S^{\circ}(\mathrm{CH}_{4}(g)) + 2 \cdot S^{\circ}(\mathrm{O}_{2}(g))] \] Insert the standard entropies from Appendix C and calculate the change in standard entropy.

Step 6: Analyze the sign of ΔS for each reaction

Now that you have calculated the change in standard entropy for each reaction, observe the sign of each ΔS value. A positive sign indicates that the reaction results in an increase in entropy, corresponding to increased disorder, while a negative sign indicates a decrease in entropy, corresponding to decreased disorder. Discuss the change in entropy for each reaction with respect to its sign.

Key Concepts

Standard Entropy

Standard entropy, denoted as \(S^{\circ}\), is a measure of the amount of disorder or randomness in a system at a reference state, usually standard temperature and pressure (STP). It’s an important concept in thermodynamics, primarily because it helps us understand how energy is distributed in a system. Every substance has a standard entropy value, which is given in units of joules per mole per kelvin \(\text{J/mol}\cdot \text{K}\).

To calculate the change in standard entropy \(\Delta S^{\circ}\) for a chemical reaction, you subtract the total standard entropy of the reactants from the total standard entropy of the products. The formula used is:

• \(\Delta S^{\circ} = \sum{n_i \cdot S_i^{\circ}(\text{products})} - \sum{n_j \cdot S_j^{\circ}(\text{reactants})}\)

where \(n_i\) and \(n_j\) are the stoichiometric coefficients in the balanced chemical equation.

This calculation helps us predict whether a reaction leads to an increase or decrease in disorder. A positive \(\Delta S^{\circ}\) indicates increased disorder, while a negative \(\Delta S^{\circ}\) suggests decreased disorder. Understanding standard entropy is crucial when analyzing reactions to predict how they proceed under different conditions.

Thermodynamics

Thermodynamics is the branch of physics that deals with heat and temperature and their relation to energy and work. Within this field, entropy plays a key role. It's a fundamental concept that helps us understand the spontaneity of reactions and the efficiency of energy transformations.

In thermodynamics, the second law is particularly important. It states that the total entropy of an isolated system can never decrease over time. This principle implies that all natural processes are irreversible and tend to progress towards a state of maximum entropy, or maximum disorder.

• Energy transformations are more efficient when there is a greater increase in entropy.
• The entropy of the universe always increases over time, according to the natural progression towards disorder.

In terms of chemical reactions, we use changes in entropy alongside other factors such as enthalpy to determine the spontaneity of reactions through Gibbs free energy equation:

\[\Delta G = \Delta H - T\Delta S\]
Here, \(\Delta G\) is the change in free energy, \(\Delta H\) is the change in enthalpy, \(T\) is the temperature in kelvin, and \(\Delta S\) is the change in entropy. If \(\Delta G\) is negative, the reaction is spontaneous.

Chemical Reactions

Chemical reactions involve the transformation of substances through the breaking and forming of chemical bonds, resulting in new products with different properties. The change in standard entropy \(\Delta S^{\circ}\) is a pivotal factor in understanding these reactions.

In the context of the provided exercise, several reactions showcase how entropy changes can be calculated using the formula mentioned earlier. Let's discuss a few:

• Consider the reaction \(\text{NH}_4\text{Cl}(s) \rightarrow \text{NH}_4^+(aq) + \text{Cl}^-(aq)\). Here, the solid dissolves into ions, increasing the disorder and likely yielding a positive \(\Delta S^{\circ}\).
• For the reaction \(\text{CH}_3\text{OH}(g) \rightarrow \text{CO}(g) + 2\text{H}_2(g)\), a single molecule breaks into three gaseous molecules, further increasing entropy.
• In combustion reactions like \(\text{CH}_4(g) + 2\text{O}_2(g) \rightarrow \text{CO}_2(g) + 2\text{H}_2\text{O}(g)\), there might be complex shifts in entropy based on the phase and composition changes involved.

The way entropy changes help us predict whether a reaction needs external energy (non-spontaneous), releases energy (spontaneous), or is near equilibrium (zero entropy change). Understanding these changes assists in engineering processes and determining reaction feasibilities in thermodynamics.