Problem 89 - Chemistry: The Central Science in SI Units, Global Edition

Question

The reaction \(2 \mathrm{Mg}(s)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{MgO}(s)\) is highly spontaneous. A classmate calculates the entropy change for this reaction and obtains a large negative value for \(\Delta S^{\circ}\). Did your classmate make a mistake in the calculation? Explain.

Step-by-Step Solution

Answer

The classmate did not make a mistake in calculating the entropy change, ΔS°, for the reaction \(2 \mathrm{Mg}(s)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{MgO}(s)\). The large negative value for ΔS° is consistent with the expected decrease in entropy since the reaction involves converting gaseous molecules (greater disorder) to solid molecules (less disorder). Furthermore, the reaction is highly spontaneous, which means ΔG° must be negative. This also supports the obtained negative value for ΔS°.

Step 1: Examine the given reaction.

Observe the given reaction: \(2 \mathrm{Mg}(s)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{MgO}(s)\). We can see that magnesium (Mg) and oxygen (O₂) combine to form magnesium oxide (MgO). This reaction is highly spontaneous, which means the Gibbs free energy change (ΔG° < 0) for this reaction should be negative.

Step 2: Relationship between ΔG°, ΔH°, and ΔS°

Recall the formula for Gibbs free energy change: ΔG° = ΔH° - TΔS° A negative ΔG° indicates a spontaneous reaction. ΔH° represents the change in enthalpy (heat) of the reaction, and ΔS° represents the change in entropy (disorder) in the system during the reaction.

Step 3: Analyze the given reaction in terms of entropy

Looking at the reaction, \(2 \mathrm{Mg}(s)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{MgO}(s)\), we see that one mole of oxygen gas is being converted into two moles of solid magnesium oxide. Gas molecules have more entropy (disorder) than solid molecules since they occupy more volume and have more random movement compared to solids. As the number of gaseous molecules decreases and the number of solid molecules increases in the reaction, the overall entropy (ΔS°) should decrease, meaning ΔS° is negative.

Step 4: Determine if the classmate made a mistake

Since the reaction is spontaneous (highly negative ΔG°), and ΔS° is also negative due to the decrease in entropy, the only possibility for the reaction to be spontaneous is that ΔH° is negative and its absolute value is greater than TΔS°. So, if the classmate obtained a large negative value for ΔS°, it is consistent with the expected decrease in entropy for this reaction, and the classmate did not make a mistake in the calculation.

Key Concepts

Gibbs Free Energy

The concept of Gibbs Free Energy is crucial for understanding whether or not a chemical reaction is spontaneous. It is denoted by the symbol \( \Delta G \) and defined by the equation:
\[\Delta G = \Delta H - T\Delta S\]Here, \( \Delta H \) is the change in enthalpy, \( T \) is the temperature in Kelvin, and \( \Delta S \) is the change in entropy. Gibbs Free Energy essentially tells us the balance between these factors during a chemical process.

• A negative \( \Delta G \) means the reaction is spontaneous.
• A positive \( \Delta G \) indicates that the reaction is non-spontaneous.
• If \( \Delta G \) is zero, the system is in equilibrium, meaning there is no net change in the composition of reactants and products.

When we apply this to our example reaction \(2 \mathrm{Mg}(s)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{MgO}(s)\), a negative \( \Delta G \) confirms that the reaction is highly spontaneous. This spontaneity emerges even though there might be a decrease in entropy (\( \Delta S \) is negative), provided that the enthalpy change \( \Delta H \) has a larger magnitude and is also negative.

Entropy Change

Entropy change, denoted by \( \Delta S \), measures the degree of disorder or randomness in a system. In the reaction \(2 \mathrm{Mg}(s)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{MgO}(s)\), there is a transition from a more disordered state (with gaseous oxygen) to a more ordered state (with solid magnesium oxide).
When gaseous molecules transform into solid molecules, entropy generally decreases because:

• Gases have higher entropy due to more molecular motion and space to occupy.
• Solids have lower entropy because their particles are fixed in place with limited movement.

For our reaction, the shift from one mole of gaseous oxygen to two moles of solid magnesium oxide results in a negative \( \Delta S \). This decrease in entropy aligns with your classmate’s findings, indicating no calculation mistake on their part given the nature of the reaction.

Spontaneous Reaction

A spontaneous reaction is one that occurs naturally without needing extra energy once the reaction is initiated. These reactions are characterized by a negative Gibbs Free Energy \( \Delta G \).
For the reaction given, \(2 \mathrm{Mg}(s)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{MgO}(s)\), spontaneity seems evident due to the highly negative \( \Delta G \) value. Various factors contribute:

• The reaction releases a significant amount of energy, visible in its exothermic enthalpy change (\( \Delta H \) is negative).
• The system shifts from a higher entropy state (with gas) to a lower entropy state (with solid product), contributing less to spontaneity but requires compensation through enthalpy.

Despite a negative entropy change, the overall reaction remains spontaneous as the energy release (\( \Delta H \)) outweighs the decrease in disorder allowing even reactions with such conditions to proceed naturally.