Question

The reaction \(2 \mathrm{Mg}(s)+\mathrm{O}_{2}(\mathrm{~g}) \longrightarrow 2 \mathrm{MgO}(s)\) is highly spontaneous. A classmate calculates the entropy change for this reaction and obtains a large negative value for \(\Delta S^{5}\). Did your classmate make a mistake in the calculation? Explain.

Short answer

The classmate's calculation of a large negative value for ∆S° is correct since the reaction transforms a gaseous molecule (O₂) into solid molecules (2 MgO) with a decrease in entropy. Despite the negative entropy change, the reaction is highly spontaneous, which can be explained by a sufficiently large negative value of ∆H° that leads to a negative ∆G° at the given conditions.

Step-by-step solution

Understanding spontaneity and entropy

For a reaction to be spontaneous at constant temperature and pressure, the change in Gibbs free energy (∆G°) should be negative. The relationship between Gibbs free energy, enthalpy change (∆H°) and entropy change (∆S°) is given by the equation: \[∆G° = ∆H° - T∆S°\] Where T is the temperature in Kelvin. Spontaneity of a reaction depends on both enthalpy and entropy changes. Generally, a reaction with an increase in entropy (∆S° > 0) and a decrease in enthalpy (∆H° < 0) is spontaneous. However, there are cases where a reaction with a decrease in entropy may still be spontaneous if the decrease in enthalpy is large enough.

Analyzing given reaction in terms of reactants and products

Let's analyze the given reaction: \[2 \mathrm{Mg}(s)+\mathrm{O}_{2}(\mathrm{~g}) \longrightarrow 2 \mathrm{MgO}(s)\] In the reactant side, we have 2 moles of solid magnesium (Mg) and 1 mole of gaseous oxygen (O₂). In the product side, we have 2 moles of solid magnesium oxide (MgO).

Evaluating entropy change

From the given reaction, we can see that solid reactants are forming solid products with the conversion of gaseous oxygen to solid magnesium oxide. Generally, gas molecules have more degrees of freedom and higher entropy than solid molecules. The reaction also involves the decrease in the number of gaseous molecules from 1 mole of O₂ to none in the products. As such, the reaction should result in a decrease in entropy, i.e., a negative value for ∆S°.

Determining if there is a mistake in the calculation

Considering our analysis, the classmate's calculation of a large negative value for ∆S° seems to be correct, since the reaction leads to a decrease in entropy. The fact that the reaction is highly spontaneous despite having a negative entropy change can be explained by a sufficiently large negative value of ∆H° that can still lead to a negative ∆G°, making the reaction spontaneous at the given conditions.

Key concepts

Entropy Change

Entropy change (\( \Delta S \)) is a measure of how the disorder or randomness of a system changes during a reaction. Generally, reactions where gases form from solids or liquids lead to an increase in entropy because gases have more ways to arrange themselves.
In the given reaction, 2 magnesium atoms (\( \text{Mg} \)) and oxygen gas (\( \text{O}_2 \)) react to form magnesium oxide (\( \text{MgO} \)) which is solid. Here, a gaseous molecule (\( \text{O}_2 \)) changes to a solid (\( \text{MgO} \)). This transformation decreases the system's entropy.
- **Gases have higher entropy** because gas particles move more freely.- When a gas reacts to form a solid, \( \Delta S \) is negative.
In this reaction, the negative entropy change is expected because we're going from a state with more disorder (containing a gas) to one with less disorder (only solids). Therefore, your classmate's negative value for \( \Delta S \) is indeed correct.

Enthalpy Change

Enthalpy change (\( \Delta H \)) reflects the heat absorbed or released by a system at constant pressure. It's often used to determine whether a reaction is exothermic (releases heat, \( \Delta H < 0 \)) or endothermic (absorbs heat, \( \Delta H > 0 \)).
In the context of the given reaction, it is important to note that forming magnesium oxide from magnesium and oxygen is highly exothermic. It means that a considerable amount of energy is released to the surroundings, which is why the reaction still proceeds spontaneously even when the entropy change is negative.
- **Exothermic reactions** release energy, causing \( \Delta H \) to be negative.- The large negative \( \Delta H \) can compensate for a less favorable \( \Delta S \) in the overall spontaneity of the reaction.
Thus, the large negative enthalpy change ensures that the Gibbs free energy (\( \Delta G \)) of the system is negative, confirming the reaction's spontaneity under standard conditions.

Spontaneity of Reaction

The spontaneity of a reaction is determined by the Gibbs free energy change (\( \Delta G \)), defined by the equation:
\[\Delta G = \Delta H - T\Delta S \]
A reaction is spontaneous if \( \Delta G \) is negative. Both enthalpy and entropy changes influence this. - If \( \Delta H \) is negative and \( \Delta S \) is positive, the reaction is spontaneous.- If both \( \Delta H \) and \( \Delta S \) are negative, the reaction might still be spontaneous if the enthalpy change outweighs the entropy change.
In our case, although \( \Delta S \) for the reaction of magnesium with oxygen is negative, \( \Delta H \) is sufficiently large and negative. This results in a negative \( \Delta G \), implying the process is spontaneous at common temperatures.
Therefore, the key to understanding spontaneity lies in managing the balance between \( \Delta H \) and \( \Delta S \) and knowing that sometimes, a negative entropy change doesn’t prevent a reaction from being spontaneous as long as the enthalpy change can drive the system forward.