Question

Consider the following reaction at \(298 \mathrm{~K}:\) $$ 2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(l) \quad \Delta H^{\circ}=-571.6 \mathrm{~kJ} / \mathrm{mol} $$ Calculate \(\Delta S_{\mathrm{svs}}, \Delta S_{\text {surr }}\), and \(\Delta S_{\text {univ }}\) for the reaction.

Short answer

\( \Delta S_{sys} = -327 \ \text{J/mol K}, \Delta S_{surr} = 1918 \ \text{J/mol K}, \Delta S_{univ} = 1591 \ \text{J/mol K} \).

Step-by-step solution

Calculate the Entropy Change of the System

To find the entropy change of the system \( \Delta S_{sys} \), we need to look up standard molar entropy values for the reactants and products from a reliable source (e.g., a chemistry handbook or database). Then use the equation \( \Delta S_{sys} = \sum S_{products} - \sum S_{reactants} \).Suppose the entropies are as follows: \( S_{H_2} = 131 \text{ J/mol K} \), \( S_{O_2} = 205 \text{ J/mol K} \), and \( S_{H_2O(l)} = 70 \text{ J/mol K} \).The equation for \( \Delta S_{sys} \) becomes:\[ \Delta S_{sys} = [2 \times 70] - [(2 \times 131) + (1 \times 205)] \ \text{J/mol K}\]Calculating this gives:\[ \Delta S_{sys} = 140 - 467 = -327 \ \text{J/mol K} \]

Calculate the Entropy Change of the Surroundings

The entropy change of the surroundings \( \Delta S_{surr} \) is associated with the heat exchanged with the environment. Given \( \Delta H^{\circ} = -571.6 \ \text{kJ/mol} \), use the relation:\[ \Delta S_{surr} = - \frac{\Delta H^{\circ}}{T} \] \[ \] where \( T \) is the temperature in Kelvin.For this reaction at \( 298 \ \text{K} \):\[ \Delta S_{surr} = - \frac{-571.6 \times 10^3 \ \text{J/mol}}{298} \approx 1918 \ \text{J/mol K} \]

Calculate the Entropy Change of the Universe

The entropy change of the universe \( \Delta S_{univ} \) is the sum of the system and the surroundings. Use the equation:\[ \Delta S_{univ} = \Delta S_{sys} + \Delta S_{surr} \].Substitute the calculated values:\[ \Delta S_{univ} = -327 + 1918 = 1591 \ \text{J/mol K} \]

Interpret the Results

The calculated \( \Delta S_{univ} \) is positive, which is consistent with the second law of thermodynamics. This suggests that the reaction is spontaneous under the given conditions.

Key concepts

Second Law of Thermodynamics

The Second Law of Thermodynamics helps us understand why certain processes occur or do not occur naturally. It states that the total entropy of an isolated system will always tend to increase over time or remain unchanged for a reversible process. This can be summarized as:

• The universe's entropy either increases or stays consistent, never decreases.
• If a process increases the total entropy, it is likely to be spontaneous.

In the original exercise, the calculation of the total entropy change of the universe (\( \Delta S_{univ} \)) was positive, suggesting the reaction is indeed spontaneous. This aligns perfectly with the second law, reinforcing the idea that reactions aligning with a net increase in entropy are more natural. This is how the universe prefers to "lean" towards disorder or randomness. This makes understanding entropy critical to predicting the feasibility of reactions.

Spontaneous Reactions

Spontaneous reactions are processes that occur without any input of external energy once started. These reactions move the system from a state of lower entropy to higher entropy, making them compatible with the Second Law of Thermodynamics.

• Spontaneity does not indicate speed. Some spontaneous reactions can take a long time.
• A reaction being spontaneous implies it is thermodynamically favorable.

In the exercise, the positive change in the entropy of the universe (\( \Delta S_{univ} \)) indicates that the reaction runs spontaneously. This spontaneous nature is often driven by an overall increase in the disorder of the surroundings or system, as observed here.

Enthalpy

Enthalpy (\( \Delta H \)) is another key factor in determining reaction spontaneity and involves the heat absorbed or released during a reaction at constant pressure. It aids in understanding the heat exchange aspect of a chemical reaction.

• A negative enthalpy value means the reaction is exothermic, releasing heat.
• A positive value indicates an endothermic reaction that absorbs heat.

The enthalpy change for the given reaction in the exercise is negative (\( \Delta H^{\circ} = -571.6 \ \text{kJ/mol} \)), indicating it releases heat to its environment. Exothermic reactions often align with spontaneous processes, adding to the favorability of the reaction being spontaneous in this scenario.

Gibbs Free Energy

Gibbs Free Energy (\( \Delta G \)) helps combine both entropy and enthalpy to predict if a reaction will be spontaneous. It is expressed by:\[\Delta G = \Delta H - T \Delta S\]where:

• \( \Delta G < 0 \): Reaction is spontaneous.
• \( \Delta G = 0 \): Reaction is at equilibrium.
• \( \Delta G > 0 \): Reaction is non-spontaneous.

With enthalpy and entropy values from the exercise, \( \Delta G \) can confirm the spontaneous nature without just relying on individual entropy or enthalpy values. Essentially, Gibbs Free Energy offers a complete picture to assess reaction spontaneity, providing a holistic view in scenarios of chemical reactions.