Question

From the values of \(\Delta H\) and \(\Delta S\), predict which of the following reactions would be spontaneous at \(25^{\circ} \mathrm{C}:\) reaction A: \(\Delta H=10.5 \mathrm{~kJ} / \mathrm{mol}, \Delta S=30 \mathrm{~J} / \mathrm{K} \cdot \mathrm{mol} ;\) reaction B: \(\Delta H=1.8 \mathrm{~kJ} / \mathrm{mol}, \Delta S=-113 \mathrm{~J} / \mathrm{K} \cdot \mathrm{mol}\). If either of the reactions is nonspontaneous at \(25^{\circ} \mathrm{C},\) at what temperature might it become spontaneous?

Short answer

Both reactions are non-spontaneous at 25°C; Reaction A becomes spontaneous above 350 K, while Reaction B is always non-spontaneous.

Step-by-step solution

Understand Spontaneous Reaction Conditions

For a reaction to be spontaneous, the Gibbs free energy change (\(\Delta G\)) must be negative. The equation to calculate \(\Delta G\) is given by:\[\Delta G = \Delta H - T \Delta S\]where \(T\) is the temperature in Kelvin. Spontaneity depends on the signs of \(\Delta H\) and \(\Delta S\) and the temperature.

Convert Units if Necessary

Since \(\Delta S\) is given in \(\text{J/mol}\cdot\text{K}\) and \(\Delta H\) is in \(\text{kJ/mol}\), convert \(\Delta S\) to \(\text{kJ/mol}\cdot\text{K}\) by dividing by 1000. For reaction A: \(\Delta S = 0.03\, \text{kJ/mol}\cdot\text{K}\).For reaction B: \(\Delta S = -0.113\, \text{kJ/mol}\cdot\text{K}\).

Evaluate Reaction A at 25°C

Determine if Reaction A is spontaneous at \(25^\circ\text{C}\), where \(T = 298\, \text{K}\). Calculate \(\Delta G\):\[\Delta G = 10.5\, \text{kJ/mol} - 298\, \text{K} \times 0.03\, \text{kJ/mol}\cdot\text{K} = 10.5\, \text{kJ/mol} - 8.94\, \text{kJ/mol} = 1.56\, \text{kJ/mol}\]Since \(\Delta G > 0\), Reaction A is non-spontaneous at \(25^\circ\text{C}\).

Evaluate Reaction B at 25°C

Determine if Reaction B is spontaneous at \(25^\circ\text{C}\). Use the formula for \(\Delta G\):\[\Delta G = 1.8\, \text{kJ/mol} - 298\, \text{K} \times \left(-0.113\right)\, \text{kJ/mol}\cdot\text{K} = 1.8\, \text{kJ/mol} + 33.674\, \text{kJ/mol} = 35.474\, \text{kJ/mol}\]Since \(\Delta G > 0\), Reaction B is also non-spontaneous at \(25^\circ\text{C}\).

Determine Spontaneity Condition for Reaction A

For Reaction A to become spontaneous, \(\Delta G < 0\) and thus:\[\Delta H < T \Delta S \10.5\, \text{kJ/mol} < T \times 0.03\, \text{kJ/mol}\cdot\text{K}\]Solving for \(T\):\[T > \frac{10.5}{0.03}\, \text{K} = 350\, \text{K}\]Reaction A becomes spontaneous at temperatures above \(350\, \text{K}\).

Determine Spontaneity Condition for Reaction B

Since Reaction B has both \(\Delta H > 0\) and \(\Delta S < 0\), it remains non-spontaneous at all temperatures. Therefore, Reaction B cannot become spontaneous at any temperature.

Key concepts

Spontaneous Reaction

A reaction is considered spontaneous if it can proceed on its own without requiring external energy once it has started. In thermodynamics, this means the Gibbs free energy change, represented as \(\Delta G\), must be negative. The equation linking these conditions is:

• \(\Delta G = \Delta H - T \Delta S\)
• Where \(T\) is the temperature in Kelvin, \(\Delta H\) is the change in enthalpy, and \(\Delta S\) is the change in entropy.

If \(\Delta G < 0\), the reaction is spontaneous. However, if \(\Delta G > 0\), the reaction is non-spontaneous. Whether a reaction becomes spontaneous can depend on the temperature, so it’s crucial to evaluate \(T\) in the Gibbs free energy equation.

Enthalpy

Enthalpy, denoted as \(\Delta H\), represents the total heat content of a system. It is an important factor in determining the spontaneity of a reaction.

• A negative \(\Delta H\) indicates an exothermic reaction, which releases heat to the surroundings.
• A positive \(\Delta H\) means an endothermic reaction, which absorbs heat from the surroundings.

In our examples:

• Reaction A has \(\Delta H = 10.5 \text{ kJ/mol}\), indicating it absorbs heat.
• Reaction B has \(\Delta H = 1.8 \text{ kJ/mol}\), also absorbing heat, although to a lesser extent.

Both reactions being endothermic affect their potential spontaneity, as they require heat input to proceed.

Entropy

Entropy, symbolized as \(\Delta S\), measures the disorder or randomness in a system. Increased randomness is often favorable in chemical reactions.

• A positive \(\Delta S\) suggests an increase in disorder, which generally supports spontaneity.
• A negative \(\Delta S\) denotes a decrease in disorder, which can oppose spontaneity.

For the reactions considered:

• Reaction A has \(\Delta S = 30 \text{ J/mol}\cdot\text{K}\), indicating increased randomness.
• Reaction B's \(\Delta S = -113 \text{ J/mol}\cdot\text{K}\) shows decreased randomness, making spontaneity unlikely.

Entropy changes are key in deciding whether a reaction can become spontaneous, particularly when coupled with temperature considerations.

Reaction Temperature

Temperature is vital in determining the spontaneity of a reaction, acting as a multiplier for entropy in the Gibbs free energy equation.

• In Reaction A, increasing the temperature can make \(T \Delta S\) enough to overcome a positive \(\Delta H\), making \(\Delta G\) negative.
• The calculation shows Reaction A needs a temperature above \(350 \text{ K}\) to become spontaneous.
• For Reaction B, since increasing \(T\) cannot offset the double disadvantage of a positive \(\Delta H\) and negative \(\Delta S\), it remains non-spontaneous at all temperatures.

Understanding how temperature influences Gibbs free energy helps predict at which conditions a non-spontaneous reaction might become spontaneous.

Thermodynamics

Thermodynamics provides the framework for understanding energy changes and balance within chemical reactions. It encompasses the principles of enthalpy, entropy, and temperature.

• The first law of thermodynamics covers energy conservation; energy in a closed system remains constant.
• The second law suggests that spontaneous processes increase overall entropy, aligning with the concept that reactions naturally progress toward greater randomness and lower energy states.

Using thermodynamic principles, you can analyze reactions to predict whether they'll proceed spontaneously under given conditions. By calculating and interpreting \(\Delta G\), one applies these laws to determine the energetic feasibility of reactions.