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

Short answer

At \(25^{\circ} C\), both the reactions A and B are non-spontaneous. Reaction A becomes spontaneous at temperatures greater than roughly \(76.85^{\circ} C\). Reaction B will remain non-spontaneous at all temperatures due to the negative value of entropy.

Step-by-step solution

Convert to correct units

Before starting, ensure all the values are in the correct units for calculation. Here, \(\Delta H\) is given in kJ/mol, which is perfect, but \(\Delta S\) is in J/K.mol. The latter needs to be converted into kJ/K.mol by dividing by 1000. For Reaction A: \(\Delta S_{A} = \dfrac{30}{1000} = 0.03 \, kJ/K \cdot mol \), and for Reaction B: \(\Delta S_{B} = \dfrac{-113}{1000} = -0.113 \, kJ/K \cdot mol \).

Calculate \(\Delta G\) at \(25^{\circ} \mathrm{C}\) for both reactions

Use the formula \(\Delta G = \Delta H – T\Delta S\) , where \(T\) is the temperature in Kelvin. Hence, first convert the temperature to Kelvin: \(T = 25^{\circ} C + 273.15 = 298.15 \, K\). Now calculate \(\Delta G\) for both reactions A and B:For Reaction A: \(\Delta G_{A} = \Delta H_{A} − T\Delta S_{A} = 10.5 - 298.15 \times 0.03 \approx 1.6 \, kJ/mol\).For Reaction B: \(\Delta G_{B} = \Delta H_{B} – T\Delta S_{B} = 1.8 - 298.15 \times -0.113 \approx 35.8 \, kJ/mol\).

Determine Spontaneity

A reaction is spontaneous if \(\Delta G < 0\), non-spontaneous if \(\Delta G > 0\), and at equilibrium if \(\Delta G = 0\). From Step 2, it's clear that both Reaction A and B are not spontaneous at \(25^{\circ} C\) since \(\Delta G_{A} > 0 \) and \(\Delta G_{B} > 0 \).

Calculate the temperatures for spontaneous reaction

For a reaction to be spontaneous, \(\Delta G\) must be less than 0. Hence, set \(\Delta G = 0\), and solve the equation \(\Delta H – T\Delta S = 0\) for \(T\), i.e., \(T = \dfrac{\Delta H}{\Delta S}\).For Reaction A: \(T_{A} = \dfrac{10.5}{0.03} \approx 350 \, K \,or\, 76.85^{\circ} C\).For Reaction B: Since \(\Delta S_{B}\) is negative , Reaction B will never be spontaneous.

Key concepts

Spontaneous Reactions

In chemistry, a spontaneous reaction is one that occurs on its own without needing any external energy input. It doesn't mean the reaction happens quickly; it just means it naturally tends to happen. The key to understanding spontaneity lies in Gibbs Free Energy ( G), which is calculated as G = H - T S, where H is the change in enthalpy, S is the change in entropy, and T is the temperature in Kelvin.

• If G < 0, the reaction is spontaneous.
• If G > 0, the reaction is non-spontaneous.
• If G = 0, the system is in equilibrium.

From our calculations in the exercise, both reactions were non-spontaneous at 25°C as their G values were greater than zero. This suggests that at this temperature, no reaction naturally favors proceeding without extra energy.

Enthalpy

Enthalpy, represented by the symbol H, refers to the total heat content of a system. It includes the internal energy plus the product of pressure and volume. In chemical reactions, the change in enthalpy ( H) indicates whether heat is absorbed or released.

• A negative H ( H < 0) means the reaction is exothermic, releasing heat to the surroundings.
• A positive H ( H > 0) means the reaction is endothermic, absorbing heat from the surroundings.

In our exercise, both reactions had positive H values, showing they are endothermic. Reaction A had a H of 10.5 kJ/mol and Reaction B had a H of 1.8 kJ/mol, which means both need heat for them to proceed.

Entropy

Entropy, symbolized as S, measures the disorder or randomness in a system. The second law of thermodynamics states that any spontaneous process will lead to an increase in the total entropy of the universe.

• An increase in entropy ( S > 0) signifies a system becoming more disordered, often favorable for spontaneity.
• A decrease in entropy ( S < 0) indicates a system becoming more ordered, which may hinder spontaneity.

In the exercise, Reaction A exhibited positive entropy ( S = 0.03 kJ/K⋅mol), suggesting increased disorder. However, reaction B had negative entropy ( S = -0.113 kJ/K⋅mol), indicating increased order. This negative entropy makes it challenging for reaction B to become spontaneous because it naturally wants to resist further disorder.