Question

Given the values of \(\Delta H\) and \(\Delta S,\) which of the following changes will be spontaneous at constant \(T\) and \(P ?\) a. \(\Delta H=+25 \mathrm{kJ}, \Delta S=+5.0 \mathrm{J} / \mathrm{K}, T=300 . \mathrm{K}\) b. \(\Delta H=+25 \mathrm{kJ}, \Delta S=+100 . \mathrm{J} / \mathrm{K}, T=300 . \mathrm{K}\) c. \(\Delta H=-10 . \mathrm{kJ}, \Delta S=+5.0 \mathrm{J} / \mathrm{K}, T=298 \mathrm{K}\) d. \(\Delta H=-10 . \mathrm{kJ}, \Delta S=-40 . \mathrm{J} / \mathrm{K}, T=200 . \mathrm{K}\)

Short answer

Options b, c, and d represent spontaneous reactions at constant T and P, with ∆G values of -5000 J, -11490 J, and -2000 J, respectively.

Step-by-step solution

Recall the Gibbs free energy equation

Recall the formula for Gibbs free energy change: ∆G = ∆H - T∆S, where ∆G is Gibbs free energy change, ∆H is enthalpy change, T is temperature in Kelvin, and ∆S is entropy change. Step 2: Calculate ∆G for each option

Calculate ∆G for each option

Plug the given values for ∆H, ∆S, and T into the formula and calculate ∆G for each of the four cases. a) ∆H = +25 kJ, ∆S = +5.0 J/K, T = 300 K ∆G = (25*1000) - 300*(5) = 25000 - 1500 = 23500 J b) ∆H = +25 kJ, ∆S = +100 J/K, T = 300 K ∆G = (25*1000) - 300*(100) = 25000 - 30000 = -5000 J c) ∆H = -10 kJ, ∆S = +5.0 J/K, T = 298 K ∆G = (-10*1000) - 298*(5) = -10000 - 1490 = -11490 J d) ∆H = -10 kJ, ∆S = -40 J/K, T = 200 K ∆G = (-10*1000) - 200*(-40) = -10000 + 8000 = -2000 J Step 3: Determine if the reaction is spontaneous

Determine if the reaction is spontaneous

Analyze the calculated ∆G values to determine if the reaction is spontaneous or not. If ∆G is negative, the reaction is spontaneous. a) ∆G = 23500 J - The reaction is NOT spontaneous. b) ∆G = -5000 J - The reaction IS spontaneous. c) ∆G = -11490 J - The reaction IS spontaneous. d) ∆G = -2000 J - The reaction IS spontaneous. As a result, options b, c, and d represent spontaneous reactions at constant T and P.

Key concepts

Enthalpy Change

Enthalpy change, represented by \(\Delta H\), is the measure of heat absorbed or released during a chemical reaction at constant pressure. It helps us understand whether a reaction is endothermic or exothermic. - **Endothermic Reactions**: These reactions absorb heat from the surroundings, leading to a positive \(\Delta H\). For example, in option a and b of the exercise, \(\Delta H\) is +25 kJ, indicating an endothermic reaction.- **Exothermic Reactions**: These reactions release heat to the surroundings, resulting in a negative \(\Delta H\). In options c and d of the exercise, \(\Delta H\) is -10 kJ, indicating an exothermic reaction.Understanding enthalpy change is crucial because it gives insight into the energy dynamics of a reaction, guiding predictions about the spontaneity and feasibility of reactions under specific conditions.

Entropy Change

Entropy change, denoted \(\Delta S\), reflects the degree of disorder or randomness in a system during a chemical reaction. It is a crucial part of predicting reaction spontaneity because it shows how the system's order changes. - **Positive Entropy Change**: When \(\Delta S\) is positive, the disorder increases. This is observed in the exercise's options a, b, and c. For example, in option b, \(\Delta S\) is +100 J/K, signifying increased disorder, which often favors spontaneity, especially at higher temperatures.- **Negative Entropy Change**: Here, \(\Delta S\) is negative, indicating decreased disorder. Option d in the exercise illustrates this, with a \(\Delta S\) of -40 J/K. Decreased disorder might make a reaction non-spontaneous unless other factors like negative \(\Delta H\) contribute to spontaneity.Entropy change helps us see the bigger picture of a reaction, allowing us to assess spontaneous potential when paired with enthalpy change and temperature.

Spontaneity of Reactions

The spontaneity of a chemical reaction is determined by the Gibbs free energy change, \(\Delta G\). A reaction is considered spontaneous if \(\Delta G < 0\). We calculate \(\Delta G\) through the equation:\[ \Delta G = \Delta H - T \Delta S \]- **Calculating \(\Delta G\)**: We take the given \(\Delta H\), \(\Delta S\), and temperature \(T\) and plug them into the formula. For example, in the exercise, option b gives \(\Delta G = -5000 \text{ J}\), indicating a spontaneous reaction.- **Temperature's Role**: Since \(\Delta S\) is temperature-dependent, temperature can influence spontaneity. For instance, in option a, even though \(\Delta S\) is positive, the temperature factor results in a positive \(\Delta G\), making the reaction non-spontaneous.Understanding spontaneity helps us predict whether a reaction will occur without external intervention, making \(\Delta G\) an essential tool for chemists.