Question

For a particular reaction, \(\Delta H=-32 \mathrm{~kJ}\) and \(\Delta S=-98 \mathrm{~J} / \mathrm{K}\). Assume that \(\Delta H\) and \(\Delta S\) do not vary with temperature. (a) At what temperature will the reaction have \(\Delta G=0 ?(\mathbf{b})\) If \(T\) is increased from that in part (a), will the reaction be spontaneous or nonspontaneous?

Short answer

The temperature at which ΔG equals 0 for this reaction is approximately 326.53 K. If the temperature is increased from 326.53 K, the reaction will be non-spontaneous since it is exothermic and has a decrease in entropy.

Step-by-step solution

(Step 1: Calculate the temperature for ΔG = 0)

To find the temperature at which ΔG equals 0, we need to set the Gibbs free energy equation to zero and solve for T. 0 = -32 kJ - T(-98 J/K)

(Step 2: Convert units to match)

Since our ΔH is in kJ and ΔS is in J/K, we need to make sure that all of our units are consistent before moving forward. Our goal is to express ΔH in J: -32 kJ = -32,000 J 0 = -32,000 J - T(-98 J/K)

(Step 3: Solve for T)

Now, we can solve for the temperature T. 0 = -32,000 J + 98 J/K * T 32,000 J = 98 J/K * T T = 32,000 J / (98 J/K) = 326.53 K Thus, the temperature at which ΔG equals 0 is approximately 326.53 K.

(Step 4: Determine spontaneity of the reaction for increased T)

With temperature T = 326.53 K, we know ΔG = 0. When the temperature is increased, we can analyze the signs of ΔH and ΔS to predict whether the reaction will be spontaneous or non-spontaneous. ΔH = -32 kJ (exothermic reaction) ΔS = -98 J/K (decrease in entropy) For an exothermic reaction with a decrease in entropy, when the temperature increases, the reaction will become less spontaneous because the TΔS term in the Gibbs free energy equation will become more positive, causing ΔG to become more positive. Therefore, if the temperature is increased from 326.53 K, the reaction will be non-spontaneous.

Key concepts

Enthalpy

Enthalpy, represented as \( \Delta H \), is a measure of the total heat content of a system. It reflects the energy required to create the system and is often associated with chemical reactions as the amount of heat absorbed or released.

• In our problem, \( \Delta H \) is given as \(-32 \mathrm{~kJ} \), indicating an exothermic reaction.
• An exothermic reaction releases heat to its surroundings, which is why relative to the system, \( \Delta H \) has a negative value.

Understanding this helps you predict the behavior of chemical processes, particularly in relation to temperature changes. In the context of Gibbs Free Energy, \( \Delta H \) plays a crucial role as it can decide whether a reaction absorbs or releases heat, impacting its spontaneity.

Entropy

Entropy, denoted as \( \Delta S \), is a measure of the disorder or randomness in a system. It describes how energy is spread out in a process and affects the feasibility and direction of a reaction.

• In this case, \( \Delta S \) is \(-98 \mathrm{~J/K} \), meaning the system becomes more ordered as the reaction proceeds.
• A negative entropy change implies that the products of the reaction possess a more orderly arrangement than the reactants.

Entropy change directly influences Gibbs Free Energy. Since \( \Delta S \) is negative here, as temperature increases, the \( T\Delta S \) term becomes more negative, eventually making Gibbs Free Energy positive. Therefore, the reaction tends to be less spontaneous when entropy decreases.

Reaction Spontaneity

Reaction spontaneity tells us whether a reaction will proceed on its own without external input. The spontaneity of a reaction is determined by the sign of its Gibbs Free Energy change \( \Delta G \).

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

From the exercise, we calculate the temperature at which \( \Delta G = 0 \) using the relation \( \Delta G = \Delta H - T\Delta S \). At 326.53 K, \( \Delta G = 0 \), indicating the system is at equilibrium. As the temperature rises, the effect of \( -T \Delta S \) grows larger because \( \Delta S \) is negative. This makes \( \Delta G \) positive, turning the reaction non-spontaneous, as energy in the form of heat is absorbed to maintain randomness.