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 \(\Delta G=0\) ? (b) If \(T\) is increased from that in part (a), will the reaction be spontaneous or nonspontaneous?

Short answer

(a) The temperature at which the reaction will have \(\Delta G = 0\) is approximately \(T \approx 326.53\,\text{K}\). (b) As the temperature increases from \(T \approx 326.53\,\text{K}\), the reaction will become nonspontaneous.

Step-by-step solution

Set up the equation for the Gibbs free energy change

We have the equation for Gibbs free energy change: \[ \Delta G = \Delta H - T\Delta S \] We're given \(\Delta H = -32\,\text{kJ}\) and \(\Delta S = -98\,\text{J/K}\). Note that we must convert \(\Delta H\) to J so that the units are consistent in the equation: \(-32\,\text{kJ} = -32,000\,\text{J}\). Now we can write the equation with the given values: \[ \Delta G = -32,000\,\text{J} - T(-98\,\text{J/K}) \]

Solve for the temperature at which \(\Delta G = 0\)

We want to find the temperature \(T\) at which the free energy change \(\Delta G = 0\). Thus, we can set the equation to zero and solve for \(T\): \[ 0 = -32,000\,\text{J} + 98\,\text{J/K} \cdot T \] Now, isolate the temperature on one side of the equation: \[ T = \frac{32{,}000\,\text{J}}{98\,\text{J/K}} \] Divide to find the temperature: \[ T \approx 326.53\,\text{K} \]

Determine if the reaction is spontaneous or nonspontaneous when the temperature increases

To find if the reaction is spontaneous or nonspontaneous, we must first look at the sign of \(\Delta G\) as the temperature increases. Recall that for this problem, both \(\Delta H\) and \(\Delta S\) are negative. When the temperature increases from \(T \approx 326.53\,\text{K}\), the magnitude of the term \(T\Delta S\) will also increase, because the product of two negative numbers will be positive. Since \(\Delta H\) is already negative, adding more positive values (increasing the magnitude of \(T\Delta S\)) will make \(\Delta G\) more positive. When \(\Delta G\) is positive, the reaction is nonspontaneous. In summary, as the temperature increases from \(T \approx 326.53\,\text{K}\), the reaction will become nonspontaneous.

Key concepts

Understanding Thermodynamics

Thermodynamics is a branch of physics that studies how energy moves and changes in the form of heat and work. It's essential for understanding chemical reactions and how they occur. One of the primary ideas in thermodynamics is the concept of energy conservation, which states that energy can neither be created nor destroyed.
In chemical reactions, we often deal with changes in energy states. This is where Gibbs Free Energy, \( \Delta G \), comes into play. \( \Delta G \) helps us determine if a reaction occurs spontaneously. A negative value for \( \Delta G \) implies the reaction will happen on its own, while a positive value suggests it won't.
In our example, the relationship \( \Delta G = \Delta H - T\Delta S \) links enthalpy (\( \Delta H \)), entropy (\( \Delta S \)), and temperature (\( T \)). Each of these components are critical in predicting the direction of a reaction.

• Enthalpy (\( \Delta H \)): Measures heat absorbed or released.
• Entropy (\( \Delta S \)): Reflects disorder or randomness.
• Temperature (\( T \)): Affects energy distribution.

Understanding these components helps clarify why some reactions are spontaneous and others are not.

Exploring Enthalpy

Enthalpy, denoted as \( \Delta H \), is a measure of the total heat content of a system. In simple terms, it's the energy stored within the chemical bonds of a substance. During a reaction, enthalpy indicates whether heat is absorbed or released to the environment.
Reactions can either be exothermic or endothermic:
\( \bullet \) **Exothermic**: The reaction releases heat, making \( \Delta H \) negative. This means energy is leaving the system.
\( \bullet \) **Endothermic**: The reaction absorbs heat, resulting in a positive \( \Delta H \). Energy enters the system.
In our exercise, \( \Delta H = -32 \,\text{kJ} \) reveals an exothermic process. Understanding whether a reaction is exothermic or endothermic helps predict its behavior under different conditions, especially temperature changes.
Temperature plays a vital role, as evident in the Gibbs Free Energy equation. As the temperature increases, it can alter the spontaneity of the reaction by changing the balance between \( \Delta H \) and the \( T\Delta S \) term.

Understanding Entropy

Entropy, represented as \( \Delta S \), measures the disorder or randomness within a system. In thermodynamics, it reflects how energy is dispersed among particles. Generally, the greater the entropy, the more disordered the system is.
\( \bullet \) **Positive \( \Delta S \):** Represents an increase in disorder (more chaos).
\( \bullet \) **Negative \( \Delta S \):** Indicates a decrease in disorder (less chaos).
In our reaction, \( \Delta S = -98 \,\text{J/K} \) shows that as the reaction proceeds, the system becomes more ordered. This decrease in entropy suggests more significant organization in the products compared to the reactants.
Entropy is crucial in determining the favorability of a reaction, particularly in the Gibbs Free Energy context. As temperature rises, the term \( T\Delta S \) increases, significantly affecting whether a reaction remains spontaneous or not. Understanding entropy helps us anticipate how a reaction may behave under different temperature conditions, emphasizing its role in the equation affecting spontaneity.