Problem 69
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?
Step-by-Step Solution
Verified Answer
The reaction will have \(\Delta G=0\) at approximately 326.53 K. If the temperature is increased from 326.53 K, the reaction will be non-spontaneous.
1Step 1: Write down the given values and equation
We are given:
\(\Delta H = -32 \mathrm{~kJ} = -32000 \mathrm{~J}\)
\(\Delta S = -98 \mathrm{~J/K}\)
We are going to use the Gibbs-Helmholtz equation:
\(\Delta G = \Delta H - T\Delta S\)
2Step 2: Calculate the temperature for \(\Delta G=0\)
Set \(\Delta G = 0\) in the equation and solve for temperature:
\(0 = -32000\mathrm{~J} - T(-98 \mathrm{~J/K})\)
Rearrange the equation to isolate T:
\(T = \frac{-32000 \mathrm{~J}}{-98 \mathrm{~J/K}}\)
3Step 3: Solve for temperature
Now, we'll simplify the equation:
\(T = \frac{32000}{98} \approx 326.53\mathrm{~K}\)
So the reaction will have \(\Delta G=0\) at approximately 326.53 K.
4Step 4: Determine spontaneity when temperature is increased
We know that when the reaction is spontaneous, \(\Delta G < 0\). If the temperature is increased, we can analyze the Gibbs-Helmholtz equation to determine the effect on \(\Delta G\):
\(\Delta G = \Delta H - T\Delta S\)
Given \(\Delta H < 0\) and \(\Delta S < 0\), increasing the temperature makes the term \(T\Delta S\) less negative (since a negative multiplied by a positive is negative, and increasing the positive value decreases the absolute value of the result). Since both terms have the same sign (negative), decreasing the absolute value of \(T\Delta S\) will result in a less negative \(\Delta G\). As \(\Delta G\) approaches zero or becomes positive, the reaction becomes non-spontaneous.
Therefore, if the temperature is increased from 326.53 K, the reaction will be non-spontaneous.
Key Concepts
EnthalpyEntropyReaction Spontaneity
Enthalpy
Enthalpy is a measure of the total energy in a thermodynamic system. It combines the system's internal energy with the work required to make room for it in the environment, often visualized as the energy needed to maintain pressure and volume. Here are some key points about enthalpy:
- Enthalpy, denoted as \( \Delta H \), reflects the heat change at constant pressure.
- If \( \Delta H \) is negative, the process releases heat, known as an exothermic reaction. This means the system loses energy to its surroundings, like in our exercise where \( \Delta H = -32 \mathrm{~kJ} \).
- Conversely, if \( \Delta H \) is positive, heat is absorbed from the surroundings, marking an endothermic process.
Entropy
Entropy is a measure of randomness or disorder in a system. The second law of thermodynamics tells us that entropy tends to increase in isolated systems. Here’s what you need to know about entropy:
- Entropy is denoted by \( \Delta S \). In our exercise, \( \Delta S = -98 \mathrm{~J/K} \), indicating a reduction in disorder during the reaction.
- A positive \( \Delta S \) suggests increased disorder or chaos, while a negative \( \Delta S \) implies a more ordered system.
- Entropy helps predict the direction of spontaneous processes: more disorder naturally occurs, so processes that increase entropy are more likely to be spontaneous.
Reaction Spontaneity
Reaction spontaneity describes whether a chemical reaction will proceed on its own without any input of external energy. This is where Gibbs Free Energy (\( \Delta G \)) comes into play:
- Gibbs Free Energy combines enthalpy, temperature, and entropy effects to predict the spontaneity of a reaction.
- If \( \Delta G < 0 \), the reaction is spontaneous, occurring without added energy input.
- If \( \Delta G > 0 \), the reaction is non-spontaneous, requiring external energy to proceed.
- At \( \Delta G = 0 \), the system is in equilibrium, meaning no net change happens over time, as seen with the reaction temperature at about 326.53 K in our exercise before increasing the temperature even more.
Other exercises in this chapter
Problem 64
Sulfur dioxide reacts with strontium oxide as follows: $$ \mathrm{SO}_{2}(g)+\mathrm{SrO}(g) \longrightarrow \mathrm{SrSO}_{3}(s) $$ (a) Without using thermoche
View solution Problem 66
From the values given for \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\), calculate \(\Delta G^{*}\) for each of the following reactions at \(298 \mathrm{~K}\).
View solution Problem 74
(a) Using data in Appendix C, estimate the temperature at which the free- energy change for the transformation from \(I_{2}(s)\) to \(I_{2}(g)\) is zero. What a
View solution Problem 77
Indicate whether \(\Delta G\) increases, decreases, or stays the same for each of the following reactions as the partial pressure of \(\mathrm{O}_{2}\) is incre
View solution