Problem 46
Question
Determine whether each of the following reactions is spontaneous. a. \(\Delta H_{\text { system }}=-75.9 \mathrm{k} J, T=273 \mathrm{K}, \Delta S_{\text { system }}=138 \mathrm{J} / \mathrm{K}\) b. \(\Delta H_{\text { system }}=-27.6 \mathrm{k} \mathrm{J}, T=535 \mathrm{K}, \Delta S_{\mathrm{system}}=-55.2 \mathrm{J} / \mathrm{K}\) c. \(\Delta H_{\text { system }}=365 \mathrm{kJ}, T=388 \mathrm{K}, \Delta S_{\text { system }}=-55.2 \mathrm{J} / \mathrm{K}\) d. \(\Delta H_{\text { system }}=452 \mathrm{kJ}, T=165 \mathrm{K}, \Delta S_{\text { system }}=55.7 \mathrm{J} / \mathrm{K}\)
Step-by-Step Solution
Verified Answer
a. Reaction is spontaneous (\(\Delta G < 0\))
b. Reaction is spontaneous (\(\Delta G < 0\))
c. Reaction is non-spontaneous (\(\Delta G > 0\))
d. Reaction is non-spontaneous (\(\Delta G > 0\))
1Step 1: Calculate Gibbs free energy change for reaction a
For the first reaction, we have the following given values:
- \(\Delta H_{\text { system }} = -75.9 \mathrm{k} J\)
- \(T = 273 \mathrm{K}\)
- \(\Delta S_{\text { system }} = 138 \mathrm{J} / \mathrm{K}\)
Using the Gibbs free energy change formula, we can calculate \(\Delta G\) for reaction a.
\[ \Delta G = \Delta H - T \Delta S = -75.9 \times 10^{3} \mathrm{J} - (273 \mathrm{K})(138 \mathrm{J} / \mathrm{K}) \]
2Step 2: Calculate Gibbs free energy change for reaction b
For the second reaction, we have the following given values:
- \(\Delta H_{\text { system }} = -27.6 \mathrm{k} \mathrm{J}\)
- \(T = 535 \mathrm{K}\)
- \(\Delta S_{\text { system }} = -55.2 \mathrm{J} / \mathrm{K}\)
Using the Gibbs free energy change formula, we can calculate \(\Delta G\) for reaction b.
\[ \Delta G = \Delta H - T \Delta S = -27.6 \times 10^3 \mathrm{J} - (535 \mathrm{K})(-55.2 \mathrm{J} / \mathrm{K}) \]
3Step 3: Calculate Gibbs free energy change for reaction c
For the third reaction, we have the following given values:
- \(\Delta H_{\text { system }} = 365 \mathrm{kJ}\)
- \(T = 388 \mathrm{K}\)
- \(\Delta S_{\text { system }} = -55.2 \mathrm{J} / \mathrm{K}\)
Using the Gibbs free energy change formula, we can calculate \(\Delta G\) for reaction c.
\[ \Delta G = \Delta H - T \Delta S = 365 \times 10^3 \mathrm{J} - (388 \mathrm{K})(-55.2 \mathrm{J} / \mathrm{K}) \]
4Step 4: Calculate Gibbs free energy change for reaction d
For the fourth reaction, we have the following given values:
- \(\Delta H_{\text { system }} = 452 \mathrm{kJ}\)
- \(T = 165 \mathrm{K}\)
- \(\Delta S_{\text { system }} = 55.7 \mathrm{J} / \mathrm{K}\)
Using the Gibbs free energy change formula, we can calculate \(\Delta G\) for reaction d.
\[ \Delta G = \Delta H - T \Delta S = 452 \times 10^3 \mathrm{J} - (165 \mathrm{K})(55.7 \mathrm{J} / \mathrm{K}) \]
5Step 5: Determine spontaneity of each reaction
Calculate the values of \(\Delta G\) for each of the reactions in steps 1 through 4. Then, determine whether each reaction is spontaneous, non-spontaneous, or at equilibrium based on the sign of \(\Delta G\). If \(\Delta G<0\), the reaction is spontaneous; if \(\Delta G>0\), the reaction is non-spontaneous; if \(\Delta G=0\), the reaction is at equilibrium.
Key Concepts
Enthalpy ChangeEntropy ChangeSpontaneity of ReactionsThermodynamics
Enthalpy Change
In thermodynamics, enthalpy change is a measurement of heat absorbed or released in a system at constant pressure. Represented by the symbol \( \Delta H \), it helps us understand whether a chemical reaction is endothermic (absorbing heat) or exothermic (releasing heat).
Typically, when \( \Delta H \) is negative, the reaction is exothermic, meaning heat is released as bonds are formed. This often indicates a potential for spontaneity, depending on other factors.
Typically, when \( \Delta H \) is negative, the reaction is exothermic, meaning heat is released as bonds are formed. This often indicates a potential for spontaneity, depending on other factors.
- Exothermic reactions: \( \Delta H < 0 \)
- Endothermic reactions: \( \Delta H > 0 \)
Entropy Change
Entropy is a measure of disorder or randomness in a system. In a chemical reaction, the entropy change, denoted as \( \Delta S \), indicates how the disorder changes from reactants to products.
When \( \Delta S \) is positive, the system becomes more disordered, which often favors spontaneity, especially at higher temperatures. Conversely, a negative \( \Delta S \) implies the system becomes more ordered.
When \( \Delta S \) is positive, the system becomes more disordered, which often favors spontaneity, especially at higher temperatures. Conversely, a negative \( \Delta S \) implies the system becomes more ordered.
- Increasing disorder: \( \Delta S > 0 \)
- Decreasing disorder: \( \Delta S < 0 \)
Spontaneity of Reactions
To predict whether a chemical reaction is spontaneous, we rely on the Gibbs free energy change (\( \Delta G \)). The formula \( \Delta G = \Delta H - T\Delta S \) helps determine spontaneity by combining enthalpy, temperature, and entropy.
If \( \Delta G \) is negative, the reaction occurs spontaneously. This means at the given temperature, the system's free energy decreases, favoring product formation.
If \( \Delta G \) is negative, the reaction occurs spontaneously. This means at the given temperature, the system's free energy decreases, favoring product formation.
- Spontaneous reaction: \( \Delta G < 0 \)
- Non-spontaneous reaction: \( \Delta G > 0 \)
- Equilibrium: \( \Delta G = 0 \)
Thermodynamics
Thermodynamics is the branch of physics dealing with energy and its transformations. It's essential in evaluating chemical reactions for energy changes, particularly focusing on enthalpy and entropy.
- The first law of thermodynamics, the principle of conservation of energy, states that energy cannot be created or destroyed, only transformed.
- The second law establishes that entropy in an isolated system always increases over time.
Other exercises in this chapter
Problem 44
Predict the sign of \(\Delta S_{\text { system for each of the following changes. }}\) a. \(C I F(g)+F_{2}(g) \rightarrow C | F_{3}(g)\) c. \(C H_{3} O H(1) \ri
View solution Problem 45
Challenge Comment on the sign of \(\Delta S_{\text { system for the following reaction. }}\) \({Fe}({s})+{Zn}^{2+}({aq}) \rightarrow {Fe}^{2+}({aq})+{Zn}({s})\)
View solution Problem 48
Compare and contrast spontaneous and nonspontaneous reactions.
View solution Problem 49
Describe how a system’s entropy changes if the system becomes more disordered during a process.
View solution