Problem 53
Question
For a certain chemical reaction, \(\Delta H^{\circ}=-35.4 \mathrm{~kJ}\) and \(\Delta S^{\circ}=-85.5 \mathrm{~J} / \mathrm{K} .(\mathrm{a}) \mathrm{ls}\) the reaction exothermic or en- dothermic? (b) Does the reaction lead to an increase or decrease in the randomness or disorder of the system? (c) Calculate \(\Delta G^{\circ}\) for the reaction at \(298 \mathrm{~K}\). (d) Is the reaction spontaneous at \(298 \mathrm{~K}\) under standard conditions?
Step-by-Step Solution
Verified Answer
(a) The reaction is exothermic since ∆H is negative. (b) The reaction leads to a decrease in the randomness or disorder of the system because ∆S is negative. (c) To calculate ∆G at 298 K, use the formula ∆G = ∆H - T∆S, which yields ∆G = -9921 J. (d) The reaction is spontaneous at 298 K under standard conditions, as ∆G is negative.
1Step 1: (a) Determine whether the reaction is exothermic or endothermic
Check the sign of ∆H. Exothermic reactions have a negative ∆H, while endothermic reactions have a positive ∆H. In this case, ∆H = -35.4 kJ, which is negative. Therefore, the reaction is exothermic.
2Step 2: (b) Determine whether the reaction increases or decreases randomness
Check the sign of ∆S. An increase in randomness or disorder has a positive ∆S, while a decrease has a negative ∆S. In this case, ∆S = -85.5 J/K, which is negative. Therefore, the reaction leads to a decrease in the randomness or disorder of the system.
3Step 3: (c) Calculate ∆G for the reaction at 298 K
To calculate ∆G, use the formula ∆G = ∆H - T∆S. First, we need to convert ∆H to J by multiplying by 1000 (since 1 kJ = 1000 J):
∆H = -35.4 kJ × 1000 = -35400 J
Now, we can calculate ∆G at 298 K:
∆G = ∆H - T∆S = -35400 J - (298 K × -85.5 J/K) = -35400 J + 25479 J = -9921 J
4Step 4: (d) Determine whether the reaction is spontaneous at 298 K
To determine whether the reaction is spontaneous at 298 K under standard conditions, check the sign of ∆G calculated in part (c). A negative ∆G indicates a spontaneous reaction, while a positive ∆G indicates a non-spontaneous reaction. In this case, ∆G = -9921 J, which is negative. Therefore, the reaction is spontaneous at 298 K under standard conditions.
Key Concepts
EnthalpyEntropySpontaneous ReactionThermodynamics
Enthalpy
Enthalpy is a crucial concept in thermodynamics that helps us understand the heat exchange in chemical reactions. It is denoted by the symbol \(\Delta H\). Enthalpy change, \(\Delta H\), tells us whether a reaction is exothermic or endothermic. Exothermic reactions release heat to the surroundings, leading to a temperature increase. These reactions have a negative \(\Delta H\) because the system loses heat. On the other hand, endothermic reactions absorb heat, causing the surroundings to cool down.These reactions have a positive \(\Delta H\).
- An exothermic reaction: \(\Delta H < 0\)
- An endothermic reaction: \(\Delta H > 0\)
Entropy
Entropy reflects the level of disorder or randomness in a system, represented by the symbol \(\Delta S\). A fundamental idea in thermodynamics, entropy determines the direction in which energy is dispersed in a system. When \(\Delta S\) is positive, it implies increased disorder as systems evolve toward chaos, a common natural process. Conversely, a negative \(\Delta S\) signals a decrease in randomness, meaning the system becomes more ordered.
- Increase in disorder: \(\Delta S > 0\)
- Decrease in disorder: \(\Delta S < 0\)
Spontaneous Reaction
The spontaneity of a reaction indicates its ability to proceed without external energy input. This nature is evaluated through Gibbs Free Energy Change (\(\Delta G\)), calculated as the difference between the system's enthalpy and the temperature-adjusted entropy change:\[\Delta G = \Delta H - T\Delta S\]A negative \(\Delta G\) suggests a spontaneous reaction, meaning the process naturally progresses on its own. A positive \(\Delta G\) means the reaction needs additional energy to occur. In our setup, at a temperature of 298 K, \(\Delta G\) was calculated as -9921 J, emphasizing the spontaneous nature of the reaction.
- Spontaneous reaction: \(\Delta G < 0\)
- Non-spontaneous reaction: \(\Delta G > 0\)
Thermodynamics
Thermodynamics is the study of energy transformations and their impact on matter, providing insight into how and why changes occur. It connects fundamental concepts like heat, work, entropy, and enthalpy, explaining the nature of reactions.In our case, by solving for \(\Delta G\), which determines spontaneity, thermodynamics showcases its guiding role in predicting reaction pathways under standard conditions.
- The First Law involves conservation of energy: energy cannot be created nor destroyed, only transformed.
- The Second Law predicts that entropy, or disorder, always increases in an isolated system.
- The concept of Gibbs Free Energy relates these laws to calculate potential work and spontaneous reaction progression.
Other exercises in this chapter
Problem 51
(a) For a process that occurs at constant temperature, express the change in Gibbs free energy in terms of changes in the enthalpy and entropy of the system. (b
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(a) What is the meaning of the standard free-energy change, \(\Delta G^{\circ}\), as compared with \(\Delta G\) ? (b) For any process that occurs at constant te
View solution Problem 54
A certain reaction has \(\Delta H^{\circ}=-19.5 \mathrm{~kJ}\) and \(\Delta S^{\circ}=\) \(+42.7 \mathrm{~J} / \mathrm{K} .\) (a) Is the reaction exothermic or
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Using data in Appendix \(C\), calculate \(\Delta H^{\circ}, \Delta S^{\circ}\), and \(\Delta G^{\circ}\) at \(298 \mathrm{~K}\) for each of the following reacti
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