Chapter 17: Problem 94
For Exercises 17.87 to 17.102 , assume that \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) do not change with temperature. Suppose you have an endothermic reaction with \(\Delta H^{\circ}=+15 \mathrm{~kJ}\) and a \(\Delta S^{\circ}\) of \(-150 \mathrm{~J} / \mathrm{K} .\) Calculate \(\Delta G^{\circ}\) and \(K_{\mathrm{eq}}\) at 10,100 , and \(1000 \mathrm{~K}\).
Short Answer
Step by step solution
Understanding Endothermic Reactions
Calculating \( \Delta G^{\circ} \)
Compute \( \Delta G^{\circ} \) at \( 10 \mathrm{~K} \)
Compute \( \Delta G^{\circ} \) at \( 100 \mathrm{~K} \)
Compute \( \Delta G^{\circ} \) at \( 1000 \mathrm{~K} \)
Calculating \( K_{\mathrm{eq}} \)
Calculate \( K_{\mathrm{eq}} \) at \( 10 \mathrm{~K} \)
Calculate \( K_{\mathrm{eq}} \) at \( 100 \mathrm{~K} \)
Calculate \( K_{\mathrm{eq}} \) at \( 1000 \mathrm{~K} \)
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Gibbs Free Energy
- \( \Delta G^{\circ} = \Delta H^{\circ} - T \Delta S^{\circ} \)
- If \( \Delta G^{\circ} < 0 \), the reaction is spontaneous.
- If \( \Delta G^{\circ} > 0 \), the reaction is non-spontaneous.
Endothermic Reactions
In the context of the exercise, with \( \Delta H^{\circ} = +15 \) kJ, the reaction absorbs 15,000 J of energy. Additionally, the entropy change \( \Delta S^{\circ} \) is negative \((-150 \, \text{J/K})\), signifying a decrease in disorder or randomness in the system during the reaction.
Endothermic reactions can often be observed in everyday processes such as:
- Melting of ice into water
- Evaporation of water
- Photosynthesis in plants
Equilibrium Constant
- A high \( K_{\text{eq}} \) indicates the equilibrium position lies towards the products, meaning the reaction proceeds almost to completion.
- A low \( K_{\text{eq}} \) suggests the equilibrium position favors the reactants.
- \( \Delta G^{\circ} = -RT \ln K_{\text{eq}} \)
- \( K_{\text{eq}} = e^{-\frac{\Delta G^{\circ}}{RT}} \)
- \( R = 8.314 \, \text{J/mol K} \) is the universal gas constant
- \( T \) is the temperature in Kelvin