Chapter 6: Problem 145
For the reaction, \(\mathrm{Ag}_{2} \mathrm{O}(\mathrm{s}) \rightleftharpoons 2 \mathrm{Ag}(\mathrm{s})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{~g})\) \(\Delta \mathrm{H}, \Delta \mathrm{S}\) and \(\mathrm{T}\) are \(40.657 \mathrm{~kJ} \mathrm{~mol}^{-1}, 109 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\) and \(373 \mathrm{~K}\) respectively. Find the free energy change \((\Delta \mathrm{G})\) of the reaction.
Short Answer
Step by step solution
Understanding the Formula
Substituting Known Values
Converting Units
Calculating \( T \Delta S \)
Finding \( \Delta G \)
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Enthalpy Change
- **Endothermic reactions:** Absorb heat, resulting in a positive \( \Delta H \).
- **Exothermic reactions:** Release heat, resulting in a negative \( \Delta H \).
In our example, a positive \( \Delta H \) indicates that the decomposition of \( \mathrm{Ag}_2\mathrm{O} \) occurs by absorbing heat from its surroundings, making it endothermic.
Entropy Change
Here, the decomposition of \( \mathrm{Ag}_2\mathrm{O} \) increases entropy because a solid is converted into a more disordered gas phase. This change represents a natural tendency of systems to move towards higher entropy.
- **Positive \( \Delta S \):** Indicates an increase in disorder.
- **Negative \( \Delta S \):** Indicates a decrease in disorder.
Understanding entropy gives you insight into the feasibility of a reaction, as higher entropy often favors a reaction proceeding under given conditions.
Thermodynamics
- **First Law of Thermodynamics:** Energy is conserved; energy lost by a system is gained by its surroundings, and vice versa.
- **Second Law of Thermodynamics:** Entropy tends to increase; a system is more likely to favor states with higher entropy.
The third law establishes the impossibility of reaching absolute zero, where entropy of a pure crystalline substance is zero. With Gibbs free energy, which combines entropy and enthalpy into a single value (\( \Delta G \)), we can determine the spontaneity of a reaction:
- **If \( \Delta G < 0 \):** The reaction is spontaneous.
- **If \( \Delta G > 0 \):** The reaction is non-spontaneous.
Thus, a balance of enthalpy and entropy changes, along with temperature (373 K in our case), helps evaluate reaction feasibility using the formula: \( \Delta G = \Delta H - T\Delta S \).
Unit Conversion
Here's how to perform the necessary conversions:
- **Converting J to kJ:** Divide by 1000.
By converting \( \Delta S = 109 \text{ J K}^{-1}\text{ mol}^{-1} \) to \( 0.109 \text{ kJ K}^{-1}\text{ mol}^{-1} \), we can accurately compute Gibbs free energy using \( \Delta G = \Delta H - T\Delta S \). The same principle applies whenever dealing with unit conversions: convert all units to desired or common units before computations.