To determine the temperature at which a reaction becomes spontaneous, we use the Gibbs free energy equation:

$\Delta G = \Delta H - T \Delta S$

A reaction becomes spontaneous when $\Delta G < 0$. Therefore, for spontaneity:

$\Delta H - T \Delta S < 0 \quad \text{or} \quad T > \frac{\Delta H}{\Delta S}$

Where:

• $\Delta H$ is the enthalpy change of the reaction.
• $\Delta S$ is the entropy change of the reaction.
• $T$ is the temperature at which the reaction becomes spontaneous (in Kelvin).

Now, for the reaction:

$\text{CaO(s)} + \text{H}_2\text{O(l)} \rightarrow \text{Ca(OH)}_2\text{(s)}$

We need the values of $\Delta H$ and $\Delta S$, which are typically found in thermodynamic tables.

Step 1: Gathering Thermodynamic Data

• Standard enthalpy of formation ($\Delta H_f^\circ$) and entropy ($S^\circ$) for each substance at standard conditions (298 K):
  • CaO(s):
    • $\Delta H_f^\circ = -635.1 \, \text{kJ/mol}$
    • $S^\circ = 39.8 \, \text{J/(mol·K)}$
  • H2O(l):
    • $\Delta H_f^\circ = -285.8 \, \text{kJ/mol}$
    • $S^\circ = 69.9 \, \text{J/(mol·K)}$
  • Ca(OH)2(s):
    • $\Delta H_f^\circ = -986.1 \, \text{kJ/mol}$
    • $S^\circ = 83.4 \, \text{J/(mol·K)}$

Step 2: Calculate $\Delta H$ and $\Delta S$

$\Delta H = \sum \Delta H_f^\circ \, \text{(products)} - \sum \Delta H_f^\circ \, \text{(reactants)}$

$\Delta S = \sum S^\circ \, \text{(products)} - \sum S^\circ \, \text{(reactants)}$

Enthalpy change ($\Delta H$):

[ \Delta H = [-986.1 , \text{kJ/mol}] - [-635.1 , \text{kJ/mol} + (-285.8 , \text{kJ/mol})] ]

$\Delta H = -986.1 - (-920.9) = -65.2 \, \text{kJ/mol}$

Entropy change ($\Delta S$):

[ \Delta S = [83.4 , \text{J/(mol·K)}] - [39.8 , \text{J/(mol·K)} + 69.9 , \text{J/(mol·K)}] ]

$\Delta S = 83.4 - 109.7 = -26.3 \, \text{J/(mol·K)}$

Step 3: Calculate the temperature at which the reaction becomes spontaneous

To find the temperature where the reaction becomes spontaneous, we set $\Delta G = 0$:

$T = \frac{\Delta H}{\Delta S}$

Convert $\Delta H$ to J/mol:

$\Delta H = -65.2 \, \text{kJ/mol} = -65200 \, \text{J/mol}$

Now, calculate the temperature:

$T = \frac{-65200}{-26.3} \, \text{K}$

$T = 2479 \, \text{K}$

Conclusion:

The reaction becomes spontaneous at a temperature above 2479 K.

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Would you like further details on this or have any other questions?

Here are 5 related questions to explore:

1. What is the significance of $\Delta G = 0$ in thermodynamics?
2. How do enthalpy and entropy influence the spontaneity of a reaction?
3. What are the typical temperature ranges for various industrial reactions?
4. How is Gibbs free energy used in biological systems?
5. What methods can be used to experimentally determine $\Delta H$ and $\Delta S$?

Tip: Always ensure that your thermodynamic data is consistent, using the same units when performing calculations.