Calculate the temperature at which each of the following reactions becomes spontaneous.


(a)    CaO(*s*) + H2O(*l*)  Ca(OH)2(*s*)

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**.

---

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.

Calculate the temperature at which each of the following reactions becomes spontaneous. (a) CaO(s) + H2O(l) → Ca(OH)2(s)

Determine the temperature at which the reaction CaO(s) + H2O(l) → Ca(OH)2(s) becomes spontaneous using the Gibbs free energy equation ΔG = ΔH - TΔS. The problem involves calculating ΔH, ΔS, and determining the temperature where T > ΔH/ΔS. Suitable for college-level students studying thermodynamics.

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