Concerete is produced from a mixture of cement , water, sand and small stones. It consists primarily of calcium silicates and calcium aluminates formed by heating and grinding of clay and limestone. In later steps of cement production a small amount of gypsym , `CaSO_(4).2H_(2)O` is added to improve subsequent hardening of concrete . The use of elevated temperature during the final production may lead to formation of unwanted hemihydrate, `CaSO_(4).1//2H_(2)O.` Consider the following reaction:
`CaSO_(4).2H_(2)O(s) rarr CaSO_(4).(1)/(2) H_(2)O(s)+(3)/(2)H_(2)O(g)`
The following themodynamic data apply at `25^(@)C` standard pressure : 1 bar
`{:("Compound","DeltaH_(f(KJ//mol))^(@),S^(@)(JK^(-1)mol^(-1))),(CaSO_(4).2H_(2)O(s),-2021.0,194.0),(CaSO_(4).(1)/(2)H_(2)O(s),-1575.0,130.5),(H_(2)O(g),-242.8,188.6):}`
R=8.314 `JK^(-1)mol^(-1)`
Temperature at which the equillibrium water vapour pressure is 1.00 bar

Options

A. `107^(@)` C
B. `380^(@)C`
C. `215^(@) C`
D. `240^(@)C`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Write the reaction and identify the components
The reaction given is:
\[ \text{CaSO}_4 \cdot 2\text{H}_2\text{O}(s) \rightarrow \text{CaSO}_4 \cdot \frac{1}{2}\text{H}_2\text{O}(s) + \frac{3}{2}\text{H}_2\text{O}(g) \]

### Step 2: Calculate the change in enthalpy (\( \Delta H^\circ \))
Using the standard enthalpy of formation values provided:
- \( \Delta H_f^\circ(\text{CaSO}_4 \cdot 2\text{H}_2\text{O}(s)) = -2021.0 \, \text{kJ/mol} \)
- \( \Delta H_f^\circ(\text{CaSO}_4 \cdot \frac{1}{2}\text{H}_2\text{O}(s)) = -1575.0 \, \text{kJ/mol} \)
- \( \Delta H_f^\circ(\text{H}_2\text{O}(g)) = -242.8 \, \text{kJ/mol} \)

The change in enthalpy for the reaction is calculated as:
\[ \Delta H^\circ = [\Delta H_f^\circ(\text{CaSO}_4 \cdot \frac{1}{2}\text{H}_2\text{O}) + \Delta H_f^\circ(\frac{3}{2}\text{H}_2\text{O})] - [\Delta H_f^\circ(\text{CaSO}_4 \cdot 2\text{H}_2\text{O})] \]
\[ = \left[-1575.0 + \left(-242.8 \times \frac{3}{2}\right)\right] - (-2021.0) \]
\[ = \left[-1575.0 - 364.2\right] + 2021.0 \]
\[ = -1939.2 + 2021.0 = 81.8 \, \text{kJ/mol} \]

### Step 3: Calculate the change in entropy (\( \Delta S^\circ \))
Using the standard entropy values provided:
- \( S^\circ(\text{CaSO}_4 \cdot 2\text{H}_2\text{O}(s)) = 194.0 \, \text{J/(K mol)} \)
- \( S^\circ(\text{CaSO}_4 \cdot \frac{1}{2}\text{H}_2\text{O}(s)) = 130.5 \, \text{J/(K mol)} \)
- \( S^\circ(\text{H}_2\text{O}(g)) = 188.6 \, \text{J/(K mol)} \)

The change in entropy for the reaction is calculated as:
\[ \Delta S^\circ = [S^\circ(\text{CaSO}_4 \cdot \frac{1}{2}\text{H}_2\text{O}) + S^\circ(\frac{3}{2}\text{H}_2\text{O})] - [S^\circ(\text{CaSO}_4 \cdot 2\text{H}_2\text{O})] \]
\[ = \left[130.5 + \left(188.6 \times \frac{3}{2}\right)\right] - 194.0 \]
\[ = \left[130.5 + 282.9\right] - 194.0 \]
\[ = 413.4 - 194.0 = 219.4 \, \text{J/(K mol)} \]

### Step 4: Convert \( \Delta H^\circ \) to J/mol
\[ \Delta H^\circ = 81.8 \, \text{kJ/mol} = 81800 \, \text{J/mol} \]

### Step 5: Calculate the temperature (\( T \)) at equilibrium
Using the relationship \( \Delta G^\circ = \Delta H^\circ - T \Delta S^\circ \) and knowing that at equilibrium \( \Delta G^\circ = 0 \):
\[ 0 = \Delta H^\circ - T \Delta S^\circ \]
\[ T = \frac{\Delta H^\circ}{\Delta S^\circ} \]
Substituting the values:
\[ T = \frac{81800 \, \text{J/mol}}{219.4 \, \text{J/(K mol)}} \approx 372.5 \, \text{K} \]

### Step 6: Convert temperature to Celsius
\[ T_{Celsius} = T_{Kelvin} - 273.15 \approx 372.5 - 273.15 \approx 99.35 \, ^\circ C \]

### Step 7: Identify the temperature at which the equilibrium water vapor pressure is 1.00 bar
The temperature at which the equilibrium water vapor pressure is 1.00 bar is approximately 99.35 °C.

### Final Answer
The temperature at which the equilibrium water vapor pressure is 1.00 bar is approximately **99.35 °C**.

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