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)`
`DeltaH^(@)` for the formation of 1.00 kg of `CaSO_(4).(1)/(2) H_(2)O(s)` from `CaSO_(4).2H_(2)O(s)` is

Options

A. `+446 KJ`
B. `+484 KJ`
C. `-446 KJ`
D. `-484 KJ`

Explanation

To solve the problem, we need to calculate the enthalpy change (ΔH) for the formation of 1.00 kg of CaSO₄·(1/2)H₂O(s) from CaSO₄·2H₂O(s) using the given thermodynamic data.

### Step-by-Step Solution:

1. **Write the Reaction**:
The reaction we are considering is:
\[ \text{CaSO}_4 \cdot 2\text{H}_2\text{O}(s) \rightarrow \text{CaSO}_4 \cdot \frac{1}{2}\text{H}_2\text{O}(s) + \frac{3}{2}\text{H}_2\text{O}(g) \]

2. **Identify the Enthalpy Values**:
From the provided data:
- ΔH_f (CaSO₄·2H₂O) = -2021.0 kJ/mol
- ΔH_f (CaSO₄·(1/2)H₂O) = -1575.0 kJ/mol
- ΔH_f (H₂O(g)) = -242.8 kJ/mol

3. **Calculate the ΔH for the Reaction**:
Using the formula for the enthalpy change of the reaction:
\[ \Delta H_{\text{reaction}} = \Delta H_f(\text{products}) - \Delta H_f(\text{reactants}) \]
Substitute the values:
\[ \Delta H_{\text{reaction}} = \left(-1575.0 \, \text{kJ/mol} + 0.5 \times (-242.8) \, \text{kJ/mol}\right) - (-2021.0 \, \text{kJ/mol}) \]
\[ = -1575.0 - 121.4 + 2021.0 \]
\[ = 324.6 \, \text{kJ/mol} \]

4. **Calculate for 1 kg of CaSO₄·(1/2)H₂O**:
First, we need to find the molar mass of CaSO₄·(1/2)H₂O:
- Molar mass of CaSO₄ = 40.08 (Ca) + 32.07 (S) + 64.00 (O₄) = 136.15 g/mol
- Molar mass of (1/2)H₂O = 0.5 × (2.02) = 1.01 g/mol
- Total molar mass = 136.15 + 1.01 = 137.16 g/mol

Now, calculate the number of moles in 1 kg (1000 g):
\[ \text{Number of moles} = \frac{1000 \, \text{g}}{137.16 \, \text{g/mol}} \approx 7.29 \, \text{mol} \]

5. **Calculate the Total ΔH for 1 kg**:
Now, we can find the total ΔH for the formation of 1 kg of CaSO₄·(1/2)H₂O:
\[ \Delta H_{\text{total}} = \Delta H_{\text{reaction}} \times \text{Number of moles} \]
\[ = 324.6 \, \text{kJ/mol} \times 7.29 \, \text{mol} \approx 2363.5 \, \text{kJ} \]

### Final Answer:
The ΔH for the formation of 1.00 kg of CaSO₄·(1/2)H₂O from CaSO₄·2H₂O is approximately **2363.5 kJ**.