Calculate the heat produced (in kJ) when 224 gm of CaO is completely converted to `CaCO_(3)` by reaction with `CO_(2)` at `27^(@)` in a container of fixed volume.
Given : `DeltaH_(f)^(@)(CaCO_(3),s)=-1207kJ//mol," "DeltaH_(f)^(@)(CaO,s)=-635kJ//mol`
`DeltaH_(f)^(@)(CO_(2),g)=-394kJ//mol,["Use R"=8.3JK^(-1)mol^(-1)]`

Options

A. `702.04` kJ
B. `721.96` kJ
C. 712 kJ
D. 721 kJ

Explanation

To calculate the heat produced when 224 g of CaO is completely converted to CaCO₃ by reaction with CO₂ at 27°C, we will follow these steps:

### Step 1: Write the balanced chemical reaction
The reaction between calcium oxide (CaO) and carbon dioxide (CO₂) to form calcium carbonate (CaCO₃) can be written as:
\[ \text{CaO (s)} + \text{CO}_2 \text{(g)} \rightarrow \text{CaCO}_3 \text{(s)} \]

### Step 2: Calculate the number of moles of CaO
To find the number of moles of CaO, we use its molar mass. The molar mass of CaO is:
- Ca: 40 g/mol
- O: 16 g/mol
- Molar mass of CaO = 40 + 16 = 56 g/mol

Now, calculate the number of moles in 224 g of CaO:
\[ \text{Number of moles of CaO} = \frac{\text{mass}}{\text{molar mass}} = \frac{224 \text{ g}}{56 \text{ g/mol}} = 4 \text{ moles} \]

### Step 3: Use the enthalpy of formation values
The enthalpy change (ΔH) for the reaction can be calculated using the enthalpy of formation values given:
- ΔH_f°(CaCO₃, s) = -1207 kJ/mol
- ΔH_f°(CaO, s) = -635 kJ/mol
- ΔH_f°(CO₂, g) = -394 kJ/mol

The enthalpy change for the reaction is given by:
\[ \Delta H = \Delta H_f^\circ \text{(products)} - \Delta H_f^\circ \text{(reactants)} \]
Substituting the values:
\[ \Delta H = [-1207 \text{ kJ/mol}] - [-635 \text{ kJ/mol} + (-394 \text{ kJ/mol})] \]
Calculating the enthalpy change:
\[ \Delta H = -1207 + 635 + 394 = -178 \text{ kJ/mol} \]

### Step 4: Calculate the total heat produced for 4 moles
Since we have 4 moles of CaO, the total heat produced (Q) can be calculated as:
\[ Q = \Delta H \times \text{number of moles} = -178 \text{ kJ/mol} \times 4 \text{ moles} = -712 \text{ kJ} \]

### Step 5: Adjust for internal energy at constant volume
At constant volume, the heat produced (Q) is related to the change in internal energy (ΔU) as follows:
\[ \Delta U = \Delta H - \Delta N_g RT \]
Where:
- ΔN_g = moles of gaseous products - moles of gaseous reactants
- R = 8.314 J/(mol·K) = 0.008314 kJ/(mol·K)
- T = 27°C = 300 K

In this reaction:
- Moles of gaseous products = 0 (CaCO₃ is solid)
- Moles of gaseous reactants = 1 (from CO₂)

Thus:
\[ \Delta N_g = 0 - 1 = -1 \]

Now substituting into the equation:
\[ \Delta U = -712 \text{ kJ} - (-1) \times (0.008314 \text{ kJ/(mol·K)}) \times (300 \text{ K}) \]
Calculating the second term:
\[ \Delta U = -712 \text{ kJ} + 2.494 \text{ kJ} = -709.506 \text{ kJ} \]

### Final Answer
The heat produced when 224 g of CaO is completely converted to CaCO₃ is approximately:
\[ \Delta U \approx -709.51 \text{ kJ} \]