For a reaction,
`CaCO_(3(s)) rarr CaO_((s))+CO_(2(g))`
`Delta_(f)H^(@)(CaO)=-"631.1 kJ mol"^(-1)`
`Delta_(f)H^(@)(CO_(2))=-"393.5 kJ mol"^(-1) and`
`Delta_(f)H^(@)(CaCO_(3))=-"1206.9 kJ mol"^(-1)`
Which of the following is a correct statement? (A) A large amount of heat is evolved during the decomposition of C a C O 3 . (B)Decomposition of C a C O 3 is an endothermic process and heat is provided for decomposition (C) The amount of heat evolved cannot be calculated from the data provided . (D) Δ f H ∘ = Σ Δ f H ∘ (reactants) − Σ Δ f H ∘ (products)

Options

A. A large amount of heat is evolved during the decomposition of `CaCO_(3)`.
B. Decomposition of `CaCO_(3)` is an endothermic process and heat is provided for decomposition
C. The amount of heat evolved cannot be calculated from the data provided .
D. `Delta_(f)H^(@)=SigmaDelta_(f)H^(@)" (reactants)"-Sigma Delta_(f)H^(@)" (products)"`

Explanation

To solve the problem, we need to calculate the enthalpy change (ΔH) for the reaction given:

\[ \text{CaCO}_{3(s)} \rightarrow \text{CaO}_{(s)} + \text{CO}_{2(g)} \]

We will use the standard enthalpy of formation values provided in the question:

- \( \Delta_f H^\circ (\text{CaO}) = -631.1 \, \text{kJ/mol} \)
- \( \Delta_f H^\circ (\text{CO}_2) = -393.5 \, \text{kJ/mol} \)
- \( \Delta_f H^\circ (\text{CaCO}_3) = -1206.9 \, \text{kJ/mol} \)

### Step 1: Write the formula for the enthalpy change of the reaction

The enthalpy change for the reaction can be calculated using the following formula:

\[ \Delta H = \sum \Delta_f H^\circ (\text{products}) - \sum \Delta_f H^\circ (\text{reactants}) \]

### Step 2: Identify the products and reactants

- **Products**:
- CaO
- CO2

- **Reactants**:
- CaCO3

### Step 3: Substitute the values into the formula

Now we substitute the values into the formula:

\[ \Delta H = [\Delta_f H^\circ (\text{CaO}) + \Delta_f H^\circ (\text{CO}_2)] - \Delta_f H^\circ (\text{CaCO}_3) \]

Substituting the values:

\[ \Delta H = [(-631.1) + (-393.5)] - (-1206.9) \]

### Step 4: Calculate the sum of the products' enthalpies

Calculating the sum of the products:

\[ \Delta H = [-631.1 - 393.5] + 1206.9 \]

\[ \Delta H = -1024.6 + 1206.9 \]

### Step 5: Final calculation

Now, we perform the final calculation:

\[ \Delta H = 182.3 \, \text{kJ} \]

### Conclusion

Since the enthalpy change (ΔH) is positive, this indicates that the reaction is endothermic. Therefore, the correct statement is:

(B) Decomposition of CaCO3 is an endothermic process and heat is provided for decomposition.