For `CaCO_(3)(s)rarr CaO(s)+CO_(2)(g)` at `977^(@)C, Delta H = 174` KJ/mol , then `Delta E` is :-

Options

A. 163 kJ
B. 163.6 Kj
C. 186.4 kJ
D. 180 kJ

Answer

To solve the problem, we need to find the change in internal energy (ΔE) for the reaction:

\[ \text{CaCO}_3(s) \rightarrow \text{CaO}(s) + \text{CO}_2(g) \]

Given:
- ΔH = 174 kJ/mol
- Temperature = 977 °C

### Step-by-Step Solution:

1. **Convert ΔH to Joules:**
\[ \Delta H = 174 \, \text{kJ/mol} = 174 \times 1000 \, \text{J/mol} = 174000 \, \text{J/mol} \]

2. **Calculate ΔN_g (Change in moles of gas):**
- In the products, we have 1 mole of CO₂ (g).
- In the reactants, we have 0 moles of gas (since CaCO₃ is a solid).
\[ \Delta N_g = \text{moles of gas in products} - \text{moles of gas in reactants} = 1 - 0 = 1 \]

3. **Convert the temperature to Kelvin:**
\[ T = 977 \, \text{°C} + 273 = 1250 \, \text{K} \]

4. **Use the relation between ΔH and ΔE:**
\[ \Delta H = \Delta E + \Delta N_g \cdot R \cdot T \]
Where R (the universal gas constant) is:
\[ R = 8.314 \, \text{J/(K·mol)} \]

5. **Calculate ΔN_g \cdot R \cdot T:**
\[ \Delta N_g \cdot R \cdot T = 1 \cdot 8.314 \, \text{J/(K·mol)} \cdot 1250 \, \text{K} = 10392.5 \, \text{J/mol} \]

6. **Substitute the values into the equation:**
\[ 174000 \, \text{J/mol} = \Delta E + 10392.5 \, \text{J/mol} \]

7. **Rearranging to find ΔE:**
\[ \Delta E = 174000 \, \text{J/mol} - 10392.5 \, \text{J/mol} = 163607.5 \, \text{J/mol} \]

8. **Convert ΔE back to kJ:**
\[ \Delta E = \frac{163607.5 \, \text{J}}{1000} = 163.6075 \, \text{kJ/mol} \approx 163.6 \, \text{kJ/mol} \]

### Final Answer:
\[ \Delta E \approx 163.6 \, \text{kJ/mol} \]