Given the reaction at `975^(@)C` and 1 atm.
`CaCO_(3)(s)toCaO(s)+CO_(2)(g),DeltaH=176kJ`
Then `DeltaE` is equal to

Options

A. 186.4kJ
B. 162kJ
C. 165.63kJ
D. 180kJ

Explanation

To find the value of ΔE for the reaction:

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

at \( 975^\circ C \) and 1 atm, we will use the relationship between ΔH and ΔE.

### Step-by-Step Solution:

1. **Convert Temperature to Kelvin**:
\[ T(K) = 975 + 273 = 1248 \, K \]

2. **Identify the Given Values**:
- ΔH = 176 kJ
- The reaction involves 1 mole of CO2 gas produced.

3. **Calculate ΔnG**:
- ΔnG = moles of gaseous products - moles of gaseous reactants
- In this reaction, we have:
- Products: 1 mole of CO2 (g)
- Reactants: 0 moles of gaseous reactants (since CaCO3 is solid)
\[ \Delta n_G = 1 - 0 = 1 \]

4. **Use the Formula**:
The relationship between ΔH and ΔE is given by:
\[ \Delta H = \Delta E + \Delta n_G RT \]
Rearranging gives:
\[ \Delta E = \Delta H - \Delta n_G RT \]

5. **Substitute the Values**:
- R = 8.314 J/(mol·K) = 8.314 × 10^-3 kJ/(mol·K)
- Substitute ΔH, ΔnG, R, and T into the equation:
\[ \Delta E = 176 \, kJ - (1) \times (8.314 \times 10^{-3} \, kJ/(mol \cdot K)) \times (1248 \, K) \]

6. **Calculate the Second Term**:
\[ 8.314 \times 10^{-3} \, kJ/(mol \cdot K) \times 1248 \, K \approx 10.37 \, kJ \]

7. **Final Calculation**:
\[ \Delta E = 176 \, kJ - 10.37 \, kJ \approx 165.63 \, kJ \]

Thus, the value of ΔE is approximately **165.63 kJ**.