`DeltaHgtDeltaE` for the reaction

Options

A. `CH_(4)(g)+2O_(2)(g)rarrCO_(2)(g)+2H_(2)O(g)`
B. `N_(2)(g)+3H_(2)(g)rarr2NH_(3)(g)`
C. `C_(2)H_(4)(g)+3O_(2)(g)rarr2CO_(2)(g)+2H_(2)O(g)`
D. `CaCO_(3)(s)rarrCaO(s)+CO_(2)(g)`

Answer

To determine when ΔH (enthalpy change) is greater than ΔE (internal energy change) for a given reaction, we can use the relationship between the two:

### Step-by-Step Solution:

1. **Understand the Relationship**:
The relationship between ΔH and ΔE is given by the equation:
\[ \Delta H = \Delta E + \Delta (PV) \]
where \(PV = nRT\). Therefore, we can rewrite it as:
\[ \Delta H = \Delta E + \Delta n \cdot RT \]
Here, Δn is the change in the number of moles of gas (products - reactants).

2. **Identify Δn**:
To find when ΔH is greater than ΔE, we need to analyze Δn:
\[ \Delta H > \Delta E \implies \Delta n \cdot RT > 0 \implies \Delta n > 0 \]
This means that the number of gaseous moles in the products must be greater than that in the reactants.

3. **Evaluate Each Reaction**:
We will evaluate the given reactions to find Δn:
- **Reaction 1**: Products = 2 moles (gaseous) and Reactants = 2 moles (gaseous)
\[ \Delta n = 2 - 2 = 0 \]
- **Reaction 2**: Products = 3 moles (gaseous) and Reactants = 1 mole (gaseous)
\[ \Delta n = 3 - 1 = 2 \]
- **Reaction 3**: Products = 4 moles (gaseous) and Reactants = 4 moles (gaseous)
\[ \Delta n = 4 - 4 = 0 \]
- **Reaction 4**: Products = 1 mole (gaseous) and Reactants = 0 moles (gaseous)
\[ \Delta n = 1 - 0 = 1 \]

4. **Determine the Conditions**:
- For **Reaction 1**: Δn = 0 → ΔH = ΔE
- For **Reaction 2**: Δn = 2 → ΔH > ΔE
- For **Reaction 3**: Δn = 0 → ΔH = ΔE
- For **Reaction 4**: Δn = 1 → ΔH > ΔE

5. **Conclusion**:
The reactions where ΔH is greater than ΔE are those with positive Δn, which are Reaction 2 and Reaction 4. Therefore, the correct answer is the option corresponding to these reactions.

### Final Answer:
The correct answer is **Reaction 2 and Reaction 4**.