For which reaction, `(deltaH > deltaU)`? `( delta H=` change in enthalpy,`delta U = `change in internal energy)

Options

A. `[Fe(s) + 5 CO(g) rarr Fe(CO)_5(l)]`
B. `[C(`graphite`)` `+ O_2(g) rarr Co_2(g)]`
C. `[H_2O(1) rarr H^+(aq) + OH^-(aq)]`
D. `[CaCO_3(s)rarr CaO(s) + CO_2(g)]`

Answer

To determine for which reaction ΔH (change in enthalpy) is greater than ΔU (change in internal energy), we can use the relationship between these two quantities:

\[ \Delta H = \Delta U + \Delta n_g \cdot R \cdot T \]

Where:
- Δn_g = moles of gaseous products - moles of gaseous reactants
- R = universal gas constant
- T = temperature in Kelvin

From this equation, we can see that ΔH will be greater than ΔU when Δn_g is greater than zero (i.e., when there are more moles of gaseous products than reactants).

### Step-by-Step Solution:

1. **Identify the Reaction**: We need to analyze different reactions to find Δn_g for each.

2. **Calculate Δn_g**: For each reaction, calculate Δn_g:
- Δn_g = (moles of gaseous products) - (moles of gaseous reactants)

3. **Check the Conditions**:
- If Δn_g > 0, then ΔH > ΔU.
- If Δn_g = 0, then ΔH = ΔU.
- If Δn_g < 0, then ΔH < ΔU.

4. **Evaluate Each Option**:
- For the first option, if there are 0 moles of gaseous products and 5 moles of gaseous reactants, then Δn_g = 0 - 5 = -5 (ΔH < ΔU).
- For the second option, if there is 1 mole of gaseous product and 1 mole of gaseous reactant, then Δn_g = 1 - 1 = 0 (ΔH = ΔU).
- For the third option, if there are no gaseous products and no gaseous reactants, then Δn_g = 0 (ΔH = ΔU).
- For the fourth option, if there is 1 mole of gaseous reactant (CO2) and 0 moles of gaseous products, then Δn_g = 0 - 1 = -1 (ΔH < ΔU).

5. **Conclusion**: The only reaction that satisfies ΔH > ΔU is the decomposition of calcium carbonate, where Δn_g is greater than zero.

### Final Answer:
The reaction for which ΔH > ΔU is the decomposition of calcium carbonate.

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