For the reaction
`aA(s) + bB(g) rightarrow dD(s) + cC(g)`. Then

Options

A. `Delta H - Delta E = (b-d) RT`
B. `Delta H - Delta E = (c - b) RT`
C. `Delta H - Delta E = (a + b) - (c + d) RT`
D. `Delta H - Delta E = (a - d) RT`

Answer

To solve the problem, we need to find the relationship between the change in enthalpy (ΔH) and the change in internal energy (ΔE) for the given reaction:

\[ aA(s) + bB(g) \rightarrow dD(s) + cC(g) \]

### Step-by-Step Solution:

1. **Identify the Reaction Components**:
- We have solid A and D, and gases B and C in the reaction.
- The stoichiometric coefficients are \( a \), \( b \), \( c \), and \( d \).

2. **Write the Relationship Between ΔH and ΔE**:
- The relationship between the change in enthalpy (ΔH) and the change in internal energy (ΔE) is given by the equation:
\[ \Delta H = \Delta E + \Delta N_G RT \]
where \( \Delta N_G \) is the change in the number of moles of gas.

3. **Calculate ΔN_G**:
- To find \( \Delta N_G \), we need to determine the change in the number of moles of gaseous products minus the change in the number of moles of gaseous reactants.
- From the reaction:
- Gaseous products: \( c \) moles of C
- Gaseous reactants: \( b \) moles of B
- Thus, we can calculate \( \Delta N_G \) as:
\[ \Delta N_G = \text{(moles of gaseous products)} - \text{(moles of gaseous reactants)} = c - b \]

4. **Substitute ΔN_G into the ΔH and ΔE Relationship**:
- Now, substituting \( \Delta N_G \) back into the equation:
\[ \Delta H - \Delta E = \Delta N_G RT \]
- Replacing \( \Delta N_G \) with \( c - b \):
\[ \Delta H - \Delta E = (c - b) RT \]

5. **Final Result**:
- Therefore, the relationship between ΔH and ΔE for the given reaction is:
\[ \Delta H - \Delta E = (c - b) RT \]