For the reaction, `A(s) + 3B(g) rarr 4C(g) + d(l) Delta H and Delta U` are related as :

Options

A. `Delta H = DeltaU`
B. `Delta H = DeltaU + 3RT`
C. `Delta H = DeltaU + RT`
D. `Delta H = DeltaU - 3RT`

Explanation

To solve the problem regarding the relationship between ΔH (enthalpy change) and ΔU (internal energy change) for the reaction \( A(s) + 3B(g) \rightarrow 4C(g) + D(l) \), we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Definitions**:
- ΔH is the heat of reaction at constant pressure.
- ΔU is the heat of reaction at constant volume.

2. **Identify the Reaction Components**:
- Reactants: \( A(s) \) (solid) and \( 3B(g) \) (gas).
- Products: \( 4C(g) \) (gas) and \( D(l) \) (liquid).

3. **Determine the Change in Moles of Gas (ΔN_g)**:
- Count the moles of gaseous products: \( 4C(g) \) contributes 4 moles.
- Count the moles of gaseous reactants: \( 3B(g) \) contributes 3 moles.
- Therefore, ΔN_g = moles of gaseous products - moles of gaseous reactants = \( 4 - 3 = 1 \).

4. **Use the Relation Between ΔH and ΔU**:
- The relationship between ΔH and ΔU is given by the formula:
\[ ΔU = ΔH - ΔN_g \cdot RT \]
- Substitute ΔN_g into the equation:
\[ ΔU = ΔH - 1 \cdot RT \]
- This simplifies to:
\[ ΔU = ΔH - RT \]

5. **Rearranging the Equation**:
- To express ΔH in terms of ΔU, rearrange the equation:
\[ ΔH = ΔU + RT \]

6. **Conclusion**:
- The relationship derived is:
\[ ΔH = ΔU + RT \]
- This matches with the option that states \( ΔU + RT = ΔH \).

### Final Answer:
The relationship between ΔH and ΔU for the given reaction is:
\[ ΔU + RT = ΔH \]