Statement-1. Enthalpy change at constant pressure is always greater than enthalpy change at constant volume for any reaction.
Statement -2. Work is done by the system at constant pressure for a given change in volume but no work is done by the system at constant volume.

Options

A. Statement -1 is True, Statement-2 is True, Statement-2 is a correct explanation of Statement-1
B. Statement-1 is True, Statement-2 is True, Statement-2 is NOT a correct explanation for Statement -1.
C. Statement-1 is True, Statement-2 is False.
D. Statement-1 is False, Statement-2 is True

Answer

To analyze the statements provided in the question, we need to evaluate each one based on thermodynamic principles.

### Step-by-Step Solution:

**Step 1: Evaluate Statement 1**
- Statement 1 claims that the enthalpy change at constant pressure (\( \Delta H \)) is always greater than the enthalpy change at constant volume (\( \Delta U \)) for any reaction.
- The relationship between these two quantities is given by the equation:
\[ \Delta H = \Delta U + \Delta N_{g}RT \]
where \( \Delta N_{g} \) is the change in the number of moles of gas, \( R \) is the universal gas constant, and \( T \) is the temperature.

**Step 2: Analyze the Equation**
- From the equation, we can see that \( \Delta H \) is dependent on \( \Delta U \) and the term \( \Delta N_{g}RT \).
- If \( \Delta N_{g} \) is positive (more gaseous products than reactants), then \( \Delta H \) will indeed be greater than \( \Delta U \).
- However, if \( \Delta N_{g} \) is negative (more gaseous reactants than products), the term \( \Delta N_{g}RT \) will subtract from \( \Delta U \), potentially making \( \Delta H \) less than \( \Delta U \).

**Conclusion for Statement 1:**
- Therefore, Statement 1 is **incorrect** because \( \Delta H \) is not always greater than \( \Delta U \); it depends on the change in the number of moles of gas.

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**Step 3: Evaluate Statement 2**
- Statement 2 states that work is done by the system at constant pressure for a given change in volume, but no work is done by the system at constant volume.
- The work done by a system is defined as:
\[ W = P \Delta V \]
where \( W \) is the work done, \( P \) is the pressure, and \( \Delta V \) is the change in volume.

**Step 4: Analyze Work Done**
- At constant pressure, if there is a change in volume (\( \Delta V \neq 0 \)), work is done by the system, and it is given by the formula \( W = P \Delta V \).
- Conversely, at constant volume (\( \Delta V = 0 \)), the work done is zero because there is no change in volume.

**Conclusion for Statement 2:**
- Therefore, Statement 2 is **correct** because work is indeed done at constant pressure when there is a change in volume, and no work is done at constant volume.

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### Final Conclusion:
- Statement 1 is **false**.
- Statement 2 is **true**.