The enthalpy change for the reaction of 5 litre of ethylene with 5 litre of `H_(2)` gas at 1.5 atm pressure is `Delta H= -0.5 kJ`. The value of `Delta U` will be: `( 1 atm Lt=100 J)`

Options

A. `-1.25 kJ`
B. `+1.25 kJ`
C. `0.25 kJ`
D. `-0.25 kJ`

Explanation

To find the value of \(\Delta U\) (change in internal energy) from the given information, we can use the relationship between \(\Delta H\) (change in enthalpy) and \(\Delta U\):

\[ \Delta U = \Delta H - P \Delta V \]

### Step-by-Step Solution:

1. **Identify Given Values:**
- \(\Delta H = -0.5 \, \text{kJ}\)
- Pressure \(P = 1.5 \, \text{atm}\)
- Volume of ethylene = 5 L
- Volume of hydrogen = 5 L

2. **Calculate \(\Delta V\):**
- Since both reactants (ethylene and hydrogen) are gases and they react to form ethane, the total volume of reactants is:
\[ \Delta V = V_{\text{products}} - V_{\text{reactants}} = 5 \, \text{L} - (5 \, \text{L} + 5 \, \text{L}) = 5 \, \text{L} - 10 \, \text{L} = -5 \, \text{L} \]

3. **Calculate \(P \Delta V\):**
- Using the values of \(P\) and \(\Delta V\):
\[ P \Delta V = 1.5 \, \text{atm} \times (-5 \, \text{L}) = -7.5 \, \text{L atm} \]

4. **Convert \(P \Delta V\) to kJ:**
- Given that \(1 \, \text{atm L} = 100 \, \text{J}\), we convert \(-7.5 \, \text{L atm}\) to kJ:
\[ P \Delta V = -7.5 \, \text{L atm} \times 100 \, \text{J/L atm} = -750 \, \text{J} = -0.75 \, \text{kJ} \]

5. **Substitute into the \(\Delta U\) equation:**
- Now substitute \(\Delta H\) and \(P \Delta V\) into the equation for \(\Delta U\):
\[ \Delta U = \Delta H - P \Delta V = -0.5 \, \text{kJ} - (-0.75 \, \text{kJ}) \]
\[ \Delta U = -0.5 \, \text{kJ} + 0.75 \, \text{kJ} = 0.25 \, \text{kJ} \]

### Final Answer:
\[ \Delta U = 0.25 \, \text{kJ} \]