For the real gases reaction,
`2CO(g)+O_(2)(g)to2CO_(2)(g),DeltaH=-560 kJ.` In 10 litre rigid vessel at 500 K the initial pressure is 70 bar and after the reaction it becomes 40 bar. The change in internal energy is :

Options

A. `-557 kJ`
B. `-530 kJ`
C. `-563 kJ`
D. None of these

Explanation

To find the change in internal energy (ΔU) for the reaction \(2CO(g) + O_2(g) \rightarrow 2CO_2(g)\) with the given conditions, we can use the relationship between enthalpy change (ΔH) and internal energy change (ΔU) for real gases:

\[ \Delta H = \Delta U + \Delta(PV) \]

### Step-by-Step Solution:

1. **Identify the Given Data**:
- ΔH = -560 kJ
- Initial Pressure (P1) = 70 bar
- Final Pressure (P2) = 40 bar
- Volume (V) = 10 L
- Temperature (T) = 500 K

2. **Calculate Δ(PV)**:
The change in the term PV can be calculated as:
\[ \Delta(PV) = P_2V - P_1V \]
Since the volume is constant, we can factor it out:
\[ \Delta(PV) = V(P_2 - P_1) \]

3. **Convert Units**:
We need to convert the pressure from bar to kJ. Recall that:
- 1 bar = 100 kPa
- 1 L·bar = 0.1 kJ (since 1 L·bar = 100 J)

Therefore, we can calculate:
\[ \Delta(PV) = 10 \, \text{L} \times (40 \, \text{bar} - 70 \, \text{bar}) = 10 \, \text{L} \times (-30 \, \text{bar}) = -300 \, \text{L·bar} \]

Now convert L·bar to kJ:
\[ \Delta(PV) = -300 \, \text{L·bar} \times 0.1 \, \text{kJ/L·bar} = -30 \, \text{kJ} \]

4. **Substitute into the ΔH Equation**:
Now we can substitute ΔH and Δ(PV) into the equation:
\[ \Delta U = \Delta H - \Delta(PV) \]
\[ \Delta U = -560 \, \text{kJ} - (-30 \, \text{kJ}) = -560 \, \text{kJ} + 30 \, \text{kJ} \]
\[ \Delta U = -530 \, \text{kJ} \]

### Final Answer:
The change in internal energy (ΔU) is:
\[ \Delta U = -530 \, \text{kJ} \]