For the reaction, `2CO + O_(2) to 2CO_(2), DeltaH = -560 kJ`. Two moles of CO and one mole of `O_(2)` are taken in a container of volume 1 L. They completely form two moles of `CO_(2)`, the gases deviate appreciably from ideal behaviour. If the pressure in the vessel changes from 70 to 40 atm, find the magnitude (absolute value) of `DeltaU` at 500 K.

Options

A. -557kJ
B. +557kJ
C. -5570J
D. +5570J

Answer

To solve the problem, we need to find the change in internal energy (ΔU) for the reaction given the change in pressure and the enthalpy change (ΔH). The relationship between ΔH and ΔU can be expressed as:

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

Where:
- ΔH is the change in enthalpy,
- ΔU is the change in internal energy,
- V is the volume of the gas,
- ΔP is the change in pressure.

### Step-by-Step Solution:

1. **Identify Given Values:**
- ΔH = -560 kJ (enthalpy change for the reaction)
- Initial pressure (P1) = 70 atm
- Final pressure (P2) = 40 atm
- Volume (V) = 1 L

2. **Calculate ΔP:**
\[ \Delta P = P2 - P1 = 40 \, \text{atm} - 70 \, \text{atm} = -30 \, \text{atm} \]

3. **Calculate VΔP:**
- Convert pressure change from atm to L·atm (since volume is in liters):
\[ V \Delta P = 1 \, \text{L} \times (-30 \, \text{atm}) = -30 \, \text{L·atm} \]

4. **Convert L·atm to kJ:**
- Use the conversion factor: 1 L·atm = 0.1013 kJ
\[ V \Delta P = -30 \, \text{L·atm} \times 0.1013 \, \text{kJ/L·atm} = -3.039 \, \text{kJ} \]

5. **Calculate ΔU:**
- Substitute ΔH and VΔP into the equation:
\[ \Delta U = \Delta H - V \Delta P \]
\[ \Delta U = -560 \, \text{kJ} - (-3.039 \, \text{kJ}) = -560 \, \text{kJ} + 3.039 \, \text{kJ} = -556.961 \, \text{kJ} \]

6. **Round the Result:**
- The magnitude of ΔU is approximately:
\[ |\Delta U| \approx 557 \, \text{kJ} \]

### Final Answer:
The magnitude of ΔU at 500 K is approximately **557 kJ**.