For the reaction
`2CO(g)+O_(2)(g)rarr 2CO_(2), DeltaH=-500kJ.`
Two moles of CO and one mole of `O_(2)` are taken in a container of volume 2 L. They completely from two moles of `CO_(2)`, the gas deviate appreciably from ideal behaviour. If pressure in vessel change from 35 to 20 atm. Find the magnitude of `DeltaU" at "500K`. (Assume 1 L - atom = 0.1 kJ)

Options

A. 503 kJ
B. 400 kJ
C. 480 kJ
D. 320 kJ

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).

### Step-by-Step Solution:

1. **Write the Reaction and Given Data:**
The reaction is:
\[ 2CO(g) + O_2(g) \rightarrow 2CO_2(g) \]
Given:
- ΔH = -500 kJ
- Initial pressure (P1) = 35 atm
- Final pressure (P2) = 20 atm
- Volume (V) = 2 L

2. **Use the Relation Between ΔH and ΔU:**
The relationship between ΔH and ΔU is given by:
\[ ΔH = ΔU + Δ(PV) \]
Since the volume is constant, we can express Δ(PV) as:
\[ Δ(PV) = V(P2 - P1) \]

3. **Calculate Δ(PV):**
Substitute the values into the equation:
\[ Δ(PV) = V(P2 - P1) = 2 \, \text{L} \times (20 \, \text{atm} - 35 \, \text{atm}) = 2 \, \text{L} \times (-15 \, \text{atm}) = -30 \, \text{L atm} \]

4. **Convert Δ(PV) to kJ:**
We know that 1 L·atm = 0.1 kJ, so:
\[ Δ(PV) = -30 \, \text{L atm} \times 0.1 \, \text{kJ/L atm} = -3 \, \text{kJ} \]

5. **Substitute Δ(PV) into the ΔH Equation:**
Now substitute ΔH and Δ(PV) into the equation:
\[ -500 \, \text{kJ} = ΔU - 3 \, \text{kJ} \]

6. **Solve for ΔU:**
Rearranging the equation gives:
\[ ΔU = -500 \, \text{kJ} + 3 \, \text{kJ} = -497 \, \text{kJ} \]

### Final Answer:
The magnitude of ΔU is:
\[ \boxed{497 \, \text{kJ}} \]