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. 495 kJ
D. 320 kJ

Answer

To solve the problem, we need to calculate the change in internal energy (ΔU) for the reaction given the change in enthalpy (ΔH) and the number of moles of gases involved. Here’s a step-by-step breakdown of the solution:

### Step 1: Write down the reaction and the given data
The reaction is:
\[ 2 \text{CO}(g) + \text{O}_2(g) \rightarrow 2 \text{CO}_2(g) \]
Given:
- ΔH = -500 kJ
- Initial pressure (P1) = 35 atm
- Final pressure (P2) = 20 atm
- Temperature (T) = 500 K
- Volume (V) = 2 L

### Step 2: Determine the change in the number of moles of gas (ΔNg)
ΔNg is calculated as:
\[ \Delta N_g = \text{(moles of gaseous products)} - \text{(moles of gaseous reactants)} \]
From the reaction:
- Moles of gaseous products (CO2) = 2
- Moles of gaseous reactants (CO + O2) = 2 + 1 = 3

Thus,
\[ \Delta N_g = 2 - 3 = -1 \]

### Step 3: Use the relationship between ΔH and ΔU
The relationship is given by:
\[ \Delta H = \Delta U + \Delta N_g RT \]
Where:
- R = 8.314 J/(mol·K) = 0.008314 kJ/(mol·K)
- T = 500 K

### Step 4: Substitute the known values into the equation
Substituting the values we have:
\[ -500 \text{ kJ} = \Delta U + (-1) \times (0.008314 \text{ kJ/(mol·K)}) \times (500 \text{ K}) \]

Calculating the second term:
\[ -1 \times 0.008314 \times 500 = -4.157 \text{ kJ} \]

So the equation becomes:
\[ -500 \text{ kJ} = \Delta U - 4.157 \text{ kJ} \]

### Step 5: Solve for ΔU
Rearranging the equation to solve for ΔU:
\[ \Delta U = -500 \text{ kJ} + 4.157 \text{ kJ} \]
\[ \Delta U = -495.843 \text{ kJ} \]

### Step 6: Round the answer
The magnitude of ΔU is:
\[ |\Delta U| = 495.843 \text{ kJ} \]

Thus, rounding it gives:
\[ |\Delta U| \approx 495 \text{ kJ} \]

### Final Answer
The magnitude of ΔU at 500 K is approximately **495 kJ**.