For a gaseous reaction,
`2SO_(2)+O_(2)to2SO_(3),DeltaH=-440kJ//"mole"`
at a temperature of 300 K. Calculate `DeltaU` when 1 mole of `SO_(2)` is completely reacted with 1 mole of `O_(2)` in a 10 litre rigid vessel at an initial pressure of 50 bar which decreases to 20 bar.

Options

A. `-190 kJ`
B. `-250 kJ`
C. `-410 kJ`
D. `-470 kJ`

Answer

To solve the problem step by step, we will calculate the change in internal energy (ΔU) for the given reaction.

### Step 1: Determine the enthalpy change (ΔH) for the reaction.
The reaction given is:
\[ 2SO_2 + O_2 \rightarrow 2SO_3 \]
The enthalpy change for this reaction is given as:
\[ \Delta H = -440 \, \text{kJ} \, \text{per mole of reaction} \]

Since we are reacting 1 mole of \( SO_2 \) with 1 mole of \( O_2 \), we need to find the enthalpy change for this amount. According to the stoichiometry, 2 moles of \( SO_2 \) produce 2 moles of \( SO_3 \), so for 1 mole of \( SO_2 \):
\[ \Delta H \text{ for } 1 \text{ mole of } SO_2 = \frac{-440 \, \text{kJ}}{2} = -220 \, \text{kJ} \]

### Step 2: Use the relationship between ΔH and ΔU.
The relationship between the change in enthalpy (ΔH) and the change in internal energy (ΔU) is given by:
\[ \Delta H = \Delta U + \Delta (P \cdot V) \]
Since the reaction occurs in a rigid vessel, the volume (V) is constant. Therefore, we can express the change in pressure in terms of the initial and final pressures.

### Step 3: Calculate the change in pressure (ΔP).
The initial pressure (P_initial) is 50 bar, and the final pressure (P_final) is 20 bar. Thus, the change in pressure is:
\[ \Delta P = P_{final} - P_{initial} = 20 \, \text{bar} - 50 \, \text{bar} = -30 \, \text{bar} \]

### Step 4: Convert the change in pressure to energy.
To find the energy associated with the pressure change, we use the formula:
\[ \Delta (P \cdot V) = \Delta P \cdot V \]
Where:
- V = 10 L (liters) = 0.01 m³ (cubic meters)
- 1 bar liter = 100 Joules

Now, we convert the pressure change to Joules:
\[ \Delta (P \cdot V) = -30 \, \text{bar} \times 10 \, \text{L} = -300 \, \text{bar L} \]
Converting to Joules:
\[ -300 \, \text{bar L} = -300 \times 100 \, \text{J} = -30000 \, \text{J} = -30 \, \text{kJ} \]

### Step 5: Substitute values into the ΔH equation.
Now we can substitute the values into the equation:
\[ -220 \, \text{kJ} = \Delta U - 30 \, \text{kJ} \]

### Step 6: Solve for ΔU.
Rearranging the equation gives:
\[ \Delta U = -220 \, \text{kJ} + 30 \, \text{kJ} \]
\[ \Delta U = -190 \, \text{kJ} \]

### Final Result:
The change in internal energy (ΔU) when 1 mole of \( SO_2 \) is completely reacted with 1 mole of \( O_2 \) is:
\[ \Delta U = -190 \, \text{kJ} \]

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