Calculate the wok done during combustion of 0.138 kg of ethanol, `C_2 H_5 OH_((l))` at 300 K. Given: R = 8.314 J `K^-1 mol^-1` , molar mass of ethanol = 46 g `mol^-1`.

Options

A. `-7482J`
B. 7482 J
C. `-2494 J`
D. 2494 J

Answer

To calculate the work done during the combustion of 0.138 kg of ethanol (C₂H₅OH) at 300 K, we can follow these steps:

### Step 1: Calculate the number of moles of ethanol
The number of moles (n) can be calculated using the formula:
\[ n = \frac{\text{mass}}{\text{molar mass}} \]
Given:
- Mass of ethanol = 0.138 kg = 138 g (since 1 kg = 1000 g)
- Molar mass of ethanol = 46 g/mol

Now, substituting the values:
\[ n = \frac{138 \, \text{g}}{46 \, \text{g/mol}} = 3 \, \text{mol} \]

### Step 2: Write the balanced combustion reaction
The balanced combustion reaction for ethanol is:
\[ C_2H_5OH + 3O_2 \rightarrow 2CO_2 + 3H_2O \]
In this reaction:
- Reactants: 1 mole of ethanol (C₂H₅OH) and 3 moles of oxygen (O₂)
- Products: 2 moles of carbon dioxide (CO₂) and 3 moles of water (H₂O)

### Step 3: Determine the number of gaseous moles
Identify the gaseous reactants and products:
- Gaseous reactants: 3 moles of O₂
- Gaseous products: 2 moles of CO₂

Since H₂O is in liquid state, it does not contribute to the gaseous moles.

### Step 4: Calculate ΔNG (change in number of moles of gas)
\[ \Delta N_G = \text{(moles of gaseous products)} - \text{(moles of gaseous reactants)} \]
Substituting the values:
\[ \Delta N_G = 2 - 3 = -1 \]

### Step 5: Calculate the work done (W)
The work done during the combustion can be calculated using the formula:
\[ W = -\Delta N_G \cdot R \cdot T \]
Where:
- R = 8.314 J/(K·mol)
- T = 300 K

Substituting the values:
\[ W = -(-1) \cdot 8.314 \, \text{J/(K·mol)} \cdot 300 \, \text{K} \]
\[ W = 8.314 \cdot 300 = 2494.2 \, \text{J} \]

### Final Result
The work done during the combustion of 0.138 kg of ethanol at 300 K is approximately:
\[ W \approx 2494.2 \, \text{J} \]