Given R = `8.314 J K^(-1) "mol"^(-1)`, the work done during combustion of 0.090 kg of ethane(molar mass = 30) at 300 K is

Options

A. `-18.7` kJ
B. `18.7` kJ
C. `6.234` kJ
D. `-6.234` kJ

Explanation

To solve the problem of calculating the work done during the combustion of 0.090 kg of ethane at 300 K, we can follow these steps:

### Step 1: Calculate the number of moles of ethane
Given:
- Mass of ethane = 0.090 kg = 90 g (since 1 kg = 1000 g)
- Molar mass of ethane (C2H6) = 30 g/mol

Using the formula for the number of moles (n):
\[ n = \frac{\text{mass}}{\text{molar mass}} = \frac{90 \, \text{g}}{30 \, \text{g/mol}} = 3 \, \text{moles} \]

### Step 2: Write the balanced combustion reaction for ethane
The balanced chemical equation for the combustion of ethane (C2H6) is:
\[ \text{C}_2\text{H}_6 + \frac{7}{2} \text{O}_2 \rightarrow 2 \text{CO}_2 + 3 \text{H}_2\text{O} \]

### Step 3: Calculate the change in moles of gas (Δn)
From the balanced equation:
- Moles of gaseous products = 2 (from CO2) + 3 (from H2O) = 5 moles
- Moles of gaseous reactants = 1 (from C2H6) + 3.5 (from O2) = 4.5 moles

Now, calculate Δn:
\[ \Delta n = \text{moles of products} - \text{moles of reactants} = 5 - 4.5 = 0.5 \]

### Step 4: Calculate the work done (W)
The work done during the combustion process can be calculated using the formula:
\[ W = -P \Delta V = -nRT \Delta n \]

Where:
- R = 8.314 J/(K·mol)
- T = 300 K
- n = number of moles of ethane = 3 moles

Substituting the values:
\[ W = - (3 \, \text{moles}) \times (8.314 \, \text{J/(K·mol)}) \times (300 \, \text{K}) \times (0.5) \]
\[ W = - (3 \times 8.314 \times 300 \times 0.5) = - 3741.3 \, \text{J} \]

### Step 5: Final Result
The work done during the combustion of 0.090 kg of ethane at 300 K is:
\[ W = -3741.3 \, \text{J} \]