The electolysis of acetate solution produces ethane according to reaction:
`2CH_3COO^-)toC_6H6(g)+2CO_2(g)+2e^-`
The current efficiency of the process is 80% . What volume of gases would be produced at `27^(@)` C and 740 torr, if the current of 0.5 amp is used though the solution for 96.45 min?
(a)6.0L
(b)0.60L
(c)1.365L
(d)0.91L

Options

A. 6.0L
B. 0.60L
C. 1.365L
D. 0.91L

Answer

To solve the problem, we will follow these steps:

### Step 1: Calculate the total charge passed through the solution
The total charge (Q) can be calculated using the formula:
\[ Q = I \times t \]
where:
- \( I \) = current in amperes (0.5 A)
- \( t \) = time in seconds

First, we need to convert the time from minutes to seconds:
\[ t = 96.45 \text{ min} \times 60 \text{ s/min} = 5787 \text{ s} \]

Now, calculate the total charge:
\[ Q = 0.5 \text{ A} \times 5787 \text{ s} = 2893.5 \text{ C} \]

### Step 2: Calculate the number of moles of electrons transferred
Using Faraday's constant (\( F = 96500 \text{ C/mol} \)):
\[ n_e = \frac{Q}{F} = \frac{2893.5 \text{ C}}{96500 \text{ C/mol}} \approx 0.02996 \text{ mol} \]

### Step 3: Calculate the effective moles of electrons considering current efficiency
The current efficiency is given as 80%, so we need to multiply the moles of electrons by the efficiency:
\[ n_e' = n_e \times \text{efficiency} = 0.02996 \text{ mol} \times 0.8 \approx 0.023968 \text{ mol} \]

### Step 4: Relate moles of electrons to moles of gases produced
From the reaction:
\[ 2 \text{ CH}_3\text{COO}^- \rightarrow \text{C}_2\text{H}_6(g) + 2 \text{ CO}_2(g) + 2e^- \]
2 moles of electrons produce 1 mole of \( \text{C}_2\text{H}_6 \) and 2 moles of \( \text{CO}_2 \).

Thus, the moles of \( \text{CO}_2 \) produced:
\[ n_{\text{CO}_2} = \frac{n_e'}{2} = \frac{0.023968}{2} \approx 0.011984 \text{ mol} \]

And the moles of \( \text{C}_2\text{H}_6 \) produced:
\[ n_{\text{C}_2\text{H}_6} = \frac{n_e'}{2} = \frac{0.023968}{2} \approx 0.011984 \text{ mol} \]

### Step 5: Calculate total moles of gases produced
Total moles of gases:
\[ n_{\text{total}} = n_{\text{CO}_2} + n_{\text{C}_2\text{H}_6} = 0.011984 + 0.011984 = 0.023968 \text{ mol} \]

### Step 6: Use the ideal gas law to find the volume of gases produced
Using the ideal gas law:
\[ PV = nRT \]
We can rearrange to find volume \( V \):
\[ V = \frac{nRT}{P} \]

Where:
- \( R = 0.0821 \text{ L atm/(K mol)} \)
- \( T = 27^\circ C = 300 \text{ K} \) (convert to Kelvin)
- \( P = 740 \text{ torr} = \frac{740}{760} \text{ atm} \approx 0.974 \text{ atm} \)

Now substituting the values:
\[ V = \frac{0.023968 \text{ mol} \times 0.0821 \text{ L atm/(K mol)} \times 300 \text{ K}}{0.974 \text{ atm}} \]

Calculating:
\[ V \approx \frac{0.5937}{0.974} \approx 0.609 \text{ L} \]

### Final Step: Round and compare with options
The calculated volume is approximately 0.609 L, which rounds to 0.60 L.

### Conclusion
The correct answer is (b) 0.60 L.