Calculate the quantity of electricity that would be required to reduce `12.3g` of nitrobenzene to aniline, if the current efficiency for the process is `50%`. If the potential drop across the cell is `3.0V`, how much energy will be consumed?

Explanation

To solve the problem of calculating the quantity of electricity required to reduce 12.3 g of nitrobenzene to aniline, and the energy consumed in the process, we will follow these steps:

### Step 1: Write the Reduction Reaction
The reduction of nitrobenzene (C6H5NO2) to aniline (C6H5NH2) can be represented as:
\[ \text{C}_6\text{H}_5\text{NO}_2 + 6 \text{H}^+ + 6 \text{e}^- \rightarrow \text{C}_6\text{H}_5\text{NH}_2 + 2 \text{H}_2\text{O} \]

### Step 2: Calculate the Molar Mass of Nitrobenzene
The molar mass of nitrobenzene (C6H5NO2) is calculated as follows:
- Carbon (C): 12 g/mol × 6 = 72 g/mol
- Hydrogen (H): 1 g/mol × 5 = 5 g/mol
- Nitrogen (N): 14 g/mol × 1 = 14 g/mol
- Oxygen (O): 16 g/mol × 2 = 32 g/mol

Total molar mass = 72 + 5 + 14 + 32 = 123 g/mol

### Step 3: Calculate the Number of Moles of Nitrobenzene
Using the mass of nitrobenzene given (12.3 g):
\[ \text{Number of moles} = \frac{\text{mass}}{\text{molar mass}} = \frac{12.3 \text{ g}}{123 \text{ g/mol}} = 0.1 \text{ moles} \]

### Step 4: Determine the Number of Electrons Required
From the balanced reaction, we see that 6 moles of electrons are required to reduce 1 mole of nitrobenzene. Therefore, for 0.1 moles of nitrobenzene:
\[ \text{Electrons required} = 0.1 \text{ moles} \times 6 \text{ moles of e}^- = 0.6 \text{ moles of e}^- \]

### Step 5: Calculate the Total Charge (Q)
Using Faraday's constant (96500 C/mol):
\[ Q = \text{moles of e}^- \times \text{Faraday's constant} = 0.6 \text{ moles} \times 96500 \text{ C/mol} = 57900 \text{ C} \]

### Step 6: Adjust for Current Efficiency
Given that the current efficiency is 50%, the actual quantity of electricity required is:
\[ Q_{\text{actual}} = \frac{Q}{\text{current efficiency}} = \frac{57900 \text{ C}}{0.5} = 115800 \text{ C} \]

### Step 7: Calculate the Energy Consumed
Using the potential drop across the cell (3.0 V):
\[ \text{Energy} = Q \times V = 115800 \text{ C} \times 3.0 \text{ V} = 347400 \text{ J} \]

### Step 8: Convert Energy to Kilojoules
To convert joules to kilojoules:
\[ \text{Energy in kJ} = \frac{347400 \text{ J}}{1000} = 347.4 \text{ kJ} \]

### Final Answer
The energy consumed in the process is **347.4 kJ**.

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