Question

Fluorinated organic compounds are used as herbicides, flame retardants, and fire-extinguishing agents, among other things. A reaction such as $$ \mathrm{CH}_{3} \mathrm{SO}_{2} \mathrm{F}+3 \mathrm{HF} \rightarrow \mathrm{CF}_{3} \mathrm{SO}_{2} \mathrm{F}+3 \mathrm{H}_{2} $$ is carried out electrochemically in liquid HF as the solvent. (a) If you electrolyze \(150 \mathrm{g}\) of \(\mathrm{CH}_{3} \mathrm{SO}_{2} \mathrm{F}\), what mass of HF is required, and what mass of each product can be isolated? (b) Is \(\mathrm{H}_{2}\) produced at the anode or the cathode of the electrolysis cell? (c) A typical electrolysis cell operates at \(8.0 \mathrm{V}\) and 250 A. How many kilowatt-hours of energy does one such cell consume in 24 hours?

Short answer

150 g of \(\mathrm{CH}_{3} \mathrm{SO}_{2} \mathrm{F}\) requires 91.77 g of HF to produce 229.32 g of \(\mathrm{CF}_{3} \mathrm{SO}_{2} \mathrm{F}\) and 9.27 g of \(\mathrm{H}_{2}\). \(\mathrm{H}_{2}\) is produced at the cathode, and the cell consumes 48 kWh in 24 hours.

Step-by-step solution

Calculate Moles of Reactant

First, find the molar mass of \(\mathrm{CH}_{3} \mathrm{SO}_{2} \mathrm{F}\). Its molar mass is calculated as follows:- \(\mathrm{C} (12.01 \ \text{g/mol})\),- \(\mathrm{H} (3 \times 1.01 \ \text{g/mol})\),- \(\mathrm{S} (32.07 \ \text{g/mol})\),- \(\mathrm{O} (2 \times 16.00 \ \text{g/mol})\),- \(\mathrm{F} (19.00 \ \text{g/mol})\).Calculate: \[\text{Molar mass of } \mathrm{CH}_{3} \mathrm{SO}_{2} \mathrm{F} = 12.01 + 3 \times 1.01 + 32.07 + 2 \times 16.00 + 19.00 = 98.10\ \text{g/mol}\]Next, calculate the moles of \(\mathrm{CH}_{3} \mathrm{SO}_{2} \mathrm{F}\) in 150 g:\[\text{Moles of } \mathrm{CH}_{3} \mathrm{SO}_{2} \mathrm{F} = \frac{150}{98.10} = 1.529 \ \text{mol}\]

Use Stoichiometry to Find Mass of HF Needed

According to the reaction, 1 mole of \(\mathrm{CH}_{3} \mathrm{SO}_{2} \mathrm{F}\) reacts with 3 moles of \(\mathrm{HF}\). So, for 1.529 moles of \(\mathrm{CH}_{3} \mathrm{SO}_{2} \mathrm{F}\), you need:\[\text{Moles of HF required} = 3 \times 1.529 = 4.587 \ \text{mol}\]Calculate the mass of \(\mathrm{HF}\) required using its molar mass (1.01 for H and 19.00 for F):\[\text{Molar mass of HF} = 1.01 + 19.00 = 20.01\ \text{g/mol}\]\[\text{Mass of HF required} = 4.587 \times 20.01 = 91.77 \ \text{g}\]

Calculate Mass of Products

For \(\mathrm{CF}_{3} \mathrm{SO}_{2} \mathrm{F}\), since it's produced in a 1:1 mole ratio with \(\mathrm{CH}_{3} \mathrm{SO}_{2} \mathrm{F}\), we have 1.529 moles. Calculate its molar mass:- \(\mathrm{C} (12.01)\),- \(\mathrm{F} (3 \times 19.00)\),- \(\mathrm{S} (32.07)\),- \(\mathrm{O} (2 \times 16.00)\).Calculate:\[\text{Molar mass of } \mathrm{CF}_{3} \mathrm{SO}_{2} \mathrm{F} = 12.01 + 3 \times 19.00 + 32.07 + 2 \times 16.00 = 150.07\ \text{g/mol}\]\[\text{Mass of } \mathrm{CF}_{3} \mathrm{SO}_{2} \mathrm{F} = 1.529 \times 150.07 = 229.32 \ \text{g}\]For \(\mathrm{H}_{2}\), from the reaction, you produce 3 moles of \(\mathrm{H}_{2}\) for every mole of \(\mathrm{CH}_{3} \mathrm{SO}_{2} \mathrm{F}\). Thus, for 1.529 moles of \(\mathrm{CH}_{3} \mathrm{SO}_{2} \mathrm{F}\), you produce:\[\text{Moles of } \mathrm{H}_{2} = 3 \times 1.529 = 4.587 \ \text{mol}\]Using the molar mass of \(\mathrm{H}_{2}\) (2.02 g/mol):\[\text{Mass of } \mathrm{H}_{2} = 4.587 \times 2.02 = 9.27 \ \text{g}\]

Determine Anode/Cathode for Hydrogen Production

In electrolysis, Hydrogen gas is typically produced at the cathode because it involves the reduction reaction: \(2\mathrm{H}^{+} + 2\mathrm{e}^{-} \rightarrow \mathrm{H}_{2}\).

Calculate Energy Consumption

Calculate the electrical energy consumed. Use the formula:\[\text{Energy (Wh)} = \text{Voltage (V)} \times \text{Current (A)} \times \text{Time (h)}\]Given:- Voltage = 8.0 V,- Current = 250 A,- Time = 24 h.Substitute into the formula:\[\text{Energy} = 8.0 \times 250 \times 24 = 48000 \ \text{Wh} = 48 \ \text{kWh}\]

Key concepts

Stoichiometry

Stoichiometry is the study of the quantitative relationships between reactants and products in a chemical reaction. It's essentially the recipe of chemistry, ensuring that all ingredients are used in the right amounts to get the desired product. When looking at the reaction: \( \mathrm{CH}_{3} \mathrm{SO}_{2} \mathrm{F} + 3 \mathrm{HF} \rightarrow \mathrm{CF}_{3} \mathrm{SO}_{2} \mathrm{F} + 3 \mathrm{H}_{2} \), stoichiometry helps us understand that each mole of \( \mathrm{CH}_{3} \mathrm{SO}_{2} \mathrm{F} \) requires exactly 3 moles of \( \mathrm{HF} \).
This balanced equation indicates a fixed stoichiometric ratio between all reactants and products. Using these ratios, you can calculate how much product is formed given a specific amount of reactant, or conversely, how much of a reactant is needed to produce a desired amount of product. Understanding stoichiometry is crucial for predicting the outcome of reactions and optimizing the use of reactants.

Molar Mass Calculation

Molar mass calculation is an essential part of chemistry that helps in quantifying the amount of a substance in moles, which is crucial for performing stoichiometric calculations. To find the molar mass of a compound, you sum the atomic masses of all the atoms in its molecular formula.
For example, the molar mass of \( \mathrm{CH}_{3} \mathrm{SO}_{2} \mathrm{F} \) is calculated by adding \( 12.01 \text{g/mol} \) for carbon, \( 3 \times 1.01 \text{g/mol} \) for hydrogen, \( 32.07 \text{g/mol} \) for sulfur, \( 2 \times 16.00 \text{g/mol} \) for oxygen, and \( 19.00 \text{g/mol} \) for fluorine. This gives a total of \( 98.10 \text{g/mol} \).
With the molar mass, you can convert grams of a substance to moles, aiding you in understanding how much of a substance is present in a given mass. This is foundational in further calculations for reactions, such as how much of each reactant is needed or how much product can be expected.

Electrolysis Process

Electrolysis is a fascinating process that uses electricity to drive a chemical reaction that would not occur spontaneously. In an electrolysis cell, electrical energy is used to induce reactions at different parts of the cell: the anode and the cathode. During the given reaction \( \mathrm{CH}_{3} \mathrm{SO}_{2} \mathrm{F}+3 \mathrm{HF} \rightarrow \mathrm{CF}_{3} \mathrm{SO}_{2} \mathrm{F}+3 \mathrm{H}_{2} \), hydrofluoric acid (HF) acts as the solvent.
Hydrogen gas \( \mathrm{H}_{2} \) is typically formed at the cathode through the reduction reaction \( 2\mathrm{H}^{+} + 2\mathrm{e}^{-} \rightarrow \mathrm{H}_{2} \). This is because electrons are gained at the cathode, leading to a reduction. Knowing whether a substance is oxidized or reduced is essential to understanding the electrolysis process and predicting possible products.

Energy Consumption in Electrolysis

The energy consumption in electrolysis is a critical factor, especially when considering efficiency and cost. To calculate the energy consumed in an electrolysis cell, you can use the formula: \( \text{Energy (Wh)} = \text{Voltage (V)} \times \text{Current (A)} \times \text{Time (h)} \).
With the typical conditions given: 8.0 V voltage, 250 A current, and a 24-hour operation, the cell consumes \( 8.0 \times 250 \times 24 = 48,000 \ \text{Wh} \), which translates to \( 48 \ \text{kWh} \).
Understanding the energy requirements of electrolysis processes is essential for industrial applications, where minimizing electrical consumption can lead to significant cost savings. By optimizing voltage and current, while ensuring effective electrolytic reactions, industries can enhance both product yield and energy efficiency.