Problem 86 - Chemistry The Molecular Science

Question

Fluorine, \(\mathrm{F}_{2}\), is made by the electrolysis of anhydrous \(\mathrm{HF}\). $$ 2 \mathrm{HF}(\ell) \longrightarrow \mathrm{H}_{2}(\mathrm{~g})+\mathrm{F}_{2}(\mathrm{~g}) $$ Typical electrolysis cells operate at 4000 to \(6000 \mathrm{~A}\) and 8 to \(12 \mathrm{~V}\). A large-scale plant can produce about \(9.0 \mathrm{met}-\) ric tons of \(\mathrm{F}_{2}\) gas per day. (a) Calculate the mass (g) of HF consumed. (b) Using the conversion factor of \(3.60 \times 10^{6} \mathrm{~J} / \mathrm{kWh}\), calculate how much energy in kilowatt-hours is transferred to a cell operating at \(6.0 \times 10^{3} \mathrm{~A}\) at \(12 \mathrm{~V}\) for \(24 \mathrm{~h}\).

Step-by-Step Solution

Answer

(a) Approximately 9473684.4 g of HF are consumed. (b) The energy transferred is 1728 kWh.

Step 1: Understand the Reaction

The balanced chemical equation for the process is \(2 \mathrm{HF}(\ell) \longrightarrow \mathrm{H}_{2}(\mathrm{~g})+\mathrm{F}_{2}(\mathrm{~g})\). This indicates that 2 moles of HF produce 1 mole of \(\mathrm{F}_{2}\).

Step 2: Calculate Moles of \(\mathrm{F}_{2}\) Produced

Since one mole of \(\mathrm{F}_{2}\) is approximately 38 g, the production of 9 metric tons (or 9,000,000 g) \(\mathrm{F}_{2}\) is calculated by dividing by the molar mass of \(\mathrm{F}_{2}\): \[\text{Moles of } \mathrm{F}_{2} = \frac{9,000,000 \text{ g}}{38 \text{ g/mol}}\approx 236842.1 \text{ mol}.\]

Step 3: Calculate Moles of HF Required

From the stoichiometry of the reaction, 2 moles of HF produce 1 mole of \(\mathrm{F}_{2}\). Therefore, \[\text{Moles of } \mathrm{HF} = 2 \times 236842.1 = 473684.2 \text{ mol}.\]

Step 4: Convert Moles of HF to Mass

Using the molar mass of HF (20.01 g/mol), compute the mass of HF consumed:\[\text{Mass of } \mathrm{HF} = 473684.2 \text{ mol} \times 20.01 \text{ g/mol} \approx 9473684.4 \text{ g}.\]

Step 5: Understanding the Energy Required

Energy in kilowatt-hours (kWh) for 24 hours at 6000 A and 12 V is given by the power formula: \(\text{Power } (\text{W}) = \text{Current (A)} \times \text{Voltage (V)}\).

Step 6: Calculate Power

Substitute the given values:\[\text{Power } = 6000 \text{ A} \times 12 \text{ V} = 72000 \text{ W}.\]

Step 7: Calculate Energy in Joules

Since 1 W = 1 J/s, convert power to energy over 24 hours:\[\text{Energy (J)} = 72000 \text{ W} \times 24 \text{ h} \times 3600 \text{ s/h} = 6220800000 \text{ J}.\]

Step 8: Convert Energy to kWh

Finally, convert joules to kilowatt-hours using the conversion factor:\[\text{Energy (kWh)} = \frac{6220800000 \text{ J}}{3.6 \times 10^6 \text{ J/kWh}} = 1728 \text{ kWh}.\]

Key Concepts

Stoichiometry

Stoichiometry is a fundamental concept in chemistry that involves the calculation of the quantities of reactants and products in chemical reactions. It is based on the balanced chemical equation for a reaction, which shows the precise ratios in which reactants combine and products form. In the electrolysis of HF reaction, stoichiometry tells us how much HF is needed to produce a certain amount of \(\mathrm{F}_2\).

The balanced chemical equation for the reaction is:

• \(2 \ \mathrm{HF}(\ell) \longrightarrow \mathrm{H}_2(\mathrm{g})+\mathrm{F}_2(\mathrm{g})\).

This equation reveals that 2 moles of HF produce 1 mole of \(\mathrm{F}_2\). Understanding this ratio is crucial for calculating how much HF is consumed in large-scale productions, such as when producing 9 metric tons of \(\mathrm{F}_2\). For every mole of \(\mathrm{F}_2\) that is formed, it takes double the amount of HF moles. This stoichiometric relationship ensures efficient resource management when planning chemical processes.

Energy Calculations

Energy calculations in chemical reactions are essential to understand the energy input or output involved in the process. These calculations help in assessing the feasibility and sustainability of industrial processes like electrolysis. In the example of electrolyzing HF to produce \(\mathrm{F}_2\), energy calculations are needed to determine how much electrical energy is used.

In the given scenario, the electrical parameters are specified as 6000 Amperes and 12 Volts over a period of 24 hours. To calculate the total energy consumed by the cell, we follow these steps:

• First, calculate power using the formula: \(\text{Power (W)} = \text{Current (A)} \times \text{Voltage (V)}\). With given values, this is \(72000 \text{ W}\).
• Next, calculate the energy in joules for 24 hours: \(\text{Energy (J)} = \text{Power (W)} \times \text{Time (s)} = 72000 \text{ W} \times 24 \times 3600 \text{ s}\).
• Finally, convert the total energy from joules to kilowatt-hours (kWh) using the conversion \(1 \text{ kWh} = 3.6 \times 10^6 \text{ J}\).

These steps show that the energy consumption is 1728 kWh, which is necessary for calculating the operational costs and efficiency of the process.

Chemical Reactions

Chemical reactions involve the breaking of bonds in reactants and the formation of new bonds to create products. The electrolysis of HF is a chemical reaction that involves electrical energy to drive the non-spontaneous formation of hydrogen gas and fluorine gas.

In electrolysis, an electric current is passed through a substance to cause a chemical change. For \(\mathrm{HF}\), the reaction can be described as:

• Hydrogen ions (\(\mathrm{H}^+\)) gain electrons at the cathode to form hydrogen gas (\(\mathrm{H}_2\)).
• Fluoride ions (\(\mathrm{F}^-\)) lose electrons at the anode to form fluorine gas (\(\mathrm{F}_2\)).

This decomposition requires an external power source because the reaction is not spontaneous. By understanding these electron transfer processes, chemists can better control the reaction conditions to maximize efficiency and yield. This is particularly important for industrial applications where large quantities of fluorine are produced, as the physical and chemical properties of the reactants and products influence the design and operation of the electrolysis cells.