Problem 120 - Chemistry and Chemical Reactivity

Question

ATOM ECONOMY: Ethylene oxide, \(\mathrm{C}_{2} \mathrm{H}_{4} \mathrm{O},\) is an important industrial chemical [as it is the starting place to make such important chemicals as ethylene glycol (antifreeze) and various polymers One way to make the compound is called the "chlorohydrin route." $$\mathrm{C}_{2} \mathrm{H}_{4}+\mathrm{Cl}_{2}+\mathrm{Ca}(\mathrm{OH})_{2} \rightarrow \mathrm{C}_{2} \mathrm{H}_{4} \mathrm{O}+\mathrm{CaCl}_{2}+\mathrm{H}_{2} \mathrm{O}$$ Another route is the modern catalytic reaction. $$\mathrm{C}_{2} \mathrm{H}_{4}+1 / 2 \mathrm{O}_{2} \rightarrow \mathrm{C}_{2} \mathrm{H}_{4} \mathrm{O}$$ (a) Calculate the \% atom economy for the production of \(\mathrm{C}_{2} \mathrm{H}_{4} \mathrm{O}\) in each of these reactions. Which is the more efficient method? (b) What is the percent yield of \(\mathrm{C}_{2} \mathrm{H}_{4} \mathrm{O}\) if 867 g of \(\mathrm{C}_{2} \mathrm{H}_{4}\) is used to synthesize \(762 \mathrm{g}\) of the product by the catalytic reaction?

Step-by-Step Solution

Answer

The catalytic route has 100% atom economy and approximately 55.9% yield.

Step 1: Define Atom Economy Formula

Atom economy is calculated using the formula: \[ \text{Atom Economy} = \left( \frac{\text{Molar Mass of Desired Product}}{\text{Total Molar Mass of Reactants}} \right) \times 100 \] This formula helps in understanding the efficiency of a reaction in terms of how much of the starting materials become the desired product.

Step 2: Calculate Molar Masses for Chlorohydrin Route

For the chlorohydrin route, the reactants are \(\text{C}_2 \text{H}_4\), \(\text{Cl}_2\), and \(\text{Ca(OH)}_2\). Their molar masses are approximately 28.05 g/mol, 70.90 g/mol, and 74.10 g/mol respectively. The molar mass of \(\text{C}_2 \text{H}_4 \text{O}\), the desired product, is approximately 44.05 g/mol.

Step 3: Calculate Total Molar Mass for Chlorohydrin Route

The total molar mass of reactants for the chlorohydrin route is: \[ 28.05 + 70.90 + 74.10 = 173.05 \text{ g/mol} \]

Step 4: Calculate Atom Economy for Chlorohydrin Route

Using the atom economy formula: \[ \text{Atom Economy} = \left( \frac{44.05}{173.05} \right) \times 100 \approx 25.5\% \]

Step 5: Calculate Molar Masses for Catalytic Route

For the catalytic route, the reactants are \(\text{C}_2 \text{H}_4\) and \(1/2 \text{O}_2\). Their molar masses are 28.05 g/mol and 16.00 g/mol respectively. The molar mass of \(\text{C}_2 \text{H}_4 \text{O}\), the desired product, is again 44.05 g/mol.

Step 6: Calculate Total Molar Mass for Catalytic Route

The total molar mass of reactants for the catalytic route is: \[ 28.05 + 16.00 = 44.05 \text{ g/mol} \]

Step 7: Calculate Atom Economy for Catalytic Route

Using the atom economy formula: \[ \text{Atom Economy} = \left( \frac{44.05}{44.05} \right) \times 100 = 100\% \]

Step 8: Determine More Efficient Method

Since the atom economy for the catalytic route is 100%, it is more efficient compared to the chlorohydrin route with an atom economy of 25.5%.

Step 9: Define Percent Yield Formula

The percent yield is calculated using the formula: \[ \text{Percent Yield} = \left( \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \right) \times 100 \]

Step 10: Calculate Theoretical Yield for Catalytic Route

The molar mass of \(\text{C}_2 \text{H}_4\) is 28.05 g/mol. Given 867 g of \(\text{C}_2 \text{H}_4\), we first convert this mass to moles: \[ \text{Moles of } \text{C}_2 \text{H}_4 = \frac{867 \text{ g}}{28.05 \text{ g/mol}} \approx 30.91 \text{ mol} \]Since the reaction produces \( \text{C}_2 \text{H}_4 \text{O} \) in a 1:1 molar ratio, the theoretical yield in mass is:\[ 30.91 \text{ mol} \times 44.05 \text{ g/mol} = 1362.64 \text{ g} \]

Step 11: Calculate Percent Yield for Catalytic Route

The actual yield is given as 762 g. Using the formula for percent yield: \[ \text{Percent Yield} = \left( \frac{762}{1362.64} \right) \times 100 \approx 55.9\% \]

Step 12: Conclusion

The catalytic route is more efficient with 100% atom economy. The percent yield of ethylene oxide using this method is approximately 55.9%.

Key Concepts

Chemical Reactions

Chemical reactions are processes where reactants transform into products. Understanding the details of these reactions is key for chemists. They can vary greatly, from simple combinations of molecules to complex transformations. In every reaction, bonds between atoms break and new ones form, resulting in different substances.

• Reactants: The starting materials in chemical reactions.
• Products: The substances formed as a result of the reaction.
• Reaction conditions: These can include temperature, pressure, and the presence of catalysts that can influence the rate and efficiency of the reaction.

If we look at the production of ethylene oxide (\( ext{C}_2 ext{H}_4 ext{O}\)) from ethylene (\( ext{C}_2 ext{H}_4\)), there are different methods like the chlorohydrin route and the catalytic reaction. Both are examples of industrial applications of chemical reactions. Each method has its own unique reaction conditions and efficiency in converting reactants into the desired product.

Catalytic Reaction

Catalytic reactions utilize a catalyst to increase the rate of a chemical reaction without being consumed in the process. Catalysts are substances that provide an alternative pathway for the reaction, often lowering the activation energy required. This makes the reaction more efficient and can often lead to higher yields of the desired product. Consider the modern catalytic reaction for producing ethylene oxide:

• Reactants: Ethylene (\( ext{C}_2 ext{H}_4\)) and oxygen (\(rac{1}{2} ext{O}_2\)).
• Products: Ethylene oxide (\( ext{C}_2 ext{H}_4 ext{O}\)).
• Efficiency: Catalytic reactions often have higher atom economies, meaning a greater fraction of the mass of the starting materials is incorporated into the desired product. This makes them more environmentally and economically favorable.

In the case of the catalytic route for ethylene oxide production, it achieves a 100% atom economy, meaning all the atoms in the reactants are converted into the product without producing any waste.

Percent Yield

Percent yield is a measure of the efficiency of a chemical reaction, comparing the actual yield (amount of product obtained) to the theoretical yield (amount expected based on stoichiometry). It's calculated using the formula: \[\text{Percent Yield} = \left( \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \right) \times 100\] For example, in the production of ethylene oxide using the catalytic reaction, if 867 g of ethylene results in 762 g of ethylene oxide, we need to calculate the theoretical yield first. Given the molar mass of ethylene and ethylene oxide, we can determine how many grams of ethylene oxide should have been produced. The percent yield is then approximately 55.9%, indicating that just over half of the theoretical maximum amount was actually obtained. This can happen due to several factors like incomplete reactions or side reactions. High percent yields are desirable as they indicate efficient use of materials.