Question

Ethene is converted to ethane by the reaction $$\mathrm{C}_{2} \mathrm{H}_{4}(g)+\mathrm{H}_{2}(g)\stackrel{\text { Catalyst }}{\longrightarrow} \mathrm{C}_{2} \mathrm{H}_{6}(g)$$ \(\mathrm{C}_{2} \mathrm{H}_{4}\) flows into a catalytic reactor at 25.0 \(\mathrm{atm}\) and \(300.^{\circ} \mathrm{C}\) with a flow rate of \(1000 . \mathrm{L} / \mathrm{min}\) . Hydrogen at 25.0 \(\mathrm{atm}\) and \(300 .^{\circ} \mathrm{C}\) flows into the reactor at a flow rate of \(1500 . \mathrm{L} / \mathrm{min}\) . If 15.0 \(\mathrm{kg}\) \(\mathrm{C}_{2} \mathrm{H}_{6}\) is collected per minute, what is the percent yield of the reaction?

Short answer

The percent yield of the ethene to ethane reaction is approximately 72.89%.

Step-by-step solution

Calculate the number of moles flowed in for each reactant.

First, we need to find the number of moles flowed in for both ethene (C2H4) and hydrogen (H2). Using the ideal gas law equation: \(PV = nRT\), we can find the number of moles, where P is pressure, V is volume, n is the number of moles, R is the ideal gas constant, and T is the temperature in Kelvin. First, convert the given temperature (300°C) to Kelvin: \(T_{K} = 300 + 273.15 = 573.15 \: K\) Now, we calculate the moles of ethene: Formula of ethene is \(C_{2}H_{4}\), and its molar mass is 28.05 g/mol. \(P_{C2H4} = 25\: atm\) \(V_{C2H4} = 1000\: L/min\) \(T_{C2H4} = 573.15 \: K\) \(R = 0.0821\: L.atm/mol.K\) \(n_{C2H4} = \frac{P_{C2H4}V_{C2H4}}{RT_{C2H4}} = \frac{25 \times 1000}{0.0821 \times 573.15} = 684.74\: moles/min\) Similarly, we calculate the moles of hydrogen: Formula of hydrogen is \(H_{2}\), and its molar mass is 2.02 g/mol. \(P_{H2} = 25\: atm\) \(V_{H2} = 1500\: L/min\) \(T_{H2} = 573.15 \: K\) \(n_{H2} = \frac{P_{H2}V_{H2}}{RT_{H2}} = \frac{25 \times 1500}{0.0821 \times 573.15} = 1027.11\: moles/min\)

Determine the moles of reactants left after the reaction reaches an equilibrium.

As per the balanced equation of given reaction: \(\mathrm{C}_{2} \mathrm{H}_{4}(g)+\mathrm{H}_{2}(g)\longrightarrow \mathrm{C}_{2} \mathrm{H}_{6}(g)\) Thus, the molar ratio of ethene to hydrogen in the reaction is 1:1. From Step 1, we know: \(n_{C2H4} = 684.74 \: moles/min\) \(n_{H2} = 1027.11 \: moles/min\) Since ethene is the limiting reactant, the reaction proceeds until all of the ethene is used up. After this, there will be no ethene remaining, and the unreacted hydrogen will be: \(n_{H2} - n_{C2H4} = 1027.11 - 684.74 = 342.37\: moles/min\)

Calculate the moles and mass of ethane theoretically produced.

When all ethene is used up, the theoretical moles of ethane produced are equal to the moles of ethene. \(n_{C2H6} = n_{C2H4} = 684.74\: moles/min\) Now, let's find the theoretical mass of ethane produced. Formula of ethane is \(C_{2}H_{6}\), and its molar mass is 30.07 g/mol. \(Mass_{C2H6} = n_{C2H6} * Molar\:mass_{C2H6} = 684.74 * 30.07 = 20584.20 g/min\)

Calculate the percent yield of the reaction.

Now, we have the theoretical mass of ethane produced, and we are given that the actual mass of ethane collected per minute is 15.0 kg = 15000 g. Percent yield of reaction is calculated as: \(\%\:yield = \frac{Actual\:mass\:of\:product}{Theoretical\:mass\:of\:product} * 100\) \(\%\:yield = \frac{15000}{20584.20} * 100 = 72.89\%\) The percent yield of the reaction is approximately 72.89%.

Key concepts

ideal gas law

The ideal gas law is a fundamental principle used to relate the properties of gases. It is expressed with the formula \( PV = nRT \), where \( P \) is the pressure in atm, \( V \) is the volume in liters, \( n \) is the number of moles, \( R \) is the ideal gas constant (0.0821 L.atm/mol.K), and \( T \) is the temperature in Kelvin.

In our exercise, we applied the ideal gas law to find the number of moles for ethene \( (C_2H_4) \) and hydrogen \( (H_2) \). By rearranging the equation to solve for \( n \), \( n = \frac{PV}{RT} \), we could calculate the moles by plugging in the values provided:

• Pressure \( P = 25 \, \text{atm} \)
• Volume \( V \) converted from flow rate \( \text{L/min} \)
• Temperature \( T = 573.15 \, \text{K} \) (conversion from 300°C)
• The constant \( R \)

This calculation helps determine how much of each gas is available for reaction in the reactor.

limiting reactant

The limiting reactant is the substance that is completely consumed first in a chemical reaction, determining the amount of product formed. In a balanced chemical equation, the reactants are combined in a specific proportion.

For the reaction \( \mathrm{C}_{2} \mathrm{H}_{4}(g)+\mathrm{H}_{2}(g)\rightarrow \mathrm{C}_{2} \mathrm{H}_{6}(g) \), the stoichiometry tells us that ethene and hydrogen react on a 1:1 molar basis.

From the moles calculated using the ideal gas law:

• Ethene, \( n_{C2H4} = 684.74 \, \text{moles/min} \)
• Hydrogen, \( n_{H2} = 1027.11 \, \text{moles/min} \)

Since there are fewer moles of ethene, it is the limiting reactant. This means the reaction can only produce as much ethane \( (C_2H_6) \) as the ethene moles allow. After the entire ethene is used up, no further ethane can be produced, even though some hydrogen remains unused.

molar mass

Molar mass is the mass of one mole of a substance, expressed in grams per mole \( \text{g/mol} \). It is critical in converting between moles and grams—a necessary step for understanding the quantities involved in a chemical reaction.

For this exercise, we used the molar mass to calculate the theoretical mass of ethane \( (C_2H_6) \) that could be produced:

• Ethene \( C_2H_4 \) has a molar mass of 28.05 g/mol.
• Hydrogen \( H_2 \) has a molar mass of 2.02 g/mol.
• Ethane \( C_2H_6 \) has a molar mass of 30.07 g/mol.

With these molar masses, we calculated the theoretical mass of ethane by multiplying its molar mass by the moles of ethane produced according to the chemical reaction stoichiometry. The calculation was: \( 684.74 \, \text{moles/min} \times 30.07 \, \text{g/mol} \) resulting in 20584.20 g/min.

chemical reaction stoichiometry

Chemical reaction stoichiometry involves the quantitative relationships between reactants and products in a chemical reaction.

Consider the given reaction: \( \mathrm{C}_{2} \mathrm{H}_{4}(g)+\mathrm{H}_{2}(g)\rightarrow \mathrm{C}_{2} \mathrm{H}_{6}(g) \). From the balanced equation, we see a 1:1:1 molar ratio, which indicates:

• 1 mole of ethene reacts with 1 mole of hydrogen
• 1 mole of ethane is produced

This ratio is essential for determining the quantities of reactants required and product formed. In the exercise, we applied this stoichiometry to figure out how many moles of ethane were theoretically possible if ethene is the limiting reactant.

By understanding stoichiometry, we can predict the yield of a reaction, verify our limiting reactant determination, and calculate the percent yield, which compares the actual amount produced to the theoretical maximum.