Question

Ethene is converted to ethane by the reaction $$\mathrm{C}_{2} \mathrm{H}_{4}(g)+\mathrm{H}_{2}(g) \stackrel{catalyst}{\longrightarrow} \mathrm{C}_{2} \mathrm{H}_{6}(g)$$ \(\mathrm{C}_{2} \mathrm{H}_{4}\) flows into a catalytic reactor at \(25.0\) atm and \(300 .^{\circ} \mathrm{C}\) with a flow rate of \(1000 .\) Umin. Hydrogen at \(25.0\) atm and \(300 .^{\circ} \mathrm{C}\) flows into the reactor at a flow rate of \(1500 .\) L/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 reaction for converting ethene to ethane, given the specified flow rates and conditions, is approximately 93.49%.

Step-by-step solution

Convert flow rates to moles per minute

First, we need to convert the flow rates of ethene and hydrogen into moles per minute. To do this, we will use the Ideal Gas Law equation: \(PV = nRT\) where P is the pressure (in atm), V is the volume (in L), n is the number of moles, R is the gas constant (0.0821 L⋅atm/mol⋅K), and T is the temperature (in K). We are given the flow rates in L/min, so let's first convert the temperatures from Celsius to Kelvin: \(T_{Ethene} = 300 + 273.15 = 573.15 K\) \(T_{Hydrogen} = 300 + 273.15 = 573.15 K\) Now we can rearrange the Ideal Gas Law equation to solve for n (moles per minute): \(n = \frac{PV}{RT}\) Ethene: \(n_{Ethene} = \frac{25.0 \, \mathrm{atm} × 1000 \, \mathrm{L/min}}{0.0821 \, \mathrm{L\cdot atm/mol\cdot K} × 573.15 \, \mathrm{K}}\) \(n_{Ethene} ≈ 533.39 \, \mathrm{mol/min}\) Hydrogen: \(n_{Hydrogen} = \frac{25.0 \, \mathrm{atm} × 1500 \, \mathrm{L/min}}{0.0821 \, \mathrm{L\cdot atm/mol\cdot K} × 573.15 \, \mathrm{K}}\) \(n_{Hydrogen} ≈ 800.09 \, \mathrm{mol/min}\)

Determine the limiting reactant

To determine the limiting reactant, we need to find the molar ratio between the reactants in the balanced chemical equation, which is 1:1. The molar ratio between ethene and hydrogen in the reaction: \( \frac{n_{Ethene}}{n_{Hydrogen}} = \frac{533.39}{800.09} ≈ 0.67\) Since the ratio is less than 1, ethene is the limiting reactant.

Calculate the theoretical yield of ethane

Since the balanced chemical equation shows a 1:1 ratio for the reactants and products, the theoretical yield of ethane in moles per minute will be equal to the moles of the limiting reactant (ethene). Thus, the theoretical yield of ethane is: \(n_{Ethane} = 533.39 \, \mathrm{mol/min}\) Now, convert moles of ethane to mass: \(mass_{Ethane} = n_{Ethane} \times molar\ mass_{Ethane}\) The molar mass of ethane is \(30.07 \, \mathrm{g/mol}\): \(mass_{Ethane} = 533.39 \, \mathrm{mol/min} \times 30.07 \, \mathrm{g/mol} ≈ 16042.46 \, \mathrm{g/min}\) Therefore, the theoretical yield of ethane is \(16042.46 \, \mathrm{g/min}\).

Calculate the percent yield of the reaction

Finally, we need to compare the actual yield of ethane (15.0 kg/min or 15000 g/min) to the theoretical yield of ethane (16042.46 g/min) and calculate the percent yield of the reaction: Percent yield = \(\frac{Actual\ yield}{Theoretical\ yield} \times 100\%\) Percent yield = \(\frac{15000 \, \mathrm{g/min}}{16042.46 \, \mathrm{g/min}} \times 100\% ≈ 93.49\%\) Thus, the percent yield of the reaction is approximately 93.49%.

Key concepts

Chemical Reaction

Chemical reactions are processes where reactants transform into products through the breaking and forming of chemical bonds. In our example, the reactant ethene ( C_2H_4) reacts with hydrogen gas ( H_2) to form the product ethane ( C_2H_6). This particular reaction required a catalyst to proceed, which is a substance that increases the rate of a reaction without being consumed by it.

Understanding the stoichiometry of a chemical reaction is crucial. Stoichiometry is the quantitative relationship between reactants and products in a chemical reaction. It allows us to predict how much product will be formed from given amounts of reactants, or how much of the reactants is required to create a certain amount of product. This reaction showcases a simple stoichiometry, with a molar ratio of 1:1:1 for ethene, hydrogen, and ethane, respectively.

Ideal Gas Law

The Ideal Gas Law is a fundamental equation in chemistry that relates the pressure, volume, temperature, and number of moles of a gas. The equation is represented as PV = nRT , where P is the pressure, V is the volume, n is the number of moles of gas, R is the universal gas constant, and T is the temperature in Kelvin.

For our example, the Ideal Gas Law is used to convert given flow rates of ethene and hydrogen from liters per minute to moles per minute. It's a crucial step for finding the number of moles of each reactant, which will later determine the limiting reactant and the theoretical yield of the product.

Limiting Reactant

In a chemical reaction, the limiting reactant is the substance that determines the maximum amount of product that can be formed. It is the reactant that will be completely consumed first. Identifying the limiting reactant is a key step in predicting the theoretical yield of the reaction.

In the solved exercise, we determined that ethene is the limiting reactant by comparing the mole ratio of the reactants. This is done through understanding the stoichiometry, which tells us that ethene and hydrogen react in a 1:1 ratio. Knowing which reactant limits the reaction allows us to precisely calculate the theoretical yield.

Theoretical Yield

The theoretical yield is the maximum amount of product that can be produced in a chemical reaction if everything goes perfectly and no reactants are wasted. It is calculated based on the balanced chemical equation and the amount of the limiting reactant, assuming it reacts completely to form the product.

In our exercise, we used the amount of the limiting reactant, ethene, to calculate the theoretical yield of ethane in grams per minute. By knowing the molar mass of ethane, we could convert the moles of ethane calculated from the stoichiometry to mass, resulting in the theoretical yield.

Stoichiometry

Stoichiometry involves the calculation of reactants and products in chemical reactions. It's a way to quantitatively analyze the conversion of reactants to products based on the balanced chemical equation. Understanding stoichiometric ratios is fundamental to all calculations relating to chemical reactions, including finding the limiting reactant and calculating the theoretical and actual yields.

In our scenario, stoichiometry was employed to establish a 1:1:1 molar ratio between C_2H_4, H_2, and C_2H_6. This ratio guided us in determining the theoretical yield and the percent yield of ethane. Keeping this stoichiometric ratio in mind is essential for correctly solving exercises involving yield calculations.