Problem 56 - Chemistry The Molecular Science

Question

The equilibrium constant \(K_{\mathrm{c}}\) for the cis-trans isomerization of gaseous 2 -butene has the value 1.50 at \(580 . \mathrm{K}\). C=CC(C)=C(C)C (a) Is the reaction product-favored at \(580 . \mathrm{K} ?\) Explain your answer. (b) Calculate the amount (in moles) of trans isomer produced when \(1 \mathrm{~mol}\) cis-2-butene is heated to \(580 . \mathrm{K}\) in the presence of a catalyst in a sealed, 1.00 - \(\mathrm{L}\) flask and reaches equilibrium. (c) What would the answer be if the flask had a volume of \(10.0 \mathrm{~L} ?\)

Step-by-Step Solution

Answer

(a) Yes, the reaction is product-favored. (b) 0.60 mol of trans isomer is produced. (c) 0.60 mol of trans isomer is produced in 10.0 L.

Step 1: Understanding Equilibrium Constant

The equilibrium constant \( K_c \) indicates the ratio of the concentration of products to reactants at equilibrium. If \( K_c > 1 \), the reaction is product-favored.

Step 2: Determine if Reaction is Product-favored

For the isomerization of 2-butene, \( K_c = 1.50 \). Since \( K_c > 1 \), this means the reaction is product-favored at 580 K.

Step 3: Setup Initial Conditions and Equilibrium Expression

The reaction can be represented as \( \text{cis-2-butene} \rightleftharpoons \text{trans-2-butene} \). Let \( x \) be the moles of trans-2-butene formed. The equilibrium expression for \( K_c \) is \( K_c = \frac{[\text{trans-2-butene}]}{[\text{cis-2-butene}]} = \frac{x}{1-x} = 1.50 \), considering the initial amount of cis-2-butene is 1 mol.

Step 4: Solve for Amount of Trans Isomer Produced in 1.00 L Flask

Solving \( \frac{x}{1-x} = 1.50 \):\[ x = 1.50(1-x) \]\[ x = 1.50 - 1.50x \]\[ x + 1.50x = 1.50 \]\[ 2.50x = 1.50 \]\[ x = \frac{1.50}{2.50} = 0.60 \]Thus, 0.60 mol of trans isomer is produced in a 1.00 L flask.

Step 5: Effect of 10.0 L Flask on Equilibrium

In a 10.0 L flask, the initial concentration is \( \frac{1 \, \text{mol}}{10.0 \, \text{L}} = 0.10 \, \text{mol/L} \).For equilibrium, let \( x \) be the concentration of trans isomer:\[ K_c = 1.50 = \frac{x}{0.10 - x} \]Solving:\[ x = 1.50(0.10 - x) \]\[ x = 0.15 - 1.50x \]\[ x + 1.50x = 0.15 \]\[ 2.50x = 0.15 \]\[ x = \frac{0.15}{2.50} = 0.060 \, \text{mol/L} \].The total moles of trans isomer in 10.0 L is \( 0.060 \, \text{mol/L} \times 10 \, \text{L} = 0.60 \, \text{mol} \).

Key Concepts

Equilibrium Constant

The equilibrium constant, often denoted as \( K_c \), is a crucial concept in chemical equilibrium. It indicates the extent to which the reactants are converted into products at equilibrium. This value is derived from the ratio of the concentrations of products to the concentrations of the reactants. If \( K_c > 1 \), it implies that at equilibrium, the products are favored, meaning more products are present compared to reactants.
For instance, in the isomerization of 2-butene, the equilibrium constant \( K_c \) is given as 1.50 at 580 K. This relatively high value indicates that at this temperature, the trans isomer is more stable and predominantly present when equilibrium is reached.
Understanding equilibrium constants allows chemists to predict whether a reaction will favor the formation of products based on the given conditions. It also makes it easier to calculate the concentrations of various species in a reaction mixture at equilibrium.

Isomerization

Isomerization is an intriguing process where molecules with the same molecular formula undergo structural changes to form isomers. These structural rearrangements do not change the molecular formula but can result in different physical and chemical properties. Isomerization can significantly impact the stability and reactivity of molecules.
In the context of 2-butene, isomerization involves the conversion between cis-2-butene and trans-2-butene. This change affects the geometry around the double bond, which subsequently alters the properties of the molecule. For example, the trans isomer of 2-butene is generally more stable due to less steric hindrance compared to its cis counterpart.
This equilibrium between the cis and trans forms is crucial, especially in industrial chemical processes, where the desired isomer can significantly affect the efficiency and outcome of subsequent chemical reactions.

2-butene Isomers

The compound 2-butene has two geometric isomers: cis and trans. These isomers differ in the spatial arrangement of the methyl groups around the carbon-carbon double bond.

• Cis-2-butene: In this isomer, the two methyl groups are on the same side of the double bond. This arrangement can result in steric hindrance, leading to a less stable molecule compared to its trans counterpart.
• Trans-2-butene: In contrast, the trans isomer has the methyl groups on opposite sides of the double bond. This configuration alleviates steric clashes and results in a more stable molecule. Consequently, the trans isomer is often more prevalent at equilibrium.

The physical properties of these isomers can also vary. For example, the trans isomer typically has a higher melting point due to its linear structure, allowing for more effective packing in the solid state.
Understanding the differences between cis and trans isomers is essential, as these variations can influence the energy, reactions, and application of these compounds in various chemical processes.