Problem 29 - Chemistry and Chemical Reactivity

Question

Consider the isomerization of butane with an equilibrium constant of \(K=2.5 .\) (See Study Question \(13 .\) The system is originally at equilibrium with [butane] \(=1.0 \mathrm{M}\) and [isobutane] \(=2.5 \mathrm{M}\) (a) If 0.50 mol/L. of isobutane is suddenly added and the system shifts to a new equilibrium position, what is the equilibrium concentration of each gas? (b) If 0.50 mol/L of butane is added to the original equilibrium mixture and the system shifts to a new equilibrium position, what is the equilibrium concentration of each gas?

Step-by-Step Solution

Answer

(a) [butane] = 1.143 M, [isobutane] = 2.857 M; (b) [butane] = 1.143 M, [isobutane] = 2.857 M.

Step 1: Understand the Given Data

The equilibrium constant, \( K \), for the isomerization of butane to isobutane is given as 2.5. Initially, the concentrations are \([\text{butane}] = 1.0 \ \text{M}\) and \([\text{isobutane}] = 2.5 \ \text{M}\). This satisfies the equilibrium condition \( K = \frac{[\text{isobutane}]}{[\text{butane}]} = 2.5 \).

Step 2: Calculate New Equilibrium Concentrations for part (a)

When 0.5 mol/L of isobutane is added, the new concentration of isobutane becomes \( 2.5 + 0.5 = 3.0 \ \text{M} \). Let the change in butane concentration at equilibrium be \( x \). The concentration of butane becomes \( 1.0 + x \), and isobutane becomes \( 3.0 - x \).Using the equilibrium constant: \[ K = \frac{[3.0 - x]}{[1.0 + x]} = 2.5 \]Solving for \( x \):1. Cross-multiply: \[ 3.0 - x = 2.5 (1.0 + x) \]2. Expand: \[ 3.0 - x = 2.5 + 2.5x \]3. Simplify: \[ 3.0 - 2.5 = 3.5x \]4. Solve for \( x \): \[ x = \frac{0.5}{3.5} \approx 0.143 \ \text{M} \]Final concentrations:- \([\text{butane}] = 1.0 + 0.143 = 1.143 \ \text{M}\)- \([\text{isobutane}] = 3.0 - 0.143 = 2.857 \ \text{M}\)

Step 3: Calculate New Equilibrium Concentrations for part (b)

When 0.5 mol/L of butane is added, the new concentration of butane becomes \( 1.0 + 0.5 = 1.5 \ \text{M} \). Let the change in concentration at equilibrium be \( x \). Therefore, the concentration of butane becomes \( 1.5 - x \), and isobutane becomes \( 2.5 + x \).Using the equilibrium constant:\[ K = \frac{[2.5 + x]}{[1.5 - x]} = 2.5 \]Solving for \( x \):1. Cross-multiply: \[ 2.5 (1.5 - x) = 2.5 + x \]2. Expand: \[ 3.75 - 2.5x = 2.5 + x \]3. Simplify: \[ 3.75 - 2.5 = 3.5x \]4. Solve for \( x \): \[ x = \frac{1.25}{3.5} \approx 0.357 \ \text{M} \]Final concentrations:- \([\text{butane}] = 1.5 - 0.357 = 1.143 \ \text{M}\)- \([\text{isobutane}] = 2.5 + 0.357 = 2.857 \ \text{M}\)

Key Concepts

Equilibrium Constant

The equilibrium constant, often represented by the symbol \( K \), is a crucial concept in chemical equilibrium. It tells us the ratio of the concentrations of products to reactants at equilibrium for a reversible chemical reaction. In our context, it includes butane and isobutane in an isomerization process. The equilibrium constant is calculated using the equation \( K = \frac{[\text{products}]}{[\text{reactants}]} \).

For the isomerization of butane to isobutane, the equation is \( K = \frac{[\text{isobutane}]}{[\text{butane}]} \). A \( K \) value of 2.5 suggests that the concentration of isobutane is more significant at equilibrium than that of butane.

Some important points to remember about the equilibrium constant include:

• It's temperature-specific. Changing the temperature changes \( K \).
• It provides no information on the rate of the reaction.
• It only applies to reactions at equilibrium.

Understanding \( K \) is essential for predicting how changes in concentrations affect the system, allowing us to determine the equilibrium concentrations under altered conditions.

Isomerization Reaction

An isomerization reaction involves converting one molecule into another molecule with the same atoms but arranged differently. In this case, we're dealing with butane isomerizing to form isobutane. These molecules have the same molecular formula, \( C_4H_{10} \), but differ in structure.

This reaction's significance lies in its ability to achieve balance through rearrangement rather than the loss or gain of atoms. Since it's a reversible reaction, it can go back and forth until equilibrium is achieved, where the rates of the forward and reverse reactions are equal.

In the context of chemistry, isomerization is crucial because:

• It affects the physical properties of the substances involved, such as boiling points and solubility.
• It can change the chemical behavior of the molecules.
• In our example, knowing the equilibrium constant helps calculate how many moles of each isomer are present at equilibrium.

These features make understanding isomerization reactions vital for both theoretical and practical applications in chemistry.

Le Chatelier's Principle

Le Chatelier's Principle explains how a system at equilibrium reacts to external changes. It states that if a dynamic equilibrium is disturbed by changing conditions such as concentration, temperature, or pressure, the system will adjust itself to partially counteract that change and re-establish equilibrium.

This principle is key in our exercise. When we add more isobutane, the system initially becomes unbalanced. But according to Le Chatelier's Principle, the equilibrium will shift to accommodate the excess isobutane by converting some of it back to butane to reduce the stress of adding more isobutane.

Key takeaways from Le Chatelier's Principle include:

• If you add a reactant or product, the equilibrium shifts away from the added component.
• Removing a component shifts the equilibrium towards the side where it was removed.
• Changing the temperature, particularly for reactions involving heat, will shift the equilibrium depending on whether the reaction is exothermic or endothermic.

This principle allows chemists to predict how changes in conditions can be used to control the outcome of reactions, making it a powerful tool in chemical kinetics and reaction engineering.