Problem 35 - Chemistry and Chemical Reactivity

Question

The equilibrium constant for the butane \(\rightleftarrows\) isobutane isomerization reaction is 2.5 at \(25^{\circ} \mathrm{C}\). If 1.75 mol of butane and 1.25 mol of isobutane are mixed, is the system at equilibrium? If not, when it proceeds to equilibrium, which reagent increases in concentration? Calculate the concentrations of the two compounds when the system reaches equilibrium.

Step-by-Step Solution

Answer

Isobutane concentration increases to 0.714 M when the system reaches equilibrium.

Step 1: Write the Reaction Equation

The given reaction is the isomerization of butane into isobutane, which can be written as:\[ \text{C}_4\text{H}_{10} \rightleftarrows \text{iso-C}_4\text{H}_{10} \] This is a simple equilibrium involving the conversion of one compound into another.

Step 2: Define Initial Conditions and Total Moles

We are given the initial moles of butane and isobutane: - Moles of butane = 1.75 mol - Moles of isobutane = 1.25 mol Total moles in the system initially = 1.75 + 1.25 = 3 moles.

Step 3: Calculate Initial Concentrations

Initial concentration is calculated by dividing the moles by the total volume. Assuming a constant volume where total moles are given:- Concentration of butane \( [\text{C}_4\text{H}_{10}] = \frac{1.75}{3} \approx 0.583 \text{ M} \)- Concentration of isobutane \( [\text{iso-C}_4\text{H}_{10}] = \frac{1.25}{3} \approx 0.417 \text{ M} \)

Step 4: Set Up the Reaction Quotient (Q)

The reaction quotient \( Q \) is calculated at the initial concentrations:\[ Q = \frac{[\text{iso-C}_4\text{H}_{10}]_{initial}}{[\text{C}_4\text{H}_{10}]_{initial}} = \frac{0.417}{0.583} \approx 0.714 \]

Step 5: Compare Q to the Equilibrium Constant (K)

Given the equilibrium constant \( K = 2.5 \). Since \( Q = 0.714 \) and \( Q < K \), the system is not at equilibrium. To reach equilibrium:- The concentration of isobutane must increase.- The concentration of butane must decrease.

Step 6: Set Up the Equilibrium Expressions

Let \( x \) be the change in concentration as the system shifts to equilibrium. At equilibrium:- Concentration of butane: \( [\text{C}_4\text{H}_{10}] = 0.583 - x \)- Concentration of isobutane: \( [\text{iso-C}_4\text{H}_{10}] = 0.417 + x \)

Step 7: Solve for x Using the Equilibrium Constant

Substitute into the equilibrium expression:\[ K = 2.5 = \frac{0.417 + x}{0.583 - x} \]Solving for \( x \):\[ 2.5(0.583 - x) = 0.417 + x \]\[ 1.4575 - 2.5x = 0.417 + x \]\[ 1.0405 = 3.5x \]\[ x \approx 0.297 \text{ M} \]

Step 8: Calculate Equilibrium Concentrations

Now substitute \( x \approx 0.297 \text{ M} \):- [\text{C}_4\text{H}_{10}]: \( 0.583 - 0.297 \approx 0.286 \text{ M} \)- [\text{iso-C}_4\text{H}_{10}]: \( 0.417 + 0.297 \approx 0.714 \text{ M} \)

Key Concepts

Equilibrium Constant

The equilibrium constant, often symbolized by \( K \), is a numerical value that characterizes the equilibrium state of a chemical reaction. It gives us insight into the ratio of concentrations of products to reactants when the reaction reaches equilibrium. For the equilibrium constant of a reaction, the expression typically looks like this for a simple chemical equilibrium reaction:

• \( K = \frac{[ ext{Products}]}{[ ext{Reactants}]} \).

In the isomerization reaction of butane to isobutane, the equilibrium constant \( K \) is given as 2.5 at \( 25^{\circ} \text{C} \). This tells us that, at equilibrium, the concentration of isobutane will be 2.5 times that of butane. The value of \( K \) is specific to a particular reaction and changes with temperature.

Reaction Quotient

The reaction quotient, denoted as \( Q \), is similar to the equilibrium constant \( K \), but it applies to the concentrations of the reactants and products at any point in time, not just at equilibrium. To find \( Q \), follow this formula:

• \( Q = \frac{[ ext{iso-C}_4 ext{H}_{10}]_{initial}}{[ ext{C}_4 ext{H}_{10}]_{initial}} \).

Given initial concentrations, you can quickly determine if a reaction is at equilibrium by comparing \( Q \) to \( K \). In our example, the calculation shows that \( Q \approx 0.714 \), which is less than \( K = 2.5 \). This indicates that the reaction is not yet at equilibrium and will proceed in the forward direction, increasing the concentration of isobutane.

Isomerization Reaction

An isomerization reaction involves the transformation of one isomer into another, usually through a rearrangement of atoms. Isomers have the same molecular formula but differ in the arrangement of atoms. In the exercise, butane and isobutane are structural isomers, represented by the chemical reaction:

• \( \text{C}_4 ext{H}_{10} \rightleftarrows \text{iso-C}_4 ext{H}_{10} \).

This reaction is reversible, meaning it can proceed in both forward and backward directions, reaching a state of equilibrium over time. Understanding isomerization is crucial in various chemical industries, as it can influence the properties and applications of different compounds.

Equilibrium Concentration

Equilibrium concentrations are the concentrations of reactants and products present when a chemical reaction reaches equilibrium. You can determine these by applying the equilibrium constant \( K \) and using initial concentrations. Using the values from the exercise, we found initial concentrations for butane and isobutane and calculated the changes necessary to achieve equilibrium:

• For butane, start with \(0.583\, \text{M} \) and account for a decrease by \( x = 0.297\, \text{M} \), leading to an equilibrium concentration of approximately \(0.286\, \text{M} \).
• For isobutane, start with \(0.417\, \text{M} \) and add \( x = 0.297\, \text{M} \), resulting in an equilibrium concentration of approximately \(0.714\, \text{M} \).

These calculations are vital in predicting the final state of the system and understanding how adjustments in initial conditions influence equilibrium.