Problem 50 - Chemistry The Molecular Science

Question

The hydrocarbon \(\mathrm{C}_{4} \mathrm{H}_{10}\) can exist in two gaseous forms: butane and 2 -methylpropane. The value of \(K_{\mathrm{c}}\) for conversion of butane to 2 -methylpropane is 2.5 at \(25^{\circ} \mathrm{C}\). CCCCC=CC(C)(C)CO (a) Suppose that the initial concentrations of butane and 2-methylpropane are each \(0.100 \mathrm{~mol} / \mathrm{L}\). Make up a table of initial concentrations, change in concentrations, and equilibrium concentrations for this reaction. (b) Write the equilibrium constant expression in terms of \(x\), the change in the concentration of butane, and then solve for \(x\). (c) If you place 0.017 mol butane in a \(0.50-\mathrm{L}\) sealed flask at \(25^{\circ} \mathrm{C},\) calculate the equilibrium concentration of each isomer.

Step-by-Step Solution

Answer

Butane: 0.0097 mol/L, 2-Methylpropane: 0.0243 mol/L.

Step 1: Initial Concentration Table

First, let's consider the reaction: \( \text{Butane} \rightleftharpoons \text{2-Methylpropane} \).- Initial Concentrations: Butane = 0.100 mol/L, 2-Methylpropane = 0.100 mol/L.- Change in Concentrations: Let the change in concentration for butane be \( -x \) and for 2-Methylpropane be \( +x \).- Equilibrium Concentrations: Butane = 0.100 - x, 2-Methylpropane = 0.100 + x.

Step 2: Equilibrium Expression

The equilibrium constant expression for the reaction is:\[K_c = \frac{[\text{2-Methylpropane}]}{[\text{Butane}]} = \frac{0.100 + x}{0.100 - x}\]Given that \( K_c = 2.5 \), we can set up the equation:\[2.5 = \frac{0.100 + x}{0.100 - x}\]

Step 3: Solving for x

Solve \( 2.5 = \frac{0.100 + x}{0.100 - x} \):1. Multiply both sides by \( 0.100 - x \): \( 2.5(0.100 - x) = 0.100 + x \) 2. Expand and simplify: \( 0.250 - 2.5x = 0.100 + x \) 3. Rearrange terms: \( 0.250 - 0.100 = x + 2.5x \) 4. Solve for \( x \): \( 0.150 = 3.5x \) \( x = \frac{0.150}{3.5} \approx 0.0429 \) mol/L.

Step 4: Equilibrium Concentrations

Using \( x = 0.0429 \) mol/L:- Equilibrium concentration of butane: \( [\text{Butane}] = 0.100 - 0.0429 = 0.0571 \) mol/L.- Equilibrium concentration of 2-Methylpropane: \( [\text{2-Methylpropane}] = 0.100 + 0.0429 = 0.1429 \) mol/L.

Step 5: Calculate Concentrations in Flask

For the flask:- Given initial moles of butane = 0.017 mol in a 0.50 L flask: \([\text{Butane}]_0 = \frac{0.017}{0.50} = 0.034 \text{ mol/L}\)- The equilibrium expression is: \( K_c = \frac{[\text{2-Methylpropane}]}{[\text{Butane}]} = 2.5 \) Assume change is \( x \): Equilibrium concentrations: - Butane: \( 0.034 - x \) - 2-Methylpropane: \( x \)- Substitute in terms: \( 2.5 = \frac{x}{0.034 - x} \)- Solve for \( x \) as before: Multiply both sides: \( 2.5(0.034 - x) = x \) Solve: \( 0.085 - 2.5x = x \) \( 0.085 = 3.5x \) \( x = \frac{0.085}{3.5} \approx 0.0243 \text{ mol/L} \)- Equilibrium concentrations: - Butane: \( 0.034 - 0.0243 = 0.0097 \text{ mol/L} \) - 2-Methylpropane: \( 0.0243 \text{ mol/L} \)

Key Concepts

Equilibrium Constant

The equilibrium constant, denoted as \(K_c\), is a crucial concept in chemical equilibrium. It provides a measure of the concentration of products and reactants when a reaction reaches equilibrium. For the gaseous isomerization between butane and 2-methylpropane, the equilibrium constant expression is formulated as: \[ K_c = \frac{[\text{2-Methylpropane}]}{[\text{Butane}]} \] This equation indicates that the ratio of the concentration of 2-methylpropane to butane remains constant at equilibrium. The value of \(K_c = 2.5\) at \(25^{\circ} \mathrm{C}\) implies that at equilibrium, 2-methylpropane is favored. Understanding \(K_c\) helps us predict the direction of the reaction and calculate the equilibrium concentrations when starting concentrations are given.

Reaction Quotient

The reaction quotient, \(Q\), serves as a tool to predict the direction in which a reaction will proceed to reach equilibrium. It is calculated using the same expression as the equilibrium constant, but with the initial concentrations of reactions and products. For the conversion of butane to 2-methylpropane, \[ Q = \frac{[\text{2-Methylpropane}]_{initial}}{[\text{Butane}]_{initial}} \] When \(Q = K_c\), the system is already at equilibrium. However, if \(Q < K_c\), the reaction will move forward to produce more products to reach equilibrium. Conversely, if \(Q > K_c\), the reaction will shift towards the reactants. Understanding the reaction quotient helps in anticipating changes in concentration before equilibrium is achieved.

Isomerization

Isomerization is a process where one molecule is transformed into another molecule with the same atoms but in a different arrangement. In the case of butane and 2-methylpropane, they are structural isomers. This means they have the same molecular formula, \(\text{C}_4\text{H}_{10}\), but different connectivity of their atoms. Isomerization reactions are important as they commonly occur in chemical processes, especially under certain conditions like temperature. Here, isomerization takes place in a gaseous state, and the equilibrium constant indicates which isomer is favored at \(25^{\circ} \mathrm{C}\). These reactions may involve a low amount of energy, making them practically significant in industrial chemical processes.

Reaction Stoichiometry

Reaction stoichiometry involves the quantitative relationships between reactants and products in a chemical reaction. In this exercise, it applies to the isomerization of butane to 2-methylpropane where the coefficients of the reaction are omitted, indicating a 1:1 ratio because both species are in their simplest form. This stoichiometric relationship allows us to use changes in concentration, denoted as \(x\), to calculate equilibrium concentrations. For instance, when \(x = 0.0429\ \text{mol/L}\) is determined as the change in butane's concentration, it directly relates to the increase in 2-methylpropane's concentration. Understanding stoichiometry is essential for solving many types of equilibrium problems because it connects the theoretical principles of chemistry to practical example solving through a balanced approach.