Question

Many hydrocarbons exist as structural isomers, which are compounds that have the same molecular formula but different structures. For example, both butane and isobutane have the same molecular formula of \(\mathrm{C}_{4} \mathrm{H}_{10}\) Calculate the mole percent of these molecules in an equilibrium mixture at \(25^{\circ} \mathrm{C}\), given that the standard free energy of formation of butane is \(-15.9 \mathrm{~kJ} / \mathrm{mol}\) and that of isobutane is \(-18.0 \mathrm{~kJ} / \mathrm{mol}\). Does your result support the notion that straight-chain hydrocarbons (i.e. hydrocarbons in which the \(\mathrm{C}\) atoms are joined along a line) are less stable than branch-chain hydrocarbons?

Short answer

Butane: 30%, Isobutane: 70%. Branch-chain hydrocarbons are more stable.

Step-by-step solution

Define the Reaction

Write the equation for the equilibrium between butane and isobutane. Here, \(\mathrm{C}_4\mathrm{H}_{10} (\text{butane}) \rightleftharpoons \mathrm{C}_4\mathrm{H}_{10} (\text{isobutane})\).

Calculate ΔG° for the Reaction

The standard free energy change \( \Delta G^\circ \) for the reaction can be calculated as follows: \[ \Delta G^\circ = G_f^\circ(\text{isobutane}) - G_f^\circ(\text{butane}) = (-18.0 \text{ kJ/mol}) - (-15.9 \text{ kJ/mol}) = -2.1 \text{ kJ/mol}. \]

Calculate the Equilibrium Constant (K)

Use the formula \( \Delta G^\circ = -RT \ln K \) to find \( K \). Convert \( \Delta G^\circ \) to J/mol (\(-2100 \text{ J/mol}\)). The gas constant \( R = 8.314 \text{ J/(mol·K)} \), and \( T = 298 \text{ K} \) for \(25^{\circ} \mathrm{C}\). \[ \Delta G^\circ = -RT \ln K \Rightarrow -2100 = -(8.314)(298) \ln K \]\[ \ln K = \frac{-2100}{-2477.372} \approx 0.847 \]\[ K = e^{0.847} \approx 2.333. \]

Determine Mole Percent in Mixture

The equilibrium constant \( K \) relates to the mole fraction of butane \((x)\) and isobutane \((1-x)\) as follows: \[ K = \frac{1-x}{x} \Rightarrow 2.333 = \frac{1-x}{x}. \]Solving for \( x \) gives:\[ 2.333x = 1-x \Rightarrow 3.333x = 1 \Rightarrow x = \frac{1}{3.333} \approx 0.30. \]Mole fraction of isobutane \((1-x)\) is approximately \(0.70\).

Express as Percentages

Convert the mole fractions to percentages:- Butane: \( x = 0.30 \times 100 = 30\% \) - Isobutane: \( 1-x = 0.70 \times 100 = 70\% \)

Analyze Stability

The result shows that isobutane composes 70% of the mixture, indicating it is more stable compared to butane, which composes 30%. This supports the notion that branch-chain hydrocarbons are more stable than straight-chain hydrocarbons.

Key concepts

Standard Free Energy

In the world of chemical reactions and equilibria, the concept of standard free energy is crucial to understand. It tells us how much energy is available to do work in a system at constant temperature and pressure. The standard free energy change, denoted as \( \Delta G^\circ \), is an indicator of the reaction's spontaneity. A negative \( \Delta G^\circ \) means the reaction is likely to occur spontaneously, while a positive value suggests it is not.

When we examine structural isomers like butane and isobutane, the standard free energy values for their formation shine a light on their relative stability. In our original exercise, butane had a \( \Delta G^\circ_f \) of \(-15.9 \text{ kJ/mol}\) while isobutane had \(-18.0 \text{ kJ/mol}\). This makes isobutane more stable, as its formation releases more energy, passing on this benefit to the equilibrium mixture as well.

• The more negative the \( \Delta G^\circ \), the more energy-efficient the process.
• In our exercise, \( \Delta G^\circ \) for the butane-isobutane reaction was calculated as \(-2.1 \text{ kJ/mol}\).

Understanding these energy changes helps predict how molecules interact at equilibrium.

Equilibrium Constant

The equilibrium constant, \( K \), is the ratio of the concentration of products to reactants at equilibrium. It reflects which direction a reaction favors. In our case, the equilibrium constant helps us determine whether butane or isobutane will be more prevalent in an equilibrium mixture.

To find \( K \), we used the equation \( \Delta G^\circ = -RT \ln K \). Given the values from our reaction, we calculated \( \ln K \approx 0.847 \), thus deriving \( K \approx 2.333 \). This value ultimately tells us more isobutane forms at equilibrium because the reaction favors the formation of the more stable isomer.

• \( K > 1 \) indicates products are favored at equilibrium.
• A high \( K \) for isobutane implies it forms predominantly.

Understanding \( K \) helps predict the composition of a chemical mixture, which is crucial for industrial applications and laboratory syntheses.

Mole Fraction Calculation

Mole fraction is a way of expressing the concentration of a molecule in a mixture. It is simply the number of moles of a component divided by the total number of moles in the mixture. In our exercise, we calculated the mole fractions of butane and isobutane to determine their equilibrium distribution.

Given the equilibrium constant \( K \) derived from previous calculations, \( K = \frac{1-x}{x} \), where \( x \) represents the mole fraction of butane. By solving this equation, we found \( x = 0.30 \), meaning butane comprises 30% of the mixture. Consequently, isobutane, with mole fraction \( 1-x \), makes up about 70%.

• Mole fractions provide insight into the proportions of substances in a mixture.
• Converting to percentages makes it easier to visualize the composition.

Using these calculations, we confirmed that isobutane is more prevalent, supporting the idea that branched hydrocarbons are usually more stable than their straight-chained counterparts.