Question

The equilibrium constant for the butane \(\rightleftarrows\) isobutane equilibrium at \(25^{\circ} \mathrm{C}\) is \(2.50 .\) Calculate \(\Delta_{\mathrm{r}} G^{\circ}\) at this temperature in units of \(\mathrm{kJ} / \mathrm{mol}\).

Short answer

\(-2.27 \mathrm{kJ} / \mathrm{mol}\)

Step-by-step solution

Understanding the Problem

We need to calculate the standard Gibbs free energy change, \( \Delta_{\mathrm{r}} G^\circ \), for the equilibrium between butane and isobutane. We are given the equilibrium constant \( K = 2.50 \) at \( 25^{\circ} \mathrm{C} \).

Gibbs Free Energy Equation

The standard Gibbs free energy change \( \Delta_{\mathrm{r}} G^\circ \) is related to the equilibrium constant \( K \) by the equation:\[ \Delta_{\mathrm{r}} G^\circ = -RT \ln K \]Where \( R \) is the universal gas constant \( 8.314 \frac{J}{mol\,K} \) and \( T \) is the absolute temperature in Kelvin.

Convert Temperature to Kelvin

The temperature given is \( 25^{\circ} \mathrm{C} \). Convert it to Kelvin by adding 273.15:\[ T = 25 + 273.15 = 298.15\,K \]

Calculate \( \ln K \)

Calculate the natural logarithm of the equilibrium constant \( K \):\[ \ln K = \ln(2.50) \approx 0.9163 \]

Substitute Values and Solve

Substitute \( R = 8.314 \frac{J}{mol\,K} \), \( T = 298.15\,K \), and \( \ln K \) into the Gibbs free energy equation:\[ \Delta_{\mathrm{r}} G^\circ = - (8.314 \frac{J}{mol\,K})(298.15\,K)(0.9163) \]Calculate\[ \Delta_{\mathrm{r}} G^\circ = - 2272.92\, J/mol \]Convert to \( kJ/mol \) by dividing by 1000:\[ \Delta_{\mathrm{r}} G^\circ = -2.27\, kJ/mol \]

Conclusion

The standard Gibbs free energy change for the reaction at \( 25^{\circ} \mathrm{C} \) is approximately \( -2.27 \mathrm{kJ} / \mathrm{mol} \).

Key concepts

Equilibrium Constant

The equilibrium constant, often symbolized as \( K \), plays a crucial role in understanding chemical equilibria. It represents the ratio of the concentration of products to reactants, each raised to their stoichiometric coefficients, at equilibrium. For a reaction involving butane and isobutane, the equilibrium constant tells us how much of each is present when the system has reached equilibrium.

In this context, \( K = 2.50 \) indicates a greater tendency for the formation of isobutane from butane. This value shows that at a given temperature, this is the balance at which the forward and backward reaction rates are equal. So, a higher \( K \) value generally means the reaction favors product formation more than reactants at equilibrium. Understanding the value of the equilibrium constant is key to predicting how a reaction behaves and is essential when calculating changes in Gibbs free energy.

Temperature Conversion

Temperature conversion is an important step when working with equations that involve absolute temperatures, such as those in thermodynamics and chemical kinetics. In these calculations, the common Celsius temperature scale must be converted to the Kelvin scale to ensure consistency in units.

The reason for using Kelvin is because it is the absolute temperature scale suitable for scientific calculations, where 0 Kelvin is absolute zero, the point at which particles have minimal thermal motion. The formula to convert from Celsius to Kelvin is simple: add 273.15 to the Celsius temperature.

In our problem, the given temperature is 25°C. Converting this to Kelvin:

• 25 + 273.15 = 298.15 K

This results in 298.15 K as the absolute temperature to use in further calculations involving Gibbs free energy.

Natural Logarithm

The natural logarithm, represented as \( \ln \), is a logarithm to the base \( e \), where \( e \) is approximately 2.718. This mathematical function is a vital tool in many scientific and engineering domains, providing a way to scale and linearize diverse exponential relationships such as those found in growth and decay processes.

In chemical thermodynamics, the natural logarithm comes into play when relating the equilibrium constant to Gibbs free energy. The equation \( \Delta_{\mathrm{r}} G^\circ = -RT \ln K \) shows its integration in calculating how much energy is needed for a reaction to occur. To utilize this logarithm:

• Calculate \( \ln K \) for the known equilibrium constant.
• In our exercise, \( K = 2.50 \), so \( \ln K \approx 0.9163 \).

The natural logarithm helps translate the equilibrium constant into a quantifiable energy change.

Universal Gas Constant

The universal gas constant, denoted as \( R \), is a fundamental constant in the ideal gas law equation \( PV = nRT \). Its value is \( 8.314 \frac{J}{mol \cdot K} \), providing a link between the macroscopic observations of temperature, pressure, and volume in gases, and their chemical potential energy on the microscopic scale.

\( R \) frequently appears in thermodynamic equations, including the relationship between Gibbs free energy and the equilibrium constant. In the equation \( \Delta_{\mathrm{r}} G^\circ = -RT \ln K \), \( R \) allows us to calculate energy changes across reactions at different temperatures.

Using this constant ensures our calculations are based on standardized energy units, aligning with the Kelvins used in temperature and the Joules in energy. This bridges the gap between the energy changes observed in chemical reactions and the thermal energy distributions described by gas laws.