Question

You have containers of pure \(\mathrm{O}_{2}\) and \(\mathrm{N}_{2}\) at \(298 \mathrm{K}\) and 1 atm pressure. Calculate \(\Delta G_{\text {mixing}}\) relative to the unmixed gases of a. a mixture of \(10 .\) mol of \(\mathrm{O}_{2}\) and \(10 .\) mol of \(\mathrm{N}_{2}\) b. a mixture of \(10 .\) mol of \(\mathrm{O}_{2}\) and \(20 .\) mol of \(\mathrm{N}_{2}\) c. Calculate \(\Delta G_{\text {mixing}}\) if \(10 .\) mol of pure \(\mathrm{N}_{2}\) is added to the mixture of \(10 .\) mol of \(\mathrm{O}_{2}\) and \(10 .\) mol of \(\mathrm{N}_{2}\)

Short answer

The change in Gibbs free energy (\(\Delta G_{\text {mixing}}\)) for the different mixtures are as follows: a. Mixture of 10 mol O\(_2\) and 10 mol N\(_2\): \(\Delta G_{\text {mixing}} = -5700.16 J\) b. Mixture of 10 mol O\(_2\) and 20 mol N\(_2\): \(\Delta G_{\text {mixing}} = -5082.90 J\) c. Addition of 10 mol N\(_2\) to the initial mixture of 10 mol O\(_2\) and 10 mol N\(_2\): \(\Delta G_{\text {mixing}} (step) = 617.26 J\)

Step-by-step solution

For each mixture, we need to find the mole fraction (\(x_i\)) of O\(_2\) and N\(_2\). Mole fraction is given by the ratio of the moles of a component to the total number of moles in the mixture. a. Mixture of 10 mol O\(_2\) and 10 mol N\(_2\): \(x_{O_2} = \frac{10}{10 + 10} = \frac{1}{2}\) \(x_{N_2} = \frac{10}{10 + 10} = \frac{1}{2}\) b. Mixture of 10 mol O\(_2\) and 20 mol N\(_2\): \(x_{O_2} = \frac{10}{10 + 20} = \frac{1}{3}\) \(x_{N_2} = \frac{20}{10 + 20} = \frac{2}{3}\) #Step 2: Plug mole fractions into the equation for ΔGmixing#

Now that we have calculated the mole fractions for each mixture, we can plug these values into the equation for ΔGmixing. \(\Delta G_{\text {mixing}} = nRT\left(\sum_{i = 1}^{c} x_i\ln{x_i}\right)\), where \(n\) is the total number of moles, \(R = 8.314 \frac{J}{mol \cdot K}\), and \(T = 298 K\). a. Mixture of 10 mol O\(_2\) and 10 mol N\(_2\): \[\Delta G_{\text {mixing}} = (10 + 10)(8.314)(298)\left(\frac{1}{2}\ln{\frac{1}{2}} + \frac{1}{2}\ln{\frac{1}{2}}\right) = -5700.16 J\] b. Mixture of 10 mol O\(_2\) and 20 mol N\(_2\): \[\Delta G_{\text {mixing}} = (10 + 20)(8.314)(298)\left(\frac{1}{3}\ln{\frac{1}{3}} + \frac{2}{3}\ln{\frac{2}{3}}\right) = -5082.90 J\] c. Calculate ΔGmixing when 10 mol of pure N\(_2\) is added to the mixture of 10 mol O\(_2\) and 10 mol N\(_2\): 1. First calculate the mole fractions for the final mixture: \(x_{O_2} = \frac{10}{10+10+10} = \frac{1}{3}\) \(x_{N_2} = \frac{20}{10+10+10} = \frac{2}{3}\) 2. Determine ΔGmixing for the final mixture: \[\Delta G_{\text {mixing}} = (10+10+10)(8.314)(298)\left(\frac{1}{3}\ln{\frac{1}{3}} + \frac{2}{3}\ln{\frac{2}{3}}\right) = -5082.90 J\] 3. Determine ΔGmixing for the step of adding 10 mol N\(_2\) to the initial mixture of 10 mol O\(_2\) and 10 mol N\(_2\): \[\Delta G_{\text {mixing}} (step) = -5082.90 J - (-5700.16 J) = 617.26 J \]

Key concepts

Mole Fraction

The mole fraction is a way of expressing the concentration of a component in a mixture. It is defined as the ratio of the number of moles of a given component to the total number of moles of all components in the mixture. It is represented by the symbol \( x_i \) for a component \( i \). For instance, if you have a mixture of gases A and B, with moles \( n_A \) and \( n_B \) respectively, then the mole fraction of gas A is \( x_A = \frac{n_A}{n_A + n_B} \), and similarly for gas B.

Mole fraction is a dimensionless quantity and always takes a value between 0 and 1. It's critical in determining the properties of mixtures in chemical thermodynamics. For example, in the ideal gas mixtures, the partial pressure of each gas can be calculated by multiplying its mole fraction by the total pressure of the mixture. Understanding and calculating the mole fraction is essential in the study of mixtures and solutions in various scientific fields such as chemistry, biology, and materials science.

Chemical Thermodynamics

Chemical thermodynamics is the branch of physical chemistry that deals with the interrelation of energy transfers and chemical reactions. Its core is to predict the energy changes that accompany chemical processes. One of the critical aspects of chemical thermodynamics is the concept of Gibbs free energy, denoted by \( \Delta G \).

Gibbs free energy is a thermodynamic potential that measures the 'useful' work obtainable from a thermodynamic system at a constant temperature and pressure. It combines the concepts of enthalpy (heat content), entropy (the degree of disorder), and temperature to predict the direction of chemical reactions and the balance between reactants and products at equilibrium.

In the context of mixing gases or any substances, the change in Gibbs free energy (\( \Delta G_{\text{mixing}} \)) can be used to determine whether a mixture will be spontaneous under certain conditions. If \( \Delta G_{\text{mixing}} \) is negative, the process of mixing is spontaneous, meaning it will occur without additional energy input.

Ideal Gas Mixture

An ideal gas mixture is an assumption used in chemical thermodynamics when dealing with the behavior of gas mixtures. This concept simplifies the calculations by assuming that the gases do not interact with each other and that the volume occupied by the gases themselves is negligible compared to the container volume. It's an extension of the ideal gas law to mixtures.

According to Dalton's Law of Partial Pressures, in an ideal gas mixture, the total pressure exerted is equal to the sum of the partial pressures of individual gases. Each gas in the mixture behaves independently, and its partial pressure can be calculated using its mole fraction. This principle allows us to determine the overall behavior of the mixture from knowledge of its components.

In the given exercise, we used the fact that the gases behave ideally to calculate the Gibbs free energy of mixing. The equation applied incorporates the mole fractions of the gases and the ideal gas constant, reflecting the theoretical principles of an ideal gas mixture.