Question

Consider the reaction $$ 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{NO}_{2}(g) $$ At \(430^{\circ} \mathrm{C}\), an equilibrium mixture consists of 0.020 mole of \(\mathrm{O}_{2}, 0.040\) mole of \(\mathrm{NO},\) and 0.96 mole of \(\mathrm{NO}_{2} .\) Calculate \(K_{P}\) for the reaction, given that the total pressure is 0.20 atm.

Short answer

The equilibrium constant \(K_{P}\) for the reaction is 1137.76

Step-by-step solution

Calculate the total number of moles

Add up the moles of \( \mathrm{O}_{2} \), \( \mathrm{NO} \), and \( \mathrm{NO}_{2} \) to get the total number of moles present. Which will be 0.020 + 0.040 + 0.96 = 1.02 moles

Calculate the mole fractions

The mole fraction is given by the number of moles of each component divided by the total number of moles. For \( \mathrm{O}_{2} \), \( \mathrm{NO} \), and \( \mathrm{NO}_{2} \), the mole fractions will be 0.020 / 1.02 = 0.0196, 0.040 / 1.02 = 0.0392, and 0.96 / 1.02 = 0.9412, respectively.

Calculate the partial pressures

The partial pressure is given by the mole fraction multiplied by the total pressure. For \( \mathrm{O}_{2} \), \( \mathrm{NO} \), and \( \mathrm{NO}_{2} \), the partial pressures will be 0.0196 * 0.20 atm = 0.00392 atm, 0.0392 * 0.20 atm = 0.00784 atm, and 0.9412 * 0.20 atm = 0.18824 atm, respectively.

Calculate \(K_{P}\)

\(K_{P}\) is the ratio of the product of the partial pressures of the products of the reaction (or the right hand side of the equation) raised to their stoichiometric coefficients, to the product of the partial pressures of the reactants (or the left hand side of the equation) raised to their stoichiometric coefficients. So, \( K_{P} \) = \((\text{partial pressure of } \mathrm{NO}_{2})^{2} / ((\text{partial pressure of } \mathrm{NO})^{2} \times \text{partial pressure of } \mathrm{O}_{2})\), which is (0.18824)^2 / ((0.00784)^2 * 0.00392) = 1137.76

Key concepts

Chemical Equilibrium

Chemical equilibrium represents a state in a chemical reaction where the rates of the forward and reverse reactions are equal, leading to no net change in the concentrations of reactants and products over time. At this point, it might seem like the reaction has stopped, but both reactions are still occurring; they're just perfectly balanced.

In the context of the equation \(2 \mathrm{NO}(g) + \mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{NO}_{2}(g)\), when the system reaches equilibrium at \(430^\circ \mathrm{C}\), the amount of \(\mathrm{NO}\), \(\mathrm{O}_{2}\), and \(\mathrm{NO}_{2}\) stop changing. Understanding equilibrium is crucial for chemists as it helps to predict the concentrations of each species in a reaction mixture.

Partial Pressure

Partial pressure is defined as the pressure one component of a mixture of gases would exert if it were alone in a container. It's directly proportional to its mole fraction in the mixture, meaning that the more there is of a particular gas within a mixture, the higher its partial pressure will be.

Using the equilibrium mixture from the given exercise, the partial pressures are calculated using the mole fractions and the total pressure. This concept is important when dealing with gas-phase reactions, as it allows chemists to understand how the presence of different gases at particular concentrations affects reaction behavior.

Mole Fraction

The mole fraction is a way to express the concentration of a component in a mixture. It is defined as the ratio of the number of moles of the component to the total number of moles of all components in the mixture.

It is a dimensionless number and is used in the calculation of partial pressures in gas mixtures. In the example, the mole fractions of \(\mathrm{O}_{2}\), \(\mathrm{NO}\), and \(\mathrm{NO}_{2}\) provide a basis for determining how much each gas contributes to the total pressure, which is essential for calculating the equilibrium constant.

Reaction Stoichiometry

Reaction stoichiometry involves the quantitative relationship between reactants and products in a chemical reaction. It is based on the balanced chemical equation and the law of conservation of mass.

The coefficients in the balanced equation, like the 2:1:2 ratio in \(2 \mathrm{NO}(g) + \mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{NO}_{2}(g)\), inform us of the proportional relationships necessary to retain balance in a reaction. This proportionality is key to calculating the equilibrium constant, which relies on the raised powers of the partial pressures corresponding to these stoichiometric coefficients.