Question

A reaction mixture that consisted of \(0.20 \mathrm{~mol} \mathrm{} \mathrm{N}_{2}\) and \(0.20 \mathrm{~mol} \mathrm{} \mathrm{H}_{2}\) was introduced into a \(25.0\) - \(\mathrm{L}\) reactor and heated. At equilibrium, \(5.0 \%\) of the nitrogen gas had reacted. What is the value of the equilibrium constant \(K_{c}\) for the reaction \(\mathrm{N}_{2}(\mathrm{~g})+\) \(3 \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_{3}(\mathrm{~g})\) at this temperature?

Short answer

The equilibrium constant \(K_c\) is approximately 1.7.

Step-by-step solution

Calculate Moles of N2 Reacted

Initially there were 0.20 mol of N2. If 5.0% of N2 reacted, we find the amount of N2 that has reacted by multiplying the initial moles of N2 by 0.05 (5%). \(0.20 \, \text{mol} \times 0.05 = 0.01 \, \text{mol} \text{ of } N_2\)

Calculate the Moles of N2 and H2 Left and the Moles of NH3 Formed

Subtract the moles of N2 that reacted from the initial moles to find the moles of N2 left at equilibrium and multiply the moles of N2 reacted by 3 to find the moles of H2 reacted. The stoichiometry of the reaction tells us that two moles of NH3 are formed for every mole of N2 reacted.Moles of N2 left: \(0.20 - 0.01 = 0.19 \, \text{mol} \text{ of } N_2 \) Moles of H2 left: \(0.20 - 0.01 \times 3 = 0.17 \, \text{mol} \text{ of } H_2 \) Moles of NH3 formed: \(0.01 \times 2 = 0.02 \, \text{mol} \text{ of } NH_3 \)

Calculate the Concentrations of Gases at Equilibrium

The concentration of each gas is the number of moles of that gas divided by the volume of the reactor (25.0 L). Use this to find the concentrations of N2, H2, and NH3 at equilibrium.\([N_2] = \frac{0.19 \, \text{mol}}{25.0 \, \text{L}}\), \([H_2] = \frac{0.17 \, \text{mol}}{25.0 \, \text{L}}\), \([NH_3] = \frac{0.02 \, \text{mol}}{25.0 \, \text{L}}\)

Write the Equilibrium Expression and Calculate Kc

The equilibrium constant, \(K_c\), is the ratio of the product concentrations to the reactant concentrations, each raised to the power of their stoichiometric coefficients.\[K_c = \frac{[NH_3]^2}{[N_2][H_2]^3}\]Plugging the concentrations into the equilibrium expression:\[K_c = \frac{(0.02/25.0)^2}{(0.19/25.0) \times (0.17/25.0)^3}\]Calculate this value to find \(K_c\).

Key concepts

Chemical Equilibrium

At its core, chemical equilibrium is a state in a reversible chemical reaction where the rate of the forward reaction equals the rate of the reverse reaction. Consequently, the concentrations of the reactants and products remain constant over time, although they are not equal. This point of dynamic balance is crucial to understanding how reactions proceed under specific conditions.

Understanding the law of mass action is essential. It indicates that at equilibrium, a particular ratio of concentration terms raised to the power of their coefficients in the balanced equation gives us a constant value known as the equilibrium constant. This constant is represented by the symbol K or K_c when it pertains to concentration. Different reactions have unique equilibrium constants that can tell scientists a lot about the reaction conditions, including whether the products or reactants are favored in the equilibrium mix.

Reaction Stoichiometry

Another vital concept in the context of equilibrium is reaction stoichiometry. It involves the quantitative relationship between the amounts of reactants used and products formed in a chemical reaction. The coefficients of each substance in the balanced equation provide the mole ratio, which plays an integral part in calculations.

Our example problem illustrates this. It showcased that for every one mole of N₂ that reacts, three moles of H₂ are consumed, and two moles of NH₃ are produced. Understanding this ratio is crucial when trying to calculate the amount of product from a known amount of reactant, as well as in determining the remaining reactants at equilibrium. Balancing this accounting of atoms and molecules within a reaction is a foundational skill in chemistry.

Equilibrium Concentration

The equilibrium concentration refers to the concentration of each reactant and product in a chemical system at equilibrium. Determining these concentrations is a common task when working with chemical equilibria. As with our problem, the concentrations are not simply the initial amounts, because you must account for how the reaction has progressed.

To calculate the equilibrium concentrations, we need to know the reaction volume and the amounts of reactants that have turned into products, which we then convert into molar concentration (moles per liter). In practice, we can use the concentrations to compute the equilibrium constant, which provides a concise quantitative description of the system's equilibrium position. High precision in these calculations is pivotal, as even small errors can lead to an incorrect understanding of the system behavior.