Question

The equilibrium constant \(K_{\mathrm{c}}\) for the reaction: $$2 \mathrm{NH}_{3}(g) \rightleftarrows \mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g)$$ is 0.83 at \(375^{\circ} \mathrm{C}\). A \(14.6-\mathrm{g}\) sample of ammonia is placed in a 4.00-L flask and heated to \(375^{\circ} \mathrm{C}\). Calculate the concentrations of all the gases when equilibrium is reached.

Short answer

[NH3] = 0.094 M, [N2] = 0.060 M, [H2] = 0.180 M at equilibrium.

Step-by-step solution

Convert Mass to Moles

Calculate the moles of ammonia (\(\text{NH}_3\)) using its molar mass (17.03 g/mol). The moles of ammonia is given by: \[\text{moles of } \text{NH}_3 = \frac{14.6 \text{ g}}{17.03 \text{ g/mol}} = 0.857 \text{ mol}\]

Initial Concentration of \( \text{NH}_3 \)

Calculate the initial concentration of \(\text{NH}_3\) in the 4.00 L flask: \[[\text{NH}_3]_0 = \frac{0.857 \text{ mol}}{4.00 \text{ L}} = 0.214 \text{ M}\]

Define Change in Terms of x

Let the change in concentration of \(\text{NH}_3\) that reacts be \(2x\). Thus, the change in concentration for \(\text{N}_2\) is \(x\), and for \(\text{H}_2\), it is \(3x\).

Express Equilibrium Concentrations

Express the equilibrium concentrations of all species:- \([\text{NH}_3] = 0.214 - 2x\)- \([\text{N}_2] = x\)- \([\text{H}_2] = 3x\)

Write Equilibrium Expression

The equilibrium expression for the given reaction is:\[K_c = \frac{[\text{N}_2][\text{H}_2]^3}{[\text{NH}_3]^2}\]

Substitute and Solve for x

Substitute the equilibrium concentrations into the equilibrium expression and solve for \(x\):\[0.83 = \frac{(x)(3x)^3}{(0.214 - 2x)^2}\]Simplify and solve this equation for \(x\) by trial or iteration or using a calculator. This leads to: \(x \approx 0.060\).

Calculate Equilibrium Concentrations

Substitute \(x = 0.060\) back into the expressions for equilibrium concentrations:- \([\text{NH}_3] = 0.214 - 2(0.060) = 0.094\, \text{M}\)- \([\text{N}_2] = 0.060\, \text{M}\)- \([\text{H}_2] = 3(0.060) = 0.180\, \text{M}\)

Key concepts

Chemical Equilibrium

Chemical equilibrium refers to the state of a chemical reaction where the concentrations of reactants and products do not change over time. This happens when the rates of the forward and reverse reactions are equal. In the given reaction of ammonia dissociating into nitrogen and hydrogen gases, the equilibrium is crucial. Understanding equilibrium helps to predict the concentrations of each species at a given temperature. To determine when equilibrium is reached, scientists use the equilibrium constant, denoted as \(K_c\). This constant provides insights into the ratio of the concentrations of products and reactants at equilibrium. It tells us how far the reaction proceeds before reaching stability. If \(K_c = 0.83\), it implies the reaction slightly favors the reactants at \(375^\circ C\), as \(K_c\) values less than 1 indicate that the equilibrium position leans towards the side of the reactants.

Reaction Stoichiometry

Reaction stoichiometry involves the calculation of the quantities of reactants and products in a chemical reaction. For the ammonia decomposition reaction, the stoichiometry can be seen from the balanced equation: \(2\, \text{NH}_3(g) \rightleftharpoons \text{N}_2(g) + 3\, \text{H}_2(g)\). This equation tells us several vital things: for every 2 moles of ammonia that decompose, 1 mole of nitrogen and 3 moles of hydrogen are produced. Understanding stoichiometry allows us to set up equations that describe the changes in concentration as the reaction progresses towards equilibrium. Specifically, the coefficients (2 for \(\text{NH}_3\), 1 for \(\text{N}_2\), and 3 for \(\text{H}_2\)) are used to relate the change in the amount of each species during the reaction.

Concentration Calculations

Calculating concentrations is a fundamental part of analyzing reactions at equilibrium. It usually starts by converting any given quantities into moles, as seen with ammonia in this problem. Once moles are known, concentrations in molarity (M, moles per liter) can be determined by dividing the moles by the volume of the reaction container (4.00 L in this example). This calculation gives the initial concentration of \(\text{NH}_3\) as \(0.214\, \text{M}\).

When reactions progress toward equilibrium, concentrations change according to the stoichiometry of the reaction: here, \(\text{NH}_3\)'s concentration decreases by \(2x\), while \(\text{N}_2\)'s concentration increases by \(x\) and \(\text{H}_2\)'s concentration increases by \(3x\). By setting up the equilibrium expressions, these varying concentrations over the reaction course can be precisely determined. Then, inserting these values into the equilibrium expression allows the calculation of the actual equilibrium concentrations.

Gas Reactions

Gas reactions, like the ammonia decomposition, involve reactants and products in the gaseous state. One distinguishing feature of gas reactions is the role pressure and volume play alongside concentration. As gases, the amount of substance in a defined space directly influences concentrations, hence impacting the position of the equilibrium. The ideal gas law \(PV = nRT\) is often intrinsically connected to such analyses, though it's not directly used here.

Gaseous reactions often show changes in the number of moles as the reaction proceeds— visible here as ammonia decomposes into more moles of nitrogen and hydrogen gases. This change affects the equilibrium and, thus, the calculation of \(K_c\). Understanding these dynamics helps in manipulating conditions to drive reactions in a desired direction, a critical aspect in industrial chemical processes involving gases.