Question

Two moles of gaseous \(\mathrm{NH}_{3}\) are introduced into a 1.0-liter vessel and allowed to undergo partial decomposition at high temperature according to the reaction $$ 2 \mathrm{NH}_{3}(\mathrm{~g}) \rightleftarrows \mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) $$ At equilibrium, \(1.0\) mole of \(\mathrm{NH}_{3}(\mathrm{~g})\) remains. What is the value of the equilibrium constant?

Short answer

The value of the equilibrium constant for the given reaction is \(K_c = \frac{27}{8}\).

Step-by-step solution

Write down the balanced equation

The balanced equation is already given: \[2 NH_3(g) \rightleftarrows N_2(g) + 3 H_2(g)\]

Determine the initial moles of each substance

We are given that initially, there are two moles of gaseous ammonia (\(NH_3\)) in a 1-liter vessel. There are zero moles of \(N_2(g)\) and \(H_2(g)\) to begin with.

Calculate the change in moles during the reaction

Since at equilibrium, there is 1 mole of ammonia left, and initially, we have 2 moles of ammonia, the change in moles for ammonia is: \[\Delta n_{NH_3} = -1\] Now, we need to use stoichiometry to determine the change in moles for nitrogen and hydrogen gases: \[\Delta n_{N_2} = \frac{1}{2} \Delta n_{NH_3}\] \[\Delta n_{H_2} = \frac{3}{2} \Delta n_{NH_3}\] Now substituting the value of \(\Delta n_{NH_3}\), we get: \[\Delta n_{N_2} = \frac{1}{2}(-1) = -\frac{1}{2}\] \[\Delta n_{H_2} = \frac{3}{2}(-1) = -\frac{3}{2}\] Since these values are negative, it means that the moles of nitrogen and hydrogen gases have increased by \(\frac{1}{2}\) and \(\frac{3}{2}\), respectively.

Determine the moles at equilibrium

We can now calculate the equilibrium moles for all species: \[n_{NH_3}^{eq} = n_{NH_3}^{initial} + \Delta n_{NH_3} = 2 - 1 = 1\] \[n_{N_2}^{eq} = n_{N_2}^{initial} - \Delta n_{N_2} = 0 - (- \frac{1}{2}) = \frac{1}{2}\] \[n_{H_2}^{eq} = n_{H_2}^{initial} - \Delta n_{H_2} = 0 - (- \frac{3}{2})= \frac{3}{2}\]

Calculate the equilibrium concentrations

Since the volume of the vessel is \(1.0\) L, the equilibrium concentrations are equal to the equilibrium moles since concentration = moles/volume: \[ [NH_3]_{eq} = 1 M \] \[ [N_2]_{eq} = \frac{1}{2} M \] \[ [H_2]_{eq} = \frac{3}{2} M \]

Write the expression for the equilibrium constant and solve

The equilibrium constant, \(K_c\), can be defined for this reaction as: \[K_c = \frac{[N_2]_{eq}[H_2]_{eq}^3}{[NH_3]_{eq}^2}\] Now substitute the equilibrium concentrations obtained in Step 5: \[K_c = \frac{(\frac{1}{2})(\frac{3}{2})^3}{(1)^2}\] \[K_c = \frac{27}{8}\] So, the equilibrium constant for the given reaction is: \[K_c=\boxed{\frac{27}{8}}\]

Key concepts

Equilibrium Constant

In the world of chemistry, an equilibrium constant, often denoted as \(K_c\), is a number that gets evaluated when a chemical reaction reaches a point where the rate of the forward reaction equals that of the reverse reaction. When the reaction between ammonia gas \((NH_3)\), nitrogen gas \((N_2)\), and hydrogen gas \((H_2)\) is analyzed, the equilibrium constant provides a way to represent the ratio of product concentrations to reactant concentrations at equilibrium. This can be mathematically expressed using the equation:
\[K_c = \frac{[N_2]_{eq}[H_2]_{eq}^3}{[NH_3]_{eq}^2}\]When all concentrations are in equilibrium and substituted into the equation, an equilibrium constant of\(\frac{27}{8}\) arises.
This numerical value offers insights into which direction the chemical reaction will favor, either forming more products or retaining more reactants. A larger \(K_c\) often indicates a reaction that proceeds extensively toward products. Understanding \(K_c\) helps chemists predict how a reaction behaves under various conditions, thereby assisting in controlling chemical processes.

Reaction Stoichiometry

When examining the ammonia decomposition reaction, stoichiometry involves analyzing the quantitative relationships among the reactants and products. The balanced chemical equation for this reaction is:
\[2 NH_3(g) \rightleftarrows N_2(g) + 3 H_2(g)\]From this, one can deduce how molecules interact with one another: two molecules of ammonia decompose to form one nitrogen molecule and three hydrogen molecules.
This stoichiometric relationship is at the core of calculating how moles change throughout the reaction's progression. By applying stoichiometry, the change in moles of ammonia can be transformed into changes for nitrogen and hydrogen using the conversion factors derived from their coefficients in the balanced equation.
Predicting these changes is vital for understanding how the reaction approaches and maintains equilibrium, ultimately leading to the calculation of equilibrium concentrations.

Ammonia Decomposition

Ammonia decomposition is a process where ammonia \((NH_3)\) breaks down into nitrogen \((N_2)\) and hydrogen \((H_2)\) gases. This reaction is reversible and can be influenced by various factors such as temperature, pressure, and concentration. In the equation:
\[2 NH_3(g) \rightleftarrows N_2(g) + 3 H_2(g)\]during decomposition, part of the ammonia doesn't decompose entirely, leaving some ammonia together with new products—nitrogen and hydrogen gases—at equilibrium.
This dynamic process indicates how ammonia can transform under specific conditions. Engineers and scientists use ammonia decomposition in industrial applications to produce nitrogen for fertilizers, demonstrating its fundamental role not only in chemical dynamics but also in practical applications.

Equilibrium Concentration

Equilibrium concentration refers to the concentration of each species present in a chemical reaction when it reaches equilibrium. For the decomposition of ammonia, equilibrium is achieved when the rates of forward and reverse reactions are equal, with the following concentrations:

• The equilibrium concentration of \(NH_3\) is \(1 \ M\) since only one mole remains of the initial two moles.
• The equilibrium concentration of \(N_2\) becomes \(\frac{1}{2} \ M\), calculated from the stoichiometric changes during the reaction.
• For \(H_2\), it becomes \(\frac{3}{2} \ M\), as the stoichiometry showed an increase corresponding to the decomposition of ammonia.

These equilibrium concentrations are fundamental in calculating the equilibrium constant. They depict how at equilibrium, the fixed concentrations of products and reactants reflect a balanced system, helping scientists predict and manipulate chemical reactions effectively.