The equilibrium constant \(K_{\mathrm{c}}\) for the reaction \(2 \mathrm{NH}_{3}(g) \rightleftharpoons \mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g)\) is 0.83 at \(375^{\circ} \mathrm{C} . \mathrm{A}\) 14.6-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

Expert verified
The equilibrium concentrations are obtained by solving the quadratic equation derived from the equilibrium constant expression. The detailed values depend on the results after solving the equation.

Step by step solution

01

Conversion of Mass to Moles

First, convert the given mass of NH3 to moles using its molar mass (17.03 g/mol). It results in \(14.6 \, \text{g} / 17.03 \, \text{g/mol} = 0.858 \, \text{moles of NH3}\).
02

Calculation of Initial Concentrations

Next, calculate the initial concentrations by dividing the moles of each gas by the volume of the container, which is 4.00 L. For NH3, this is \(0.858 \, \text{moles} / 4.00 \, \text{L} = 0.215 \, \text{M} \). The initial concentrations of N2 and H2 are 0, as only NH3 is present initially.
03

Set up the ICE Table

Now, set up an ICE (Initial, Change, Equilibrium) table to represent the changes in concentrations of the substances when equilibrium is reached. The change in concentration for NH3 is -2x (since 2 moles of NH3 decompose per reaction, x being the change in concentration of NH3), while for N2 and H2 the change is +x and +3x respectively (following the stoichiometry of the reaction). The equilibrium concentrations therefore are 0.215 - 2x for NH3, x for N2, and 3x for H2.
04

Solve for x using the Equilibrium Constant

Use the equilibrium expression \(K_{c} = [N_{2}][H_{2}]^{3} / [NH_{3}]^{2}\), replace the equilibrium concentrations with their expressions in terms of x. You get: \(0.83 = (x)(3x)^{3} / (0.215 - 2x)^{2}\). Solving this equation gives the value of x.
05

Calculate the Equilibrium Concentrations

Finally, substitute the value of x into the equilibrium expressions to find the equilibrium concentrations of NH3, N2, and H2.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

ICE Table
Mastering the ICE table method is a must when dealing with chemical equilibria. The acronym ICE stands for Initial, Change, and Equilibrium, which are the columns used to track the concentration of reactants and products through the process of a reaction reaching equilibrium. The table starts by listing the initial concentrations of reactants and products. After this, we factor in the stoichiometry to determine the changes that occur as substances react. The 'Change' row typically incorporates variables that represent how concentration shifts relative to the stoichiometry of the balanced reaction. Lastly, the 'Equilibrium' row tallies these changes to provide a snapshot of the system when it stabilizes. It's a systematic way to visualize the shift from initial to equilibrium states, which then sets the stage for calculations involving the equilibrium constant.

An ICE table was utilized in the textbook problem to assist in calculating the equilibrium concentrations of ammonia, nitrogen, and hydrogen. By applying stoichiometry, we can denote the interdependence of these concentrations; once the equilibrium constant is introduced, we have a solvable equation to find the unknown variables.
Le Chatelier's Principle
When dealing with equilibrium systems, it's crucial to understand Le Chatelier's Principle. This principle states that if an external condition is applied to a system at equilibrium, the system will adjust itself to partially offset the change. This could involve changes in concentration, temperature, or pressure, and the system will respond to re-establish equilibrium. For instance, if more reactants are added, the system will produce more products, and vice versa.

In the context of our textbook problem, if the system at equilibrium at 375°C for the reaction involving ammonia, nitrogen, and hydrogen were subjected to a change, like a pressure increase, Le Chatelier's Principle predicts the system's shift towards the side with fewer moles of gas to reduce pressure. Understanding this principle helps predict the behavior of the reaction under different conditions, which is invaluable for manipulating reactions to favor the desired outcome.
Stoichiometry

Why Stoichiometry Is Key in Equilibrium Calculations

Stoichiometry forms the foundation of chemical reactions. It involves the quantitative relationship between reactants and products in a reaction. This relationship follows the balanced chemical equation and dictates how much of one substance reacts with another to form products. For equilibrium calculations, stoichiometry tells us the proportionate amounts of reactants that convert to products, guiding us in setting up conversion factors that help calculate the extent of a reaction.

In our exercise, the reaction of ammonia decomposing into nitrogen and hydrogen follows a clear stoichiometric pattern: for every 2 moles of ammonia that react, 1 mole of nitrogen and 3 moles of hydrogen are produced. This ratio is pivotal as it influences the coefficients used in the ICE table and is applied in calculating equilibrium concentrations.
Equilibrium Expression
The equilibrium expression formalizes the relationship between the concentrations of reactants and products. For a given reaction, the equilibrium constant, denoted as K_c (when concentrations are used) or K_p (when partial pressures are used), remains constant at a specific temperature. The general form for a reaction \(aA + bB \rightleftharpoons cC + dD\) is \[K_c = \frac{[C]^c [D]^d}{[A]^a [B]^b}\] where the concentrations of products are raised to their respective coefficients in the balanced equation and divided by the concentrations of reactants, each also raised to their respective coefficients.

The equilibrium constant for our ammonia reaction was given as 0.83, and the equilibrium expression for the reaction was used to solve for x, which represented the equilibrium concentration changes. Understanding how to craft and apply the equilibrium expression is crucial for solving equilibrium problems, as it embodies the quantitative aspects of the chemical equilibrium.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The equilibrium constant \(K_{P}\) for the reaction $$ 2 \mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons 2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) $$ is found to be \(2 \times 10^{-42}\) at \(25^{\circ} \mathrm{C}\). (a) What is \(K_{\mathrm{c}}\) for the reaction at the same temperature? (b) The very small value of \(K_{P}\left(\right.\) and \(\left.K_{\mathrm{c}}\right)\) indicates that the reaction overwhelmingly favors the formation of water molecules. Explain why, despite this fact, a mixture of hydrogen and oxygen gases can be kept at room temperature without any change.

Consider the reaction $$ \begin{aligned} 2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{SO}_{3}(g) & \\ \Delta H^{\circ}=&-198.2 \mathrm{~kJ} / \mathrm{mol} \end{aligned} $$ Comment on the changes in the concentrations of \(\mathrm{SO}_{2}, \mathrm{O}_{2},\) and \(\mathrm{SO}_{3}\) at equilibrium if we were to \((\mathrm{a})\) increase the temperature, (b) increase the pressure, (c) increase \(\mathrm{SO}_{2},\) (d) add a catalyst, (e) add helium at constant volume.

Use Le Châtelier's principle to explain why the equilibrium vapor pressure of a liquid increases with increasing temperature.

Write the expressions for the equilibrium constants \(K_{P}\) of these thermal decompositions: (a) \(2 \mathrm{NaHCO}_{3}(s) \rightleftharpoons \mathrm{Na}_{2} \mathrm{CO}_{3}(s)+\mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g)\) (b) \(2 \mathrm{CaSO}_{4}(s) \rightleftharpoons 2 \mathrm{CaO}(s)+2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g)\)

The equilibrium constant \(K_{\mathrm{c}}\) for the decomposition of phosgene, \(\mathrm{COCl}_{2}\), is \(4.63 \times 10^{-3}\) at \(527^{\circ} \mathrm{C}\) : $$ \mathrm{COCl}_{2}(g) \rightleftharpoons \mathrm{CO}(g)+\mathrm{Cl}_{2}(g) $$ Calculate the equilibrium partial pressure of all the components, starting with pure phosgene at 0.760 atm.

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free