Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

The four gases \(\mathrm{NH}_{3}, \mathrm{O}_{2}, \mathrm{NO}\), and \(\mathrm{H}_{2} \mathrm{O}\) are mixed in a reaction vessel and allowed to reach equilibrium in the reaction \(4 \mathrm{NH}_{3}(\mathrm{~g})+5 \mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons 4 \mathrm{NO}(\mathrm{g})+6 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})\). Certain changes (see the following table) are then made to this mixture. Considering each change separately, state the effect (increase, decrease, or no change) that the change has on the original equilibrium values of the quantity in the second column (or \(K\), if that is specified). The temperature and volume are constant. \(\begin{array}{ll}\text { Change } & \text { Quantity } \\ \text { (a) add } \mathrm{NO} & \text { amount of } \mathrm{H}_{2} \mathrm{O} \\ \text { (b) add } \mathrm{NO} & \text { amount of } \mathrm{O}_{2} \\ \text { (c) remove } \mathrm{H}_{2} \mathrm{O} & \text { amount of } \mathrm{NO} \\ \text { (d) remove } \mathrm{O}_{2} & \text { amount of } \mathrm{NH}_{3} \\ \text { (e) add } \mathrm{NH}_{3} & \mathrm{~K} \\ \text { (f) remove } \mathrm{NO} & \text { amount of } \mathrm{NH}_{3} \\ \text { (g) add } \mathrm{NH}_{3} & \text { amount of } \mathrm{O}_{2}\end{array}\)

Short Answer

Expert verified
a) Decrease, b) Increase, c) Increase, d) Decrease, e) No change, f) Increase, g) Decrease

Step by step solution

01

- Analyzing Change (a)

When NO is added to the system, the reaction will shift to the left according to Le Chatelier's Principle to re-establish equilibrium. This will result in an increase in the amounts of NH3 and O2 and a decrease in the amounts of NO and H2O.
02

- Predicting Effect on H2O for Change (a)

Since the reaction shifts to the left when NO is added (from step 1), the amount of H2O produced will decrease.
03

- Analyzing Change (b)

Adding NO will not directly affect the amount of O2 since O2 isn't part of the same side of the reaction as NO. However, since the reaction shifts to the left, the overall effect will still be an increase in O2 concentration as the reaction moves to offset the added NO.
04

- Analyzing Change (c)

Removing H2O will cause the reaction to shift to the right, according to Le Chatelier's Principle, in order to produce more H2O and re-establish equilibrium, thereby increasing the amount of NO.
05

- Analyzing Change (d)

Removing O2 will cause the reaction to shift to the right in an effort to replace the O2, leading to a decrease in the amount of NH3 as it is consumed to produce more products.
06

- Analyzing Effect on Equilibrium Constant K for Change (e)

Adding more NH3 does not change the value of the equilibrium constant K because K is only dependent on temperature.
07

- Analyzing Change (f)

Removing NO will cause the reaction to shift to the left to produce more NO. This will increase the amount of NH3, as NO is a product and removing it drives the production of reactants.
08

- Analyzing Change (g)

Adding NH3 will increase the reaction quotient, Q, temporarily, causing the reaction to shift to the right to re-establish equilibrium. This will result in an increase in the amount of O2 consumed and therefore a decrease in its amount.

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.

Le Chatelier's Principle
Le Chatelier's Principle is a fundamental concept in chemistry that deals with the dynamic nature of chemical equilibrium. This principle states that if an external change, such as a change in concentration, pressure, or temperature, is applied to a system at equilibrium, the system will adjust itself to counteract the change and a new equilibrium will be established.

For instance, in the given exercise, adding NO causes the system to shift towards the reactants as the principle predicts, which affects the amounts of other substances like H2O and NH3. Similarly, when H2O is removed, the system compensates by producing more H2O, thus shifting to the right in favor of the products. This conceptual understanding helps us predict the effect of various changes on a chemical system at equilibrium.
Equilibrium Constant
The equilibrium constant, denoted as K, is a quantitative measure of the balance between products and reactants at a given temperature for a reversible reaction at equilibrium. It is a fixed value that can only change with temperature and is unaffected by changes in concentration, pressure, or presence of a catalyst.

In the provided exercise, adding NH3 does not alter the equilibrium constant (K) since it is dependent solely on the temperature of the system. Even though the concentration of reactants and products may change, the ratio of their concentrations raised to the power of their coefficients, as given in the balanced chemical equation, will remain the same at a constant temperature.
Reaction Quotient
The reaction quotient (Q) is similar in form to the equilibrium constant (K), but whereas K is the value at equilibrium, Q can be calculated at any moment of the reaction process. It indicates the direction in which a reaction needs to shift in order to reach equilibrium.

When adding NH3, as mentioned in step 8 of the solution, Q will increase because the concentration of reactants increases. This means the reaction is no longer at equilibrium and needs to adjust. A higher Q than K prompts a shift towards the products until Q equals K once again, indicating that equilibrium has been re-established.
Shifts in Equilibrium
Shifts in equilibrium refer to the way a chemical reaction responds to stresses applied to the system. These shifts occur in order to reach a new state of equilibrium in accordance with Le Chatelier's Principle.

Addition or removal of reactants or products, as demonstrated in the exercise, result in a shift in the position of equilibrium either to the right (towards more products) or to the left (towards more reactants). For instance, adding NO causes a shift to the left, decreasing the amount of H2O, while removing O2 shifts the equilibrium to the right, thus decreasing NH3 and increasing the amounts of NO and H2O.

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

At \(500 .{ }^{\circ} \mathrm{C}, K_{c}=0.061\) for \(\mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_{3}(\mathrm{~g})\). If analysis shows that the composition is \(3.00 \mathrm{~mol} \cdot \mathrm{L}^{-1} \mathrm{~N}_{2}\), \(2.00 \mathrm{~mol} \cdot \mathrm{L}^{-1} \mathrm{H}_{2}\), and \(0.500 \mathrm{~mol} \cdot \mathrm{L}^{-1} \mathrm{NH}_{3}\), is the reaction at equilibrium? If not, in which direction does the reaction tend to proceed to reach equilibrium?

Analysis of a reaction mixture showed that it had the composition \(0.624 \mathrm{~mol} \cdot \mathrm{L}^{-1} \mathrm{~N}_{2}, 0.315 \mathrm{~mol} \cdot \mathrm{L}^{-1} \mathrm{H}_{2}\), and \(0.222 \mathrm{~mol} \cdot \mathrm{L}^{-1} \mathrm{NH}_{3}\) at \(800 . \mathrm{K}\), at which temperature \(K_{c}=0.278\) for \(\mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_{3}\) (g). (a) Calculate the reaction quotient \(Q_{c}\). (b) Is the reaction mixture at equilibrium? (c) If not, is there a tendency to form more reactants or more products?

Write the reaction quotient \(Q\) for (a) \(2 \mathrm{BCl}_{3}\) (g) \(+2 \mathrm{Hg}(\mathrm{l}) \rightarrow \mathrm{B}_{2} \mathrm{Cl}_{4}\) (s) \(+\mathrm{Hg}_{2} \mathrm{Cl}_{2}\) (s) (b) \(\mathrm{P}_{4} \mathrm{~S}_{10}\) (s) \(+16 \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \longrightarrow 4 \mathrm{H}_{3} \mathrm{PO}_{4}(\mathrm{aq})+10 \mathrm{H}_{2} \mathrm{~S}(\mathrm{aq})\) (c) \(\mathrm{Br}_{2}(\mathrm{~g})+3 \mathrm{~F}_{2}(\mathrm{~g}) \longrightarrow 2 \mathrm{BrF}_{3}(\mathrm{~g})\)

For the reaction \(\mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_{3}(\mathrm{~g})\) at \(400 . \mathrm{K}\), \(K=41\). Find the value of \(K\) for each of the following reactions at the same temperature: (a) \(2 \mathrm{NH}_{3}\) (g) \(\rightleftharpoons \mathrm{N}_{2}\) (g) \(+3 \mathrm{H}_{2}\) (g) (b) \(\frac{1}{2} \mathrm{~N}_{2}(\mathrm{~g})+\frac{3}{2} \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons \mathrm{NH}_{3}(\mathrm{~g})\) (c) \(2 \mathrm{~N}_{2}(\mathrm{~g})+6 \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons 4 \mathrm{NH}_{3}(\mathrm{~g})\)

Determine whether the following statements are true or false. If false, explain why. (a) In an equilibrium reaction, the reverse reaction begins only when all reactants have been converted to products. (b) The equilibrium concentrations will be the same whether one starts with pure reactants or pure products. (c) The rates of the forward and reverse reactions are the same at equilibrium. (d) If the Gibbs free energy is greater than the standard Gibbs free energy of reaction, the reaction proceeds forward to equilibrium.

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