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

Consider the reaction $$ 4 \mathrm{NH}_{3}(g)+5 \mathrm{O}_{2}(g) \underset{4 \mathrm{NO}(g)+6 \mathrm{H}_{2} \mathrm{O}(g), \Delta H=-904.4 \mathrm{~kJ}}{\rightleftharpoons} $$ Does each of the following increase, decrease, or leave unchanged the yield of NO at equilibrium? (a) increase \(\left[\mathrm{NH}_{3}\right] ;\) (b) increase \(\left[\mathrm{H}_{2} \mathrm{O}\right] ;(\) c \()\) decrease \(\left[\mathrm{O}_{2}\right]\); (d) decrease the volume of the container in which the reaction occurs: (e) add a catalyst: (f) increase temperature.

Short Answer

Expert verified
(a) Increasing [NH3] concentration will increase the yield of NO. (b) Increasing [H2O] concentration will decrease the yield of NO. (c) Decreasing [O2] concentration will decrease the yield of NO. (d) Decreasing the volume of the container will decrease the yield of NO. (e) Adding a catalyst will not change the yield of NO. (f) Increasing temperature will decrease the yield of NO.

Step by step solution

01

(a) Increase [NH3] concentration

Increasing the concentration of NH3 will cause the equilibrium to shift to the right in order to reduce the excess NH3. This will result in an increased yield of NO as the system tries to reestablish equilibrium.
02

(b) Increase [H2O] concentration

Increasing the concentration of H2O will cause the equilibrium to shift to the left as the system tries to consume the excess H2O. This will result in a decreased yield of NO as the system returns to equilibrium.
03

(c) Decrease [O2] concentration

Decreasing the concentration of O2 will cause the equilibrium to shift to the left as the system tries to replace the reduced O2. As a result, the yield of NO will decrease as the system reestablishes equilibrium.
04

(d) Decrease the volume of the container

Decreasing the volume of the container will increase the pressure on the system. The equilibrium will therefore shift to the side with a lower number of moles of gas in order to reduce the pressure. In this case, the left side of the reaction has 9 moles of gas (4 moles NH3 and 5 moles O2), while the right side has 10 moles of gas (4 moles NO and 6 moles H2O). The equilibrium will shift to the left, resulting in a decreased yield of NO.
05

(e) Add a catalyst

Adding a catalyst will only affect the rate at which the reaction reaches equilibrium. It will not change the position of the equilibrium or the yield of NO.
06

(f) Increase temperature

Since the given reaction is exothermic (ΔH = -904.4 kJ), increasing the temperature will cause the equilibrium to shift to the endothermic side in order to counteract the added heat. In this case, the endothermic side is the left side of the reaction. As a result, the yield of NO will decrease with the increase in temperature.

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 describes how a system at equilibrium responds to changes in concentration, temperature, or pressure. When a change is applied to a system in equilibrium, the system will adjust itself to counteract that change and restore equilibrium balance.

For instance, if the concentration of a reactant is increased, the equilibrium will shift towards the products to reduce this effect. This is why in our example, increasing \(\left[\text{NH}_3\right]\) shifts the equilibrium to the right, increasing the yield of NO. Similarly, if the concentration of a product increases, the system shifts to use up this excess product, as seen with \(\left[\text{H}_2\text{O}\right]\).

Changes in volume and pressure also affect equilibrium. The system shifts towards the side with fewer moles of gas when the volume decreases to reduce pressure. In our example, this means shifting left, resulting in less NO.
  • Increase in reactant concentration shifts equilibrium toward products.
  • Increase in product concentration shifts equilibrium toward reactants.
  • Decreasing volume shifts equilibrium toward fewer moles of gas.
Exothermic Reactions
Exothermic reactions release heat into their surroundings, as indicated by a negative enthalpy change (\(\Delta H\)). When the temperature of a system with an exothermic reaction increases, the equilibrium will shift to absorb the added heat, thus moving towards the endothermic direction.

Based on our example reaction with \(\Delta H = -904.4 \, \text{kJ}\), increasing temperature causes the equilibrium to move towards the reactants, reducing the yield of NO. This shift occurs because the reaction tries to counterbalance the temperature change by favoring the direction that absorbs heat.

Exothermic reactions are quite common in everyday chemistry, contributing to various processes such as combustion and metabolism.
  • Exothermic reactions release heat (\(\Delta H < 0\)).
  • Increasing temperature shifts equilibrium towards endothermic side (consumes heat).
  • In an exothermic reaction, higher temperatures generally decrease the yield of products.
Catalysts in Chemistry
A catalyst is a substance that speeds up the rate of a chemical reaction without being consumed in the process. Importantly, while a catalyst can accelerate the achievement of equilibrium, it does not alter the position of the equilibrium itself. This means the concentrations of reactants and products at equilibrium remain unchanged in the presence of a catalyst.

In our reaction example, adding a catalyst allows the system to reach equilibrium faster but does not affect the yield of NO. This happens because catalysts lower the activation energy required for the reaction to proceed, allowing it to proceed more quickly.
  • Catalysts speed up reactions without being consumed.
  • They do not change equilibrium positions but lessen the time to reach equilibrium.
  • They work by lowering activation energy, making processes more efficient.
Catalysts are widely used in industry to enhance efficiency in chemical manufacturing processes, allowing reactions to occur under milder conditions and reducing energy costs.

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

A chemist at a pharmaceutical company is measuring equilibrium constants for reactions in which drug candidate molecules bind to a protein involved in cancer. The drug molecules bind the protein in a \(1: 1\) ratio to form a drugprotein complex. The protein concentration in aqueous solution at \(25^{\circ} \mathrm{C}\) is \(1.50 \times 10^{-6} \mathrm{M}\). Drug \(\mathrm{A}\) is introduced into the protein solution at an initial concentration of \(2.00 \times 10^{-6} \mathrm{M}\). Drug \(B\) is introduced into a separate, identical protein solution at an initial concentration of \(2.00 \times 10^{-6} \mathrm{M}\). At equilibrium, the drug A-protein solution has an A-protein complex concentration of \(1.00 \times 10^{-6} \mathrm{M}\), and the drug \(\mathrm{B}\) solution has a B-protein complex concentration of \(1.40 \times 10^{-6} \mathrm{M}\). Calculate the \(K_{c}\) value for the \(A\)-protein binding reaction and for the \(\mathrm{B}\) protein binding reaction. Assuming that the drug that binds more strongly will be more effective, which drug is the better choice for further research?

Consider the equilibrium \(\mathrm{Na}_{2} \mathrm{O}(s)+\mathrm{SO}_{2}(g) \rightleftharpoons \mathrm{Na}_{2} \mathrm{SO}_{3}(s)\). (a) Write the equilibrium-constant expression for this reaction in terms of partial pressures. (b) All the compounds in this reaction are soluble in water. Rewrite the equilibriumconstant expression in terms of molarities for the aqueous reaction

Suppose that the gas-phase reactions \(A \longrightarrow B\) and \(\mathrm{B} \longrightarrow \mathrm{A}\) are both elementary processes with rate constants of \(4.7 \times 10^{-3} \mathrm{~s}^{-1}\) and \(5.8 \times 10^{-1} \mathrm{~s}^{-1}\), respectively. (a) What is the value of the equilibrium constant for the equilibrium \(\mathrm{A}(\mathrm{g}) \rightleftharpoons \mathrm{B}(\mathrm{g})\) ? (b) Which is greater at equilibrium, the partial pressure of \(A\) or the partial pressure of \(B\) ?

Silver chloride, \(\mathrm{AgCl}(\mathrm{s})\), is an "insoluble" strong electrolyte. (a) Write the equation for the dissolution of \(\mathrm{AgCl}(s)\) in \(\mathrm{H}_{2} \mathrm{O}(l)\). (b) Write the expression for \(K_{c}\) for the reaction in part (a). (c) Based on the thermochemical data in Appendix \(C\) and Le Châtelier's principle, predict whether the solubility of \(\mathrm{AgCl}\) in \(\mathrm{H}_{2} \mathrm{O}\) increases or decreases with increasing temperature. (d) The equilibrium constant for the dissolution of \(\mathrm{AgCl}\) in water is \(1.6 \times 10^{-10}\) at \(25^{\circ} \mathrm{C}\). In addition, \(\mathrm{Ag}^{+}(a q)\) can react with \(\mathrm{Cl}^{-}(a q)\) according to the reaction $$ \mathrm{Ag}^{+}(a q)+2 \mathrm{Cl}^{-}(a q) \rightleftharpoons \mathrm{AgCl}_{2}^{-}(a q) $$ where \(K_{c}=1.8 \times 10^{5}\) at \(25^{\circ} \mathrm{C}\). Although \(\mathrm{AgCl}\) is "not soluble" in water, the complex \(\mathrm{AgCl}_{2}{ }^{\prime}\) is soluble. At \(25^{\circ} \mathrm{C}\), is the solubility of \(\mathrm{AgCl}\) in a \(0.100 \mathrm{M} \mathrm{NaCl}\) solution greater than the solubility of \(\mathrm{AgCl}\) in pure water, due to the formation of soluble \(\mathrm{AgCl}_{2}^{-}\)ions? Or is the \(\mathrm{AgCl}\) solubility in \(0.100 \mathrm{M} \mathrm{NaCl}\) less than in pure water because of a Le Châtelier-type argument? Justify your answer with calculations. (Hint: Any form in which silver is in solution counts as "solubility.")

Write the equilibrium-constant expression for the equilibrium $$ \mathrm{C}(s)+\mathrm{CO}_{2}(g) \rightleftharpoons 2 \mathrm{CO}(g) $$ The table that follows shows the relative mole percentages of \(\mathrm{CO}_{2}(g)\) and \(\mathrm{CO}(g)\) at a total pressure of 1 atm for several temperatures. Calculate the value of \(K_{p}\) at each temperature. Is the reaction exothermic or endothermic?

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