/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 55 Consider this reaction: $$\mat... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Consider this reaction: $$\mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{NO}(g) $$The equilibrium constant \(K_{P}\) for the reaction is \(1.0 \times\) \(10^{-15}\) at \(25^{\circ} \mathrm{C}\) and 0.050 at \(2200^{\circ} \mathrm{C}\). Is the formation of nitric oxide endothermic or exothermic? Explain your answer.

Short Answer

Expert verified
The formation of nitric oxide is endothermic because \(K_P\) increases with temperature.

Step by step solution

01

Understanding the Question

We are asked if the formation of nitric oxide (\(\text{NO}\)) is endothermic or exothermic. The equilibrium constants (\(K_P\)) at two different temperatures are provided. Endothermic reactions have equilibrium constants that increase with temperature, while exothermic reactions have equilibrium constants that decrease with temperature.
02

Analyzing Given Data

We have two temperatures with corresponding equilibrium constants: \(K_P = 1.0 \times 10^{-15}\) at \(25^{\circ}\text{C}\) and \(K_P = 0.050\) at \(2200^{\circ}\text{C}\). Here, \(K_P\) increases from a very small value at \(25^{\circ}\text{C}\) to a significantly larger value at \(2200^{\circ}\text{C}\).
03

Relating Equilibrium Constant Changes with Temperature

According to the van 't Hoff equation, an increase in \(K_P\) with an increase in temperature indicates the reaction absorbs heat, typical of endothermic processes.
04

Conclusion

Since \(K_P\) increases as the temperature increases from \(25^{\circ}\text{C}\) to \(2200^{\circ}\text{C}\), the formation of nitric oxide is an endothermic reaction.

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 91Ó°ÊÓ!

Key Concepts

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

Equilibrium Constant
The equilibrium constant, denoted as \(K_P\) for reactions involving gases, is a measure of the ratio of the concentrations of products to reactants at equilibrium. A reaction reaches equilibrium when the rates of the forward and reverse reactions are equal, and there is no net change in concentrations. At this point, the value of \(K_P\) is determined by the pressures of the gaseous reactants and products.
For example, in the reaction \(N_2(g) + O_2(g) \rightleftharpoons 2\text{NO}(g)\), the equilibrium constant \(K_P\) is defined as:
  • \(K_P = \frac{(\text{P}_{\text{NO}})^2}{\text{P}_{\text{N}_2} \times \text{P}_{\text{O}_2}}\)
Here, \(\text{P}_{\text{NO}}\), \(\text{P}_{\text{N}_2}\), and \(\text{P}_{\text{O}_2}\) represent the partial pressures of nitric oxide, nitrogen, and oxygen gases, respectively.
The equilibrium constant is specific to a particular reaction and temperature. A change in temperature typically results in a change in the value of \(K_P\). Understanding \(K_P\) helps predict the direction of the reaction; a larger value indicates more products at equilibrium, while a smaller value indicates more reactants.
van 't Hoff Equation
The van 't Hoff equation is a useful tool to understand how changes in temperature affect the equilibrium constant \(K\)… It is expressed as:
  • \[\ln \left( \frac{K_2}{K_1} \right) = -\frac{\Delta H^\circ}{R} \left(\frac{1}{T_2} - \frac{1}{T_1} \right)\]
In this equation:
  • \(K_1\) and \(K_2\) are equilibrium constants at temperatures \(T_1\) and \(T_2\), respectively,
  • \(\Delta H^\circ\) is the standard enthalpy change of the reaction,
  • \(R\) is the universal gas constant (\(8.314 \text{ J/mol·K}\)),
  • \(T_1\) and \(T_2\) are the absolute temperatures in Kelvin.
The van 't Hoff equation shows how \(K\) changes with temperature, providing insights into whether a reaction is exothermic or endothermic. If the value of \(K\) increases with temperature, as seen in the formation of nitric oxide, it indicates an endothermic reaction with heat being absorbed. Conversely, a decrease in \(K\) with rising temperature would suggest an exothermic process where heat is released.
Using the van 't Hoff equation can provide quantitative predictions about reaction behaviors under different thermal conditions.
Temperature Effects on Reaction Equilibrium
Temperature plays a crucial role in determining the position of reaction equilibrium. For endothermic reactions like the formation of nitric oxide, an increase in temperature shifts the equilibrium towards the products. This is because higher temperatures provide the necessary energy for the reactants to convert into products.
  • Endothermic reactions absorb heat, causing \(K\) to increase with temperature.
  • Exothermic reactions release heat, causing \(K\) to decrease as temperatures rise.
Le Chatelier’s principle further explains this behavior, suggesting that an increase in temperature will shift equilibrium to offset the change—favoring endothermic reactions.
Understanding these effects is crucial in controlling industrial processes, like the synthesis of nitric oxide. It helps in optimizing conditions to maximize yield.
In practical applications, adjusting temperature can have an economic impact, as it influences both the rate and position of equilibrium in chemical reactions.

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 dissociation of molecular iodine into iodine atoms is represented as $$\mathrm{I}_{2}(g) \rightleftharpoons 2 \mathrm{I}(g)$$At \(1000 \mathrm{~K},\) the equilibrium constant \(K_{\mathrm{c}}\) for the reac- tion is \(3.80 \times 10^{-5}\). Suppose you start with 0.0456 mole of \(\mathrm{I}_{2}\) in a \(2.30-\mathrm{L}\) flask at \(1000 \mathrm{~K}\). What are the concentrations of the gases at equilibrium?

Write the equilibrium constant expressions for \(K_{\mathrm{c}}\) and \(K_{P}\), if applicable, for these reactions: (a) \(2 \mathrm{NO}_{2}(g)+7 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g)+4 \mathrm{H}_{2} \mathrm{O}(l)\) (b) \(2 \mathrm{ZnS}(s)+3 \mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{ZnO}(s)+2 \mathrm{SO}_{2}(g)\) (c) \(\mathrm{C}(s)+\mathrm{CO}_{2}(g) \rightleftharpoons 2 \mathrm{CO}(g)\) (d) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COOH}(a q) \rightleftharpoons \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COO}^{-}(a q)+\mathrm{H}^{+}(a q)\)

Consider this equilibrium process at \(686^{\circ} \mathrm{C}\) : $$\mathrm{CO}_{2}(g)+\mathrm{H}_{2}(g) \rightleftharpoons \mathrm{CO}(g)+\mathrm{H}_{2} \mathrm{O}(g)$$The equilibrium concentrations of the reacting species are \([\mathrm{CO}]=0.050 \mathrm{M},\left[\mathrm{H}_{2}\right]=0.045 \mathrm{M},\left[\mathrm{CO}_{2}\right]=\) \(0.086 M,\) and \(\left[\mathrm{H}_{2} \mathrm{O}\right]=0.040 \mathrm{M} .\) (a) Calculate \(K_{\mathrm{c}}\) for the reaction at \(686^{\circ} \mathrm{C}\). (b) If the concentration of \(\mathrm{CO}_{2}\) were raised to \(0.50 \mathrm{~mol} / \mathrm{L}\) by the addition of \(\mathrm{CO}_{2}\), what would be the concentrations of all the gases when equilibrium is reestablished?

Explain the difference between physical equilibrium and chemical equilibrium. Give two examples of each.

Consider the potential energy diagrams for two types of reactions \(\mathrm{A} \rightleftharpoons \mathrm{B}\). In each case, answer the following questions for the system at equilibrium. (a) How would a catalyst affect the forward and reverse rates of the reaction? (b) How would a catalyst affect the energies of the reactant and product? (c) How would an increase in temperature affect the equilibrium constant? (d) If the only effect of a catalyst is to lower the activation energies for the forward and reverse reactions, show that the equilibrium constant remains unchanged if a catalyst is added to the reacting mixture.

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.