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

In the uncatalyzed reaction: $$ \mathrm{N}_{2} \mathrm{O}_{4}(g) \rightleftarrows 2 \mathrm{NO}_{2}(g) $$ the pressure of the gases at equilibrium are \(P_{\mathrm{N}_{2} \mathrm{O}_{4}}=0.377\) atm and \(P_{\mathrm{NO}_{2}}=1.56 \mathrm{~atm}\) at \(100^{\circ} \mathrm{C}\). What would happen to these pressures if a catalyst were added to the mixture?

Short Answer

Expert verified
The pressures remain unchanged: 0.377 atm for N2O4 and 1.56 atm for NO2.

Step by step solution

01

Understand the Role of a Catalyst

A catalyst speeds up the rate of a chemical reaction without being consumed in the process. It does not alter the equilibrium position of the reaction, meaning it does not change the concentrations or pressures of reactants and products at equilibrium.
02

Identify the Given Equilibrium Pressures

The given equilibrium pressures for the reaction are: - Pressure of \(_2O_4\): \(0.377\) atm - Pressure of \(O_2\): \(1.56\) atm These pressures define the state of the system at equilibrium at \(100^{\circ} C\).
03

Determine Effect of Catalyst on Equilibrium Pressures

Since a catalyst does not change the position of equilibrium, adding a catalyst to the reaction mixture will not alter the pressures of \(_2O_4\) and \(O_2\) at equilibrium. Thus, \(P_{\mathrm{N}_2O_4}\) remains \(0.377\) atm and \(P_{\mathrm{NO}_2}\) remains \(1.56\) atm.

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.

Catalyst Effect
A catalyst is a fascinating substance that can speed up a chemical reaction without being used up in the process. When it comes to equilibrium reactions, like the one involving nitrogen tetroxide ( 2 ightarrow4) and nitrogen dioxide (NO2), a catalyst increases the rate at which equilibrium is reached. However, a critical point to understand is that a catalyst does not change the equilibrium position itself.
When we say it doesn't change the equilibrium position, it means that the catalyst does not alter the concentrations or pressures of the reactants and products once equilibrium is established. In simpler words, even though a catalyst makes reactions faster, the final state at equilibrium — the pressures and concentrations of the involved gases — remains the same.
This is why, in our reaction involving N2O4 and NO2 gases, adding a catalyst will not change the pressures of these gases at equilibrium. A catalyst does not skew the balance; it simply helps the system reach that balance more quickly.
Equilibrium Pressure
Understanding equilibrium pressure is key in grasping how equilibrium reactions behave. When a chemical reaction reaches equilibrium, the rates of the forward and reverse reactions become equal, leading to constant concentrations – or pressures in the case of gases – of reactants and products. This balance is known as chemical equilibrium.
In our specific system of N2O4 converting into NO2, the pressures given, P_{N2O4} = 0.377 atm and P_{NO2} = 1.56 atm, represent this state of balance at 100°C. These pressures dictate how the substances coexist in a steady state, without any net change in their amounts.
It's important to note that changes in temperature or volume might affect these pressures at equilibrium, but the addition of a catalyst, as discussed earlier, will not. The equilibrium pressures remain as they were established, illustrating the robustness of this principle.
Reaction Rates
Reaction rates tell us how fast a reaction occurs, which is crucial to understanding why catalysts are useful. In any reaction, including those at equilibrium, different factors can influence how quickly reactants are converted to products and vice versa.
In our example of N2O4 and NO2 gases, without any intervention, the reaction would naturally progress at its specific rate. If we were to introduce a catalyst, both the forward reaction (N2O4 to NO2) and the backward reaction (NO2 to N2O4) would increase in rate equally. This simultaneous acceleration is what catalysts excel at, speeding up the approach to equilibrium.
However, the key takeaway is that while catalysts boost these rates, shifting both forward and backward reactions uniformly, they do not impact the final equilibrium point, as previously explained.
N2O4 and NO2 Gases
The chemistry of N2O4 and NO2 gases is an excellent showcase of equilibrium and how gases behave under different conditions. Nitrogen tetroxide (N2O4) and nitrogen dioxide (NO2) are two forms of nitrogen oxides that exist in dynamic equilibrium with each other. This means they continuously react in both directions.
At a given temperature, like 100°C in our scenario, these gases reach a state where their conversion into each other no longer leads to a net change. This is the equilibrium state, characterized by specific pressures (0.377 atm for N2O4 and 1.56 atm for NO2).
The interplay between these two gases highlights the beauty of equilibrium where, despite ongoing reactions, the system remains stable. This relationship is important in understanding not just chemical equilibria but also practical applications, such as in environmental chemistry and industrial processes.

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 \(1024^{\circ} \mathrm{C},\) the pressure of oxygen gas from the decomposition of copper(II) oxide \((\mathrm{CuO})\) is \(0.49 \mathrm{~atm}:\) $$4 \mathrm{CuO}(s) \rightleftarrows 2 \mathrm{Cu}_{2} \mathrm{O}(s)+\mathrm{O}_{2}(g)$$ (a) What is \(K_{P}\) for the reaction? (b) Calculate the fraction of \(\mathrm{CuO}\) that will decompose if \(0.16 \mathrm{~mol}\) of it is placed in a 2.0 -L flask at \(1024^{\circ} \mathrm{C}\). (c) What would the fraction be if a 1.0 -mol sample of \(\mathrm{CuO}\) were used? (d) What is the smallest amount of \(\mathrm{CuO}\) (in moles) that would establish the equilibrium?

The equilibrium constant \(K_{P}\) for the reaction: $$ \mathrm{PCl}_{5}(g) \rightleftarrows \mathrm{PCl}_{3}(g)+\mathrm{Cl}_{2}(g) $$ is 1.05 at \(250^{\circ} \mathrm{C}\). The reaction starts with a mixture of \(\mathrm{PCl}_{5}, \mathrm{PCl}_{3},\) and \(\mathrm{Cl}_{2}\) at pressures of \(0.177,0.223,\) and 0.111 atm, respectively, at \(250^{\circ} \mathrm{C}\). When the mixture comes to equilibrium at that temperature, which pressures will have decreased and which will have increased? Explain why.

A 2.50 -mol sample of \(\mathrm{NOCl}\) was initially in a \(1.50-\mathrm{L}\) reaction chamber at \(400^{\circ} \mathrm{C}\). After equilibrium was established, it was found that 28.0 percent of the \(\mathrm{NOCl}\) had dissociated: $$ 2 \mathrm{NOCl}(g) \rightleftarrows 2 \mathrm{NO}(g)+\mathrm{Cl}_{2}(g) $$ Calculate the equilibrium constant \(K_{\mathrm{c}}\) for the reaction.

List four factors that can shift the position of an equilibrium. Only one of these factors can alter the value of the equilibrium constant. Which one is it?

Consider the following reaction at a certain temperature: $$ \mathrm{A}_{2}+\mathrm{B}_{2} \rightleftarrows 2 \mathrm{AB} $$ The mixing of 1 mole of \(\mathrm{A}_{2}\) with 3 moles of \(\mathrm{B}_{2}\) gives rise to \(x\) mole of \(\mathrm{AB}\) at equilibrium. The addition of 2 more moles of \(\mathrm{A}_{2}\) produces another \(x\) mole of \(\mathrm{AB}\). What is the equilibrium constant for the reaction?

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