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

Does the addition of a catalyst have any effects on the position of an equilibrium?

Short Answer

Expert verified
A catalyst does not affect the equilibrium position.

Step by step solution

01

Understand the Function of a Catalyst

A catalyst speeds up the rate of a chemical reaction by providing an alternative pathway with a lower activation energy. This means that the reactions can proceed faster in both directions (forward and backward) in a reversible reaction.
02

Define Chemical Equilibrium

Chemical equilibrium is established in a reversible reaction when the rate of the forward reaction equals the rate of the backward reaction, leading to constant concentrations of the reactants and products over time.
03

Evaluate Catalyst's Role in Equilibrium

While a catalyst accelerates the speed at which equilibrium is reached, it does not change the concentrations of reactants and products at equilibrium. Therefore, the catalyst does not shift the position of the equilibrium.

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.

Catalysis
A catalyst is a substance that increases the rate of a chemical reaction without being consumed in the process. It works by providing an alternative reaction pathway with a lower activation energy. By lowering activation energy, the catalyst allows the reaction to proceed more easily, thus speeding up both the forward and reverse reactions in reversible reactions. This is because the catalyst affects the reaction kinetics, helping molecules to collide more effectively and transform into products more rapidly on the molecular level.
Since the catalyst itself is not consumed, it can be used repeatedly, making this process efficient. Importantly, while a catalyst can speed up the rate at which equilibrium is achieved, it does not influence the equilibrium position or change the concentrations of reactants and products. Thus, the catalyst only makes the reaction faster, not altering the balance between reactants and products.
Chemical Equilibrium
Chemical equilibrium occurs in a reversible reaction when the rate of the forward reaction equals the rate of the backward reaction. This leads to a state where the concentrations of reactants and products remain constant over time. Equilibrium is dynamic; even though the concentrations remain constant, the reactions themselves are still occurring at the molecular level.
At equilibrium, it might seem like the reaction has stopped, but it's actually a balance between the two opposing processes. This state can be represented by the equilibrium constant, which is determined by the ratio of the concentrations of products to reactants. However, the equilibrium position is not altered by the presence of a catalyst; it remains dependent on factors like temperature and the initial concentrations of reactants and products.
Activation Energy
Activation energy is the minimum amount of energy required for a chemical reaction to occur. It is also the energy barrier that reactants need to overcome to be converted into products. The concept of activation energy is crucial in understanding how catalysts work.
A catalyst lowers the activation energy needed for the reaction, meaning that more reactant molecules will have enough energy to react when a catalyst is present. This increases the reaction rate, as more molecules can surpass the energy barrier at a given time. By providing an alternative reaction pathway with lower activation energy, catalysts make reactions proceed faster without being part of the final products, contributing significantly to efficient chemical processes.
Reaction Rates
Reaction rates describe how quickly or slowly a chemical reaction occurs. They are influenced by several factors, including the presence of a catalyst, temperature, concentration of reactants, and surface area of reactants.
  • With a catalyst, reaction rates increase because the catalyst lowers the activation energy needed for the reaction to proceed.
  • Temperature can also affect reaction rates, as it provides energy to the reacting molecules, increasing their movement and the likelihood of collisions.
  • Higher concentrations of reactants typically lead to higher reaction rates since more molecules are available to collide and react.
  • The surface area of reactants matters as well, especially in heterogeneous reactions where reactants are in different phases.
While these factors affect how quickly equilibrium is reached, they do not alter the equilibrium position, which is determined by the energy dynamics between reactants and products.

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

Consider the following equilibrium process: \(\mathrm{PCl}_{5}(g) \rightleftarrows \mathrm{PCl}_{3}(g)+\mathrm{Cl}_{2}(g) \quad \Delta H^{\circ}=92.5 \mathrm{~kJ} / \mathrm{mol}\) Predict the direction of the shift in equilibrium when (a) the temperature is raised, (b) more chlorine gas is added to the reaction mixture, (c) some \(\mathrm{PCl}_{3}\) is removed from the mixture, (d) the pressure on the gases is increased, (e) a catalyst is added to the reaction mixture.

When a gas was heated under atmospheric conditions, its color deepened. Heating above \(150^{\circ} \mathrm{C}\) caused the color to fade, and at \(550^{\circ} \mathrm{C}\) the color was barely detectable. However, at \(550^{\circ} \mathrm{C},\) the color was partially restored by increasing the pressure of the system. Which of the following best fits the preceding description: (a) a mixture of hydrogen and bromine, (b) pure bromine, (c) a mixture of nitrogen dioxide and dinitrogen tetroxide. (Hint: Bromine has a reddish color, and nitrogen dioxide is a brown gas. The other gases are colorless.) Justify your choice.

Write equilibrium constant expressions for \(K_{\mathrm{c}}\), and for \(K_{P}\), if applicable, for the following processes: (a) \(2 \mathrm{CO}_{2}(g) \rightleftarrows 2 \mathrm{CO}(g)+\mathrm{O}_{2}(g)\) (b) \(3 \mathrm{O}_{2}(g) \rightleftarrows 2 \mathrm{O}_{3}(g)\) (c) \(\mathrm{CO}(g)+\mathrm{Cl}_{2}(g) \rightleftarrows \mathrm{COCl}_{2}(g)\) (d) \(\mathrm{H}_{2} \mathrm{O}(g)+\mathrm{C}(s) \rightleftarrows \mathrm{CO}(g)+\mathrm{H}_{2}(g)\) (e) \(\mathrm{HCOOH}(a q) \rightleftarrows \mathrm{H}^{+}(a q)+\mathrm{HCOO}^{-}(a q)\) (f) \(2 \mathrm{HgO}(s) \rightleftarrows 2 \mathrm{Hg}(l)+\mathrm{O}_{2}(g)\)

Define equilibrium. Give two examples of a dynamic equilibrium.

Baking soda (sodium bicarbonate) undergoes thermal decomposition as follows: $$ 2 \mathrm{NaHCO}_{3}(s) \rightleftarrows \mathrm{Na}_{2} \mathrm{CO}_{3}(s)+\mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g) $$ Would we obtain more \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}\) by adding extra baking soda to the reaction mixture in (a) a closed vessel or (b) an open vessel?

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