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Explain the effect of a catalyst on an equilibrium system.

Short Answer

Expert verified
A catalyst speeds up the attainment of equilibrium without changing its position.

Step by step solution

01

- Define Catalysts

A catalyst is a substance that speeds up the rate of a chemical reaction without being consumed in the process.
02

- Understanding Equilibrium

In a chemical equilibrium, the rate of the forward reaction is equal to the rate of the reverse reaction, leading to constant concentrations of reactants and products.
03

- Catalyst Effect on Reaction Rates

A catalyst increases both the forward and reverse reaction rates by lowering the activation energy required for the reactions.
04

- Impact on Equilibrium Position

Since a catalyst increases the rates of both forward and reverse reactions equally, it does not alter the equilibrium position. The ratio of product and reactant concentrations remains unchanged.
05

- Time to Reach Equilibrium

Even though the position of the equilibrium does not change, the catalyst allows the system to reach equilibrium faster by increasing the reaction rates.

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Key Concepts

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

Catalysts in Chemistry
Catalysts are fascinating substances in chemistry. They speed up chemical reactions without being consumed in the process. Here’s how they work:
  • Catalysts provide an alternate reaction pathway with a lower activation energy.
  • This means reactants convert to products faster.
  • Importantly, they do not alter the final amount of products or reactants.
For example, enzymes are biological catalysts that speed up vital reactions in our bodies. Unlike reactants, catalysts are not used up; they remain unchanged and can catalyze multiple reactions.
Chemical Equilibrium
Chemical equilibrium can be tricky but let's break it down. Imagine a chemical reaction where reactants turn into products. At equilibrium, the rate of making products equals the rate of turning those products back into reactants.
This balance means the concentrations of reactants and products stay constant over time. Here are some key points:
  • Equilibrium does not mean reactants and products are equal, just that their rates are.
  • It's a dynamic state - molecules keep reacting, but overall concentrations don't change.
  • Le Chatelier’s Principle helps predict how changes (like concentration, pressure, temperature) affect the system.
This concept helps us understand reactions in closed systems and predict how they respond to different changes.
Reaction Rates
The rate of a reaction tells us how fast reactants turn into products. Various factors affect these rates:
  • Concentration: More particles mean more collisions, hence faster reactions.
  • Temperature: Higher temperatures increase particle speed, leading to more frequent collisions.
  • Catalysts: Lower activation energy, speeding up reactions.
  • Surface Area: Smaller pieces (higher surface area) react faster because there's more area for collisions.
Understanding reaction rates is crucial, whether you're speeding up a chemical process or slowing it down for control and safety. It's all about the frequency and energy of molecular collisions.
Activation Energy
Activation energy is the energy threshold that must be overcome for a reaction to proceed. Think of it as a hill reactants must climb before forming products. Here's what to know:
  • High activation energy means a slower reaction, as fewer molecules have the needed energy to react.
  • A catalyst lowers this hill, allowing more molecules to react at a given temperature.
  • In graphs, activation energy is the peak between reactants and products.
This concept is vital in both natural processes, like enzyme function in metabolism, and industrial applications where catalysts make processes economical and efficient.

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Most popular questions from this chapter

For the formation of ammonia, the equilibrium constant is calculated to be \(5.2 \times 10^{-5}\) at \(25^{\circ} \mathrm{C}\) . After analysis, it is determined that \(\left[\mathrm{N}_{2}\right]=2.00 \mathrm{M}\) and \(\left[\mathrm{H}_{2}\right]=0.80 \mathrm{M} .\) How many grams of ammonia are in the 10. L reaction vessel at equilibrium? Use the following equilibrium equation. $$\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftarrows 2 \mathrm{NH}_{3}(g)$$

Relating ldeas Let s equal the solubility, in mol/L, of \(\mathrm{AB}_{2}\) . In terms of \(s\) , what is the molar concentration of \(\mathrm{A} ?\) of \(\mathrm{B} ?\) What is the \(K_{\mathrm{sp}}\) of \(\mathrm{A} \mathrm{B}_{2}\) ?

The value of the equilibrium constant for the reaction below is 40.0 at a specified temperature. What would be the value of that constant for the reverse reaction under the same conditions? $$\mathrm{H}_{2}(g)+\mathrm{I}_{2}(g) \rightleftarrows 2 \mathrm{HI}(g)$$

At \(25^{\circ} \mathrm{C}\) , the value of \(K\) is \(1.7 \times 10^{-13}\) for the following reaction. \(2 \mathrm{N}_{2} \mathrm{O}(g)+\mathrm{O}_{2}(g) \rightleftarrows 4 \mathrm{NO}(g)\) It is determined that \(\left[\mathrm{N}_{2} \mathrm{O}\right]=0.0035 \mathrm{mol} / \mathrm{L}\) and \(\left[\mathrm{O}_{2}\right]=0.0027 \mathrm{mol} / \mathrm{L}\) . Using this information, what is the concentration of \(\mathrm{NO}(g)\) at equilibrium?

a. Write the general expression for an equilibrium constant based on the equation \(n \mathrm{A}+m \mathrm{B}+\ldots \rightleftarrows x \mathrm{C}+y \mathrm{D}+\ldots\) b. What information is provided by the value of \(K\) for a given equilibrium system at a specified temperature?

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