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The role of a catalyst in a reversible reaction is to: (a) Alter the equilibrium constant of the reaction (b) Increase the rate of forward reaction (c) Allow the equilibrium to be achieved quickly (d) Decrease the rate of backward reaction

Short Answer

Expert verified
The correct option is (c). A catalyst allows the equilibrium to be achieved quickly.

Step by step solution

01

Identifying the Question

The question asks about the role of a catalyst in a reversible reaction. We need to determine which of the given options correctly describes how a catalyst affects such a reaction.
02

Understanding Catalysis

A catalyst is a substance that increases the rate of a chemical reaction without being consumed in the process. It works by lowering the activation energy for the reaction, which speeds up both the forward and backward reactions equally.
03

Evaluating the Equilibrium Constant

The equilibrium constant ( K_eq ) is a ratio of product concentrations to reactant concentrations at equilibrium and is only affected by changes in temperature. A catalyst does not alter this constant as it doesn't change the concentrations at equilibrium.
04

Assessing Answer Choices

Let's go through the options: - (a) Incorrect, as a catalyst does not change the equilibrium constant. - (b) Partially correct, as a catalyst increases the rate but does not specify the role in equilibrium. - (c) Correct, because a catalyst allows the equilibrium to be achieved faster by speeding up both directions. - (d) Incorrect, as the catalyst increases the rate of both forward and backward reactions equally.
05

Conclusion

Since a catalyst speeds up both the forward and backward reactions equally, it allows the system to reach equilibrium more quickly. Hence, the correct answer is (c) Allow the equilibrium to be achieved quickly.

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Key Concepts

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

Reversible Reactions
Reversible reactions are fascinating because they can go "in both directions". This means both reactants can turn into products, and products can turn back into reactants.

Imagine you are baking cookies. In a typical recipe, once they are baked, the process is done and they can't "unbake". But with reversible reactions, it's like being able to unbake those cookies and end up with the raw ingredients again. This dynamic nature is why reversible reactions are so interesting.

In chemistry, reversible reactions are represented with a double-headed arrow (↔️) between reactants and products. It signals that the reactions can happen in both ways. This is crucial in processes like the synthesis of ammonia in the Haber process.
  • They never "stop"; they reach a state where forward and backward reactions occur at the same rate.
  • This balance state is called equilibrium.
Once equilibrium is reached, the amounts of reactants and products remain constant, but the reactions keep happening. Understanding reversible reactions is key to grasping many important chemical processes in nature and industry.
Equilibrium Constant
The equilibrium constant, often denoted as \( K_{eq} \), is a crucial number in understanding reversible reactions. It helps chemists know "where the balance lies"—essentially, how much of each substance is present when the reaction is at equilibrium.

More formally, \( K_{eq} \) is the ratio of the concentrations of products to reactants when a chemical reaction is in a state of equilibrium, each raised to the power of their coefficients in the balanced equation.

Mathematically, if we have a reaction:

\( aA + bB \rightleftharpoons cC + dD \)

Then the equilibrium constant \( (K_{eq}) \) is given by:

\[ K_{eq} = \frac{[C]^c [D]^d}{[A]^a [B]^b} \]

Here's what's important about \( K_{eq} \):
  • It's a fixed value at a given temperature and doesn't change unless the temperature changes.
  • A larger \( K_{eq} \) means more products (reaction favors products).
  • A smaller \( K_{eq} \) means more reactants (reaction favors reactants).
Remember, a catalyst does not affect \( K_{eq} \) because it speeds up both the forward and backward reactions, allowing them to balance at equilibrium without shifting the position of equilibrium.
Activation Energy
Activation energy is like a hurdle that reactants need to overcome to become products. It is the minimum energy required for a chemical reaction to occur.

This concept is crucial because it helps us understand why some reactions happen quickly, while others are slow. Imagine pushing a boulder over a hill, where the top of the hill represents the activation energy. Once the boulder is over, it "rolls down" into products.

This "energy hill" must be climbed for the reaction to proceed.
  • High activation energy means a slower reaction because it's harder for reactants to get going.
  • Low activation energy means a faster reaction since the reactants have a small energy barrier to overcome.
Enter catalysts, which are incredibly useful in chemical reactions. They "lower the hill", or in technical terms, they reduce the activation energy, making the reaction happen more quickly. However, they do not change the amount of energy released or absorbed by the reaction, nor do they change the equilibrium of the reaction. They simply allow reactions to happen more easily.
Rate of Reaction
The rate of reaction is all about how fast a reaction goes from reactants to products. It's like the speed at which a story unfolds.

Certain factors can influence the rate of a chemical reaction:
  • Temperature: Higher temperatures increase kinetic energy and thus increase reaction rates.
  • Concentration: More particles can lead to more collisions and faster reactions.
  • Surface area: Smaller particle sizes offer more surface, resulting in quicker reactions.
  • Catalysts: They lower activation energy which increases the rate without being consumed.
To determine a reaction's speed, we can measure how quickly products form or how fast reactants are used up. A key concept here is the use of a catalyst. Catalysts are like speed boosters for reactions. They do not get "used up" in the process, meaning they're incredibly efficient.

Remarkably, in reversible reactions, catalysts increase the speed of both the forward and backward reactions equally. This is why catalysts are indispensable in industrial processes—they help achieve equilibrium faster, saving time and resources without altering the equilibrium itself.

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Most popular questions from this chapter

In which of the following reactions, the concentration of product is higher than the concentration of reactant at equilibrium? = equilibrium constant): (a) \(\mathrm{A} \rightleftharpoons \mathrm{B} ; \mathrm{K}=0.001\) (b) \(\mathrm{M} \rightleftharpoons \mathrm{N} ; \mathrm{K}=10\) (c) \(\mathrm{X} \rightleftharpoons \mathrm{Y} ; \mathrm{K}=0.005\) (d) \(\mathrm{R} \rightleftharpoons \mathrm{P} ; \mathrm{K}=0.01\)

In which of the following gaseous reaction, the value of \(K_{p}\) is less than \(K_{c}:\) (a) \(\mathrm{PCl}_{5} \rightleftharpoons \mathrm{PCI}_{3}+\mathrm{Cl}_{2}\) (b) \(2 \mathrm{SO}_{2}+\mathrm{O}_{2} \rightleftharpoons 2 \mathrm{SO}_{3}\) (c) \(2 \mathrm{HI} \rightleftharpoons \mathrm{H}_{2}+\mathrm{I}_{2}\) (d) \(\mathrm{N}_{2}+3 \mathrm{H}_{2} \rightleftharpoons 2 \mathrm{NH}_{3}\)

\(1.25\) moles of NOCl were placed in a \(2.50 \mathrm{~L}\) reaction chamberat \(427^{\circ} \mathrm{C}\). After equilibrium was reached, 1.10 molesofNOClremained. Calculatetheequilibrium constant \(\mathrm{K}_{\mathrm{c}}\) for the reaction, \(2 \mathrm{NOC} 1(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{~g}):\) (a) \(1.6 \times 10^{-3}\) (b) \(5.6 \times 10^{-4}\) (c) \(2.6 \times 10^{-3}\) (d) \(4.6 \times 10^{-4}\)

For the reaction: \(\mathrm{CO}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \rightleftharpoons \mathrm{CO}_{2}(\mathrm{~g})+\mathrm{H}_{2}(\mathrm{~g})\) at a given temperature, the equilibrium amount of \(\mathrm{CO}_{2}(\mathrm{~g})\) can be increased by (a) Adding a suitable catalyst (b) Adding an inert gas (c) Decreasing the volume of the container (d) Increasing the amount of \(\mathrm{CO}(\mathrm{g})\)

One mole of \(\mathrm{A}(\mathrm{g})\) is heated to \(300^{\circ} \mathrm{C}\) in a closed one litre vessel till the following equilibrium is reached. \(\mathrm{A}(\mathrm{g}) \rightleftharpoons \mathrm{B}(\mathrm{g})\) The equilibrium constant of this reaction at \(300^{\circ} \mathrm{C}\) is 4\. What is the concentration of \(\mathrm{B}\) (in \(\mathrm{mol} \mathrm{L}^{-1}\) ) at equilibrium: (a) \(0.2\) (b) \(0.6\) (c) \(0.75\) (d) \(0.1\)

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