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A reversible reaction is said to have attained equilibrium, when: (a) Backward reaction stops (b) Both backward and forward reactions take place at equal speed (c) Both backward and forward reactions stop (d) Concentration of each of the reactants and products becomes equal

Short Answer

Expert verified
Option (b): Both backward and forward reactions take place at equal speed.

Step by step solution

01

Understanding Equilibrium

In a chemical reaction, equilibrium is the state where the rate of the forward reaction equals the rate of the backward reaction. At this point, the concentrations of the reactants and products remain constant over time.
02

Analyzing the Options

The key condition for chemical equilibrium is that the forward and backward reactions occur at the same rate. This means that neither the forward nor the backward reaction stops; they just proceed at equal speeds. Therefore, options (a) and (c) are incorrect as they suggest one or both reactions stop.
03

Ruling Out Concentration Equality

In chemical equilibrium, the concentrations of reactants and products are constant, but they are not required to be equal. This eliminates option (d), which suggests that concentrations must be equal.
04

Identifying the Correct Condition

Given that equilibrium is defined by the balance of reaction rates, the correct option must reflect that both reactions occur simultaneously at equal speeds. Thus, option (b) "Both backward and forward reactions take place at equal speed" correctly describes chemical equilibrium.

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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 and fundamental in understanding chemical processes. Unlike simple reactions that go in one direction, reversible reactions can proceed forward and backward. This means the reactants transform into products and the products can revert back into the original reactants.
  • Think of it as a two-way street where traffic flows in both directions.
  • Reversible reactions are denoted by a double-headed arrow (↔) in chemical equations.
In reality, not every reaction is reversible. It's important to understand the conditions that allow reversibility. Often, certain factors like temperature and pressure must be specific for a reaction to proceed in both directions efficiently. Recognizing a reversible reaction is crucial because it sets the stage for understanding chemical equilibrium.
Reaction Rates
Chemical reactions don't happen instantly; they take time. Reaction rates describe how fast reactants convert into products. This is important for reversible reactions because the rate of the forward reaction must match the rate of the backward reaction for equilibrium to occur.
  • At the beginning of a reaction, reaction rates can change as concentrations of reactants and products shift.
  • When equilibrium is reached, these rates become equal, not zero, allowing the reactions to transit seamlessly back and forth.
In a reversible reaction, both forward and backward reactions are ongoing. They might proceed quickly or slowly, but what's critical for equilibrium is their speed equality, not their absolute speed. Understanding reaction rates helps determine how fast a reaction reaches equilibrium and predicts the dynamic nature of the chemical process.
Constant Concentrations
Chemical equilibrium is not about making reactants and products disappear. It's about reaching a state where their concentrations remain constant—and that's key to understanding equilibrium.
  • Equilibrium occurs when the forward and backward reactions happen at equal rates, keeping the concentrations of reactants and products unchanged over time.
  • "Constant concentrations" doesn't mean that the amounts of reactants and products are equal; it simply means they don't change.
When looking at a system in equilibrium, if you were to take a snapshot at any point, you'd see the same concentration levels. It's a balance, not an end state, showing the beauty of equilibrium where dynamic processes occur without altering the overall composition. Recognizing this static balance of constant concentrations helps avoid common misconceptions about chemical equilibrium.

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Most popular questions from this chapter

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

For the hypothetical reactions, the equilibrium constant \((\mathrm{K})\) values are given: \(\mathrm{A} \rightleftharpoons \mathrm{B} \mathrm{K}_{1}=2\) \(\mathrm{B} \rightleftharpoons \mathrm{C} \mathrm{K}_{2}^{1}=4\) \(\mathrm{C} \rightleftharpoons \mathrm{D} \mathrm{K}_{3}^{2}=3\) The equilibrium constant \((\mathrm{K})\) for the reaction: \(\mathrm{A} \rightleftharpoons \mathrm{D}\) is (a) 3 (b) 6 (c) 12 (d) 24

4 moles each of \(\mathrm{SO}_{2}\) and \(\mathrm{O}_{2}\) gases are allowed to react to form \(\mathrm{SO}_{3}\) in a closed vessel. At equilibrium \(25 \%\) of \(\mathrm{O}_{2}\) is used up. The total number of moles of all the gases at equilibrium is: (a) \(6.5\) (b) \(7.0\) (c) \(8.0\) (d) \(2.0\)

In which of the following cases does the reaction go farthest to completion: (a) \(\mathrm{K}=1\) (b) \(\mathrm{K}=10\) (c) \(\mathrm{K}=10^{-2}\) (d) \(\mathrm{K}=10^{2}\)

\(1.25\) 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) \(4.6 \times 10^{-4}\)

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