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

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
(b) Both backward and forward reactions take place at equal speed.

Step by step solution

01

Understanding Equilibrium

In a chemical reaction, equilibrium is a state where the concentrations of reactants and products remain constant over time. This occurs because the rate of the forward reaction equals the rate of the backward reaction, not because the reactions stop.
02

Evaluating Option (a)

Option (a) suggests that equilibrium is reached when the backward reaction stops. This is incorrect because both reactions continue to occur at equilibrium, just at equal rates.
03

Evaluating Option (b)

Option (b) suggests equilibrium occurs when both backward and forward reactions take place at equal speed. This is correct, as the rate of the forward reaction equals the backward reaction's rate, maintaining the concentrations of reactants and products.
04

Evaluating Option (c)

Option (c) suggests that equilibrium occurs when both backward and forward reactions stop. This is incorrect because reactions don't stop at equilibrium; they continue at equal pace.
05

Evaluating Option (d)

Option (d) suggests that equilibrium is achieved when the concentration of reactants and products becomes equal. This is incorrect, as equilibrium does not require equal concentrations, only constant concentrations over time due to equal reaction rates.

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.

Reversible Reactions
A reversible reaction is a chemical reaction where the products of the reaction can reform into the original reactants. This means the reaction can proceed in both forward and backward directions. In chemical notation, this is usually represented by a double-headed arrow like this: \( A + B \rightleftharpoons C + D \). Reversible reactions are a key factor in reaching equilibrium. When both the forward and backward reactions are occurring, but at equal rates, the system is said to be in chemical equilibrium. At this point, the concentrations of the reactants and products remain constant over time.

Understanding reversible reactions is crucial because it affects how we view not just single reactions but chemical processes in general. Reversible reactions emphasize the dynamic nature of chemical systems, highlighting that even when macroscopic changes are not observable, molecular-level changes are ongoing.
Reaction Rates
Reaction rates refer to how quickly or slowly reactants are converted into products in a chemical reaction. It's the speed at which a reaction happens. In the context of equilibrium, the rates of the forward and backward reactions become equal.

Several factors influence the rate of a reaction:
  • Concentration: Higher concentrations of reactants usually increase the rate of reaction.
  • Temperature: Increasing temperature typically increases reaction rates due to higher energy molecules.
  • Catalysts: These substances increase reaction rates by lowering the activation energy required for the reaction to proceed.
  • Surface Area: More surface area allows for more collisions between particles, speeding up reactions.
Understanding reaction rates is essential for predicting how a reaction will proceed over time and effectively leveraging temperature, catalysts, or other conditions to achieve desired outcomes.
Equilibrium Constant
The equilibrium constant, represented as \( K_{eq} \), is a value that indicates the ratio of product concentrations to reactant concentrations at equilibrium for a given reaction. It is defined specifically for a reversible reaction at a particular temperature and can be expressed using the equation:\[K_{eq} = \frac{[C]^c [D]^d}{[A]^a [B]^b}\]Here, \([A], [B], [C], \) and \([D]\) represent the concentrations of the reactants and products, while \(a, b, c,\) and \(d\) are their respective coefficients in the balanced chemical equation.

A large \( K_{eq} \) indicates a reaction with a high concentration of products relative to reactants, suggesting the equilibrium lies to the right. Conversely, a small \( K_{eq} \) suggests that reactants are more prevalent than products, and the equilibrium lies to the left.

The equilibrium constant is pivotal in chemistry as it provides insight into the composition of a reaction mixture at equilibrium and helps in predicting the direction and extent of chemical reactions.

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

An element \(\mathrm{X}\) being with oxygen and CO present in air as \(\mathrm{X}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons \mathrm{XO}_{2}(\mathrm{~g}) ; \mathrm{K}_{p_{1}}=27\) \(\mathrm{X}(\mathrm{g})+\mathrm{CO}(\mathrm{g}) \rightleftharpoons \mathrm{XCO}(\mathrm{g}) ; \quad \mathrm{K}_{\mathrm{P}_{2}}=10^{4}\) When \(\mathrm{X}\) at 1 atm is treated with air, \(25 \%\) of it is bound to \(\mathrm{CO}(\mathrm{g})\). The partial pressure of \(\mathrm{CO}(\mathrm{g})\) in air at equilibrium, if partial pressure of \(\mathrm{O}_{2}(\mathrm{~g})\) in air at equilibrium is \(0.2 \mathrm{~atm}\), would be (a) \(1.9 \times 10^{4} \mathrm{~atm}\) (b) \(1.9 \times 10^{-4} \mathrm{~atm}\) (c) \(2.08 \times 10^{4} \mathrm{~atm}\) (d) \(2.08 \times 10^{-4} \mathrm{~atm}\)

\(\mathrm{PCl}_{5}\) is \(50 \%\) dissociated at \(20^{\circ} \mathrm{C}\) and \(\mathrm{l}\) atm pressure. The value of \(K\) is (a) \(0.444\) (b) \(0.555\) (c) \(0.333\) (d) \(0.666\)

For a chemical reaction \(\mathrm{A} \longrightarrow\) product, the mechanism of the reaction postulated was as follows. $$ \mathrm{A} \stackrel{\mathrm{k}_{1}}{\mathrm{~g}_{2}} 3 \mathrm{~B} \frac{\mathrm{k}_{\mathrm{s}}}{\text { R.D. }}{\mathrm{\longrightarrow}} \mathrm{C}_{\mathrm{g}} $$ If the reaction occurred with individual rate constants \(\mathrm{k}_{1}, \mathrm{k}_{2}\) and \(\mathrm{k}_{3}\), determine activation energy for the overall reaction if the activation energies associated with these rate constants are \(\mathrm{E}_{a_{1}}=180 \mathrm{~kJ} \mathrm{~mol}^{-1}, \mathrm{E}_{a_{2}}=90 \mathrm{~kJ}\) \(\mathrm{mol}^{-1}\) and \(\mathrm{E}_{a_{3}}=40 \mathrm{~kJ} \mathrm{~mol}^{-1}\) (a) \(70 \mathrm{~kJ}\) (b) \(-10 \mathrm{~kJ}\) (c) \(310 \mathrm{~kJ}\) (d) \(130 \mathrm{~kJ}\)

At certain temperature compound \(\mathrm{AB}_{2}(\mathrm{~g})\) dissociates according to the reaction \(2 \mathrm{AB}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{AB}(\mathrm{g})+\mathrm{B}_{2}(\mathrm{~g})\) The value of \(K_{p}\) in terms of degree of dissociation ' \(\alpha\) ' and total pressure ' \(\mathrm{P}\) ' is (a) \(\mathrm{P} \frac{\alpha^{3}}{2}\) (b) \(\mathrm{P} \frac{\alpha^{2}}{3}\) (c) \(\mathrm{P} \frac{\alpha^{3}}{3}\) (d) \(\mathrm{P} \frac{\alpha^{2}}{2}\)

In the reaction, \(\mathrm{N}_{2}+3 \mathrm{H}_{2} \rightleftharpoons 2 \mathrm{NH}_{3}+\) heat, relation- ship between \(K_{p}\) and \(K_{c}\) is (a) \(\mathrm{K}_{\mathrm{p}}=\mathrm{K}_{\mathrm{c}}(\mathrm{RT})^{-2}\) (b) \(\mathrm{K}_{\mathrm{p}}=\mathrm{K}_{c}(\mathrm{RT})^{2}\) (c) \(\mathrm{K}_{\mathrm{p}}=\mathrm{K}_{\mathrm{c}}(\mathrm{RT})^{-3}\) (d) \(\mathrm{K}_{\mathrm{c}}=\mathrm{K}_{\mathrm{p}}(\mathrm{RT})^{3}\)

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