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) If \(Q_{c}

Short Answer

Expert verified
When \(Q_c < K_c\), the reaction will proceed in the forward direction in order to reach equilibrium. The condition that must be satisfied for \(Q_c = K_c\) is that the reaction must be at equilibrium.

Step by step solution

01

Understanding Reaction Quotient (\(Q_c\)) and Equilibrium Constant (\(K_c\))

For any given reaction, the reaction quotient, \(Q_c\), is defined as the ratio of the concentrations of the products to the concentrations of the reactants, each raised to the power of its stoichiometric coefficient. Similarly, the equilibrium constant, \(K_c\), is a measure of the extent of the reaction when the system has reached equilibrium. If a reaction is at equilibrium, the ratio of the concentrations of the products to the concentrations of the reactants remains constant, and the value of that constant is given by \(K_c\).
02

Analyzing the Given Condition (\(Q_c < K_c\))

The given condition is that the reaction quotient (\(Q_c\)) is less than the equilibrium constant (\(K_c\)). This means that the concentration of the products is relatively low compared to the concentration of the reactants.
03

Determining the Direction to Reach Equilibrium

Since \(Q_c < K_c\), the reaction needs to proceed towards increasing the concentration of products and/or reducing the concentration of reactants in order to reach equilibrium. This direction is the forward direction of the reaction. Therefore, the reaction will proceed in the forward direction to attain equilibrium when \(Q_c < K_c\).
04

(a) Answer

When \(Q_c < K_c\), the reaction will proceed in the forward direction in order to reach equilibrium.
05

Condition for \(Q_c = K_c\)

At equilibrium, the reaction reaches a state where the forward and reverse reactions proceed at the same rate. As a result, the concentrations of the products and reactants remain constant. For this constant state to be achieved, the reaction quotient (\(Q_c\)) must be equal to the equilibrium constant (\(K_c\)). In other words, the condition for \(Q_c = K_c\) is that the reaction must be at equilibrium.
06

(b) Answer

The condition that must be satisfied for \(Q_c = K_c\) is that the reaction must be at 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.

Reaction Quotient Qc
The concept of the reaction quotient, denoted as \(Q_c\), is crucial in understanding chemical equilibrium. In simple terms, \(Q_c\) represents the ratio of concentrations of the reaction's products to those of the reactants at any point in time, with each concentration raised to the power of the corresponding stoichiometric coefficient, just as you would find in the equilibrium constant expression, \(K_c\).

But why is \(Q_c\) important? Imagine you’re conducting a chemical reaction in a flask. This quotient tells you where you are on the journey to equilibrium. If \(Q_c\) is less than the equilibrium constant \(K_c\), as in our original exercise, the reaction will proceed in a forward direction to form more products until equilibrium is achieved. On the other hand, if \(Q_c\) is greater than \(K_c\), the reaction will go in reverse, consuming products to form reactants until \(Q_c\) equals \(K_c\).

During this journey to equilibrium, \(Q_c\) is always changing until it finally matches \(K_c\), signaling that the reaction mixture is in a stable state where the forward and reverse rates are equal. Therefore, measuring \(Q_c\) at different intervals can provide a snapshot of the reaction’s progress towards equilibrium.
Forward and Reverse Reactions
In chemistry, the terms 'forward' and 'reverse' reactions are used to describe the direction in which a reaction is proceeding. A forward reaction refers to the conversion of reactants into products. In contrast, a reverse reaction is the transformation of products back into reactants.

At the start of a reaction, when reactants are freshly mixed, the forward reaction is typically predominant because the concentration of the reactants is high and few, if any, products exist. Over time, as products build up, the reverse reaction becomes more significant, ultimately leading to a dynamic equilibrium when the rates of the forward and reverse reactions are equal, making the net change in reactant and product concentrations zero.

Using the exercise as a starting point, if \(Q_c < K_c\), the forward reaction rate must increase to reach equilibrium, resulting in the formation of more products. Conversely, if we were dealing with a scenario where \(Q_c > K_c\), the reverse reaction would be favored until equilibrium is re-established. This delicate balance between the forward and reverse reactions is the essence of dynamic equilibrium in chemical systems.
Reaction Concentrations
The concentrations of reactants and products play a pivotal role in both the equilibrium state of a reaction and its journey toward that equilibrium. When speaking of reaction concentrations, we are referring to the molarities or partial pressures of the substances involved in the reaction at a given moment.

Factors such as the temperature, pressure, and presence of catalysts can affect these concentrations, and by extension, influence the position of the equilibrium. As a reaction proceeds towards equilibrium, we might see the concentration of reactants decrease as they are turned into products, or we may observe the decrease of product concentrations if the reverse reaction is favored.

Bringing our exercise into context, the condition for \(Q_c = K_c\) is reached at equilibrium. At this point, the concentrations of reactants and products have stabilized in a particular ratio that will not change unless the system is disturbed. Therefore, understanding how to calculate and interpret reaction concentrations is fundamental to predicting whether a reaction will shift to form more reactants or more products under varying conditions—insight that is essential for chemical experimentation and industrial processes alike.

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

A chemist at a pharmaceutical company is measuring equilibrium constants for reactions in which drug candidate molecules bind to a protein involved in cancer. The drug molecules bind the protein in a \(1: 1\) ratio to form a drugprotein complex. The protein concentration in aqueous solution at \(25^{\circ} \mathrm{C}\) is \(1.50 \times 10^{-6} \mathrm{M}\). Drug \(\mathrm{A}\) is introduced into the protein solution at an initial concentration of \(2.00 \times 10^{-6} \mathrm{M}\). Drug \(B\) is introduced into a separate, identical protein solution at an initial concentration of \(2.00 \times 10^{-6} \mathrm{M}\). At equilibrium, the drug A-protein solution has an A-protein complex concentration of \(1.00 \times 10^{-6} \mathrm{M}\), and the drug \(\mathrm{B}\) solution has a B-protein complex concentration of \(1.40 \times 10^{-6} \mathrm{M}\). Calculate the \(K_{c}\) value for the \(A\)-protein binding reaction and for the \(\mathrm{B}\) protein binding reaction. Assuming that the drug that binds more strongly will be more effective, which drug is the better choice for further research?

If \(K_{c}=1\) for the equilibrium \(2 \mathrm{~A}(g) \rightleftharpoons \mathrm{B}(\mathrm{g})\), what is the relationship between \([A]\) and \([B]\) at equilibrium?

When \(2.00 \mathrm{~mol}\) of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) is placed in a 2.00- \(\mathrm{L}\) flask at \(303 \mathrm{~K}\), \(56 \%\) of the \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) decomposes to \(\mathrm{SO}_{2}\) and \(\mathrm{Cl}_{2}\) : $$ \mathrm{SO}_{2} \mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{SO}_{2}(g)+\mathrm{Cl}_{2}(g) $$ (a) Calculate \(K_{c}\) for this reaction at this temperature. (b) Calculate \(K_{p}\) for this reaction at \(303 \mathrm{~K}\). (c) According to Le Châtelier's principle, would the percent of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) that decomposes increase, decrease or stay the same if the mixture were transferred to a \(15.00\)-L vessel? (d) Use the equilibrium constant you calculated above to determine the percentage of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) that decomposes when \(2.00 \mathrm{~mol}\) of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) is placed in a \(15.00\) - \(\mathrm{L}\) vessel at \(303 \mathrm{~K}\).

As shown in Table 15.2, \(K_{p}\) for the equilibrium $$ \mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g) $$ is \(4.51 \times 10^{-5}\) at \(450^{\circ} \mathrm{C}\). For each of the mixtures listed here, indicate whether the mixture is at equilibrium at \(450^{\circ} \mathrm{C}\). If it is not at equilibrium, indicate the direction (toward product or toward reactants) in which the mixture must shift to achieve equilibrium. (a) \(98 \mathrm{~atm} \mathrm{} \mathrm{NH}_{3}, 45 \mathrm{~atm} \mathrm{~N}_{2}, 55 \mathrm{~atm} \mathrm{} \mathrm{H}_{2}\) (b) \(57 \mathrm{~atm} \mathrm{} \mathrm{NH}_{3}, 143 \mathrm{~atm} \mathrm{} \mathrm{N}_{2}\), no \(\mathrm{H}_{2}\) (c) \(13 \mathrm{~atm} \mathrm{NH}_{3}, 27\) atm \(\mathrm{N}_{2}, 82\) atm \(\mathrm{H}_{2}\)

The equilibrium constant \(K_{c}\) for \(\mathrm{C}(s)+\mathrm{CO}_{2}(g) \rightleftharpoons\) \(2 \mathrm{CO}(g)\) is \(1.9\) at \(1000 \mathrm{~K}\) and \(0.133\) at \(298 \mathrm{~K}\). (a) If excess \(\mathrm{C}\) is allowed to react with \(25.0 \mathrm{~g}\) of \(\mathrm{CO}_{2}\) in a \(3.00\) - \(\mathrm{L}\) vessel at 1000 \(\mathrm{K}\), how many grams of \(\mathrm{CO}\) are produced? (b) How many grams of \(\mathrm{C}\) are consumed? (c) If a smaller vessel is used for the reaction, will the yield of CO be greater or smaller? (d) Is the reaction endothermic or exothermic?

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