/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 42 (a) If \(Q_{c}>K_{c}\), how m... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

(a) If \(Q_{c}>K_{c}\), how must the reaction proceed to reach equilibrium? (b) At the start of a certain reaction, only reactants are present; no products have been formed. What is the value of \(Q_{c}\) at this point in the reaction?

Short Answer

Expert verified
(a) Reaction shifts left. (b) \(Q_{c} = 0\).

Step by step solution

01

Understand Reaction Quotient and Equilibrium Constant

The reaction quotient, \(Q_{c}\), is calculated using the concentrations of the reactants and products at any point in time during a chemical reaction. It is given by the same formula as the equilibrium constant \(K_{c}\) but is applied to concentrations at any point, not just at equilibrium. If \(Q_{c} = K_{c}\), the system is at equilibrium.
02

Compare \(Q_{c}\) and \(K_{c}\) to Determine Reaction Shift

When \(Q_{c} > K_{c}\), it means that the concentration of the products is greater than that at equilibrium, or equivalently, the reactant concentration is less than at equilibrium. Therefore, the reaction will shift to the left, favoring the formation of reactants to reach equilibrium.
03

Initial Reaction Conditions Analysis

At the start of the reaction, only reactants are present, meaning that the concentration of products is zero. Hence, the reaction quotient \(Q_{c}\) is zero because it depends on the product concentrations in the numerator, which are zero.
04

Conclusion for Both Scenarios

For part (a), if \(Q_{c} > K_{c}\), the reaction will proceed to the left to form more reactants and reach equilibrium. For part (b), at the beginning of the reaction with no products formed, \(Q_{c}\) equals zero.

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 91Ó°ÊÓ!

Key Concepts

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

Reaction Quotient
The reaction quotient, often represented as \( Q_c \), is a measure used to assess the progression of a chemical reaction at a given moment. It tells us how far a reaction has proceeded and helps predict the direction it needs to go to reach equilibrium. Unlike the equilibrium constant \( K_c \), which is calculated solely with the concentrations at equilibrium, \( Q_c \) can be calculated using the concentrations of reactants and products at any point in the reaction.

To calculate \( Q_c \), you use the same formula as for \( K_c \). However, remember that \( Q_c \) changes over time as the reaction progresses and approaches equilibrium. A helpful way to think about \( Q_c \) is as a snapshot of the reaction at a specific instant. Depending on its value relative to \( K_c \), it indicates if the reaction needs to produce more reactants or more products to reach equilibrium.

Consider a scenario where \( Q_c > K_c \). This indicates that the concentration of products is higher than it should be at equilibrium. As a result, the reaction will "shift left," favoring the production of reactants to restore equilibrium. Conversely, if \( Q_c < K_c \), the reaction will "shift right" to produce more products until equilibrium is achieved. If \( Q_c = K_c \), the reaction is exactly at equilibrium and no shift is necessary.
Equilibrium Constant
The equilibrium constant, denoted as \( K_c \), is a fundamental value that defines the ratio of concentrations of products to reactants for a reaction at equilibrium. It can be thought of as the "target" for the reaction quotient \( Q_c \). When a reaction reaches equilibrium, \( Q_c \) will equal \( K_c \).

Calculating \( K_c \) involves the concentrations of the products over the reactants, each raised to their respective coefficients as written in the balanced chemical equation. This constant value is critical as it provides insight into the extent of the reaction.

Let's break down how \( K_c \) guides us:
  • A large \( K_c \) implies that, at equilibrium, products predominate.
  • A small \( K_c \) suggests reactants are favored at equilibrium.
Essentially, \( K_c \) allows chemists to predict the concentration ratio of reactants to products once the reaction stabilizes.
Chemical Reactions
Chemical reactions involve the transformation of reactants into products and can be described by chemical equations. It's crucial to understand these processes to predict the flow and balance of reactions within a system.

At the onset of a reaction, only reactants are present with no products formed. The reaction then proceeds by rearranging atoms and molecules, which is governed by the principle to minimize free energy. This drive towards a stable arrangement is what propels the reaction forward to equilibrium.

In practice, we start with all reactants and watch for changes as some are converted into products:
  • As products form, reactant quantities decrease.
  • The reaction may stop, or equilibrium might be established where the forward and reverse reaction rates equalize.
Monitoring the concentrations of both reactants and products over time enables us to determine not only where the reaction stands but also adjust conditions to either speed it up or favor a different balance.

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

For the equilibrium $$ \mathrm{PH}_{3} \mathrm{BCl}_{3}(s) \rightleftharpoons \mathrm{PH}_{3}(g)+\mathrm{BCl}_{3}(g) $$ \(K_{p}=5.27\) at \(60^{\circ} \mathrm{C} .(\mathbf{a})\) Calculate \(K_{c} .(\mathbf{b})\) After \(3.00 \mathrm{~g}\) of solid \(\mathrm{PH}_{3} \mathrm{BCl}_{3}\) is added to a closed 1.500-L vessel at \(60^{\circ} \mathrm{C}\), the vessel is charged with \(0.0500 \mathrm{~g}\) of \(\mathrm{BCl}_{3}(g)\). What is the equilibrium concentration of \(\mathrm{PH}_{3}\) ?

Which of the following statements are true and which are false? (a) For the reaction \(2 \mathrm{~A}(g)+\mathrm{B}(g) \rightleftharpoons \mathrm{A}_{2} \mathrm{~B}(g) K_{c}\) and \(K_{p}\) are numerically the same. (b) It is possible to distinguish \(K_{c}\) from \(K_{p}\) by comparing the units used to express the equilibrium constant. (c) For the equilibrium in (a), the value of \(K_{c}\) increases with increasing pressure.

The following equilibria were measured at \(823 \mathrm{~K}\) : $$ \begin{array}{l} \mathrm{CoO}(s)+\mathrm{H}_{2}(g) \rightleftharpoons \mathrm{Co}(s)+\mathrm{H}_{2} \mathrm{O}(g) \quad K_{c}=67 \\ \mathrm{H}_{2}(g)+\mathrm{CO}_{2}(g) \rightleftharpoons \mathrm{CO}(g)+\mathrm{H}_{2} \mathrm{O}(g) \quad K_{c}=0.14 \end{array} $$ (a) Use these equilibria to calculate the equilibrium constant, \(K_{c}\), for the reaction \(\mathrm{CoO}(s)+\mathrm{CO}(g) \rightleftharpoons \mathrm{Co}(s)\) \(+\mathrm{CO}_{2}(g)\) at \(823 \mathrm{~K} .(\mathbf{b})\) Based on your answer to part (a), would you say that carbon monoxide is a stronger or weaker reducing agent than \(\mathrm{H}_{2}\) at \(T=823 \mathrm{~K} ?(\mathbf{c})\) If you were to place \(5.00 \mathrm{~g}\) of \(\mathrm{CoO}(s)\) in a sealed tube with a volume of \(250 \mathrm{~mL}\) that contains \(\mathrm{CO}(g)\) at a pressure of \(101.3 \mathrm{kPa}\) and a temperature of \(298 \mathrm{~K},\) what is the concentration of the CO gas? Assume there is no reaction at this temperature and that the CO behaves as an ideal gas (you can neglect the volume of the solid). \((\mathbf{d})\) If the reaction vessel from part (c) is heated to \(823 \mathrm{~K}\) and allowed to come to equilibrium, how much \(\operatorname{CoO}(s)\) remains?

At \(800 \mathrm{~K},\) the equilibrium constant for \(\mathrm{I}_{2}(g) \rightleftharpoons 2 \mathrm{I}(g)\) is \(K_{c}=3.1 \times 10^{-5}\). If an equilibrium mixture in a 5.00-L vessel contains \(30.5 \mathrm{mg}\) of \(\mathrm{I}(g)\), how many grams of \(\mathrm{I}_{2}\) are in the mixture?

Mercury(I) oxide decomposes into elemental mercury and elemental oxygen: \(2 \mathrm{Hg}_{2} \mathrm{O}(s) \rightleftharpoons 4 \mathrm{Hg}(l)+\mathrm{O}_{2}(g)\) (a) Write the equilibrium-constant expression for this reaction in terms of partial pressures. (b) Suppose you run this reaction in a solvent that dissolves elemental mercury and elemental oxygen. Rewrite the equilibriumconstant expression in terms of molarities for the reaction, using (solv) to indicate solvation.

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.