/*! 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 84 At \(900^{\circ} \mathrm{C}, K_{... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

At \(900^{\circ} \mathrm{C}, K_{\mathrm{C}}=0.0108\) for the reaction $$\mathrm{CaCO}_{3}(s) \rightleftharpoons \mathrm{CaO}(s)+\mathrm{CO}_{2}(g)$$ A mixture of \(\mathrm{CaCO}_{3}, \mathrm{CaO},\) and \(\mathrm{CO}_{2}\) is placed in a 10.0 - \(\mathrm{L}\) vessel at \(900^{\circ} \mathrm{C}\) . For the following mixtures, will the amount of \(\mathrm{CaCO}_{3}\) increase, decrease, or remain the same as the system approaches equilibrium? \begin{equation} \begin{array}{l}{\text { (a) } 15.0 \mathrm{g} \mathrm{CaCO}_{3}, 15.0 \mathrm{g} \mathrm{CaO}, \text { and } 4.25 \mathrm{gCO}_{2}} \\ {\text { (b) } 2.50 \mathrm{g} \mathrm{CaCO}_{3}, 25.0 \mathrm{g} \mathrm{CaO}, \text { and } 5.66 \mathrm{g} \mathrm{CO}_{2}} \\ {\text { (a) } 30.5 \mathrm{g} \mathrm{CaCO}_{3}, 25.5 \mathrm{g} \mathrm{CaO}, \text { and } 6.48 \mathrm{g} \mathrm{CO}_{2}}\end{array} \end{equation}

Short Answer

Expert verified
For the given mixtures, the amount of CaCO3 will change as follows: (a) decrease (b) increase (c) increase

Step by step solution

01

(a) Calculate moles of CO2 in the first mixture

Given, 4.25 g of CO2. To determine the number of moles, we will use the formula: moles = mass / molar mass moles = 4.25 g / 44.01 g/mol = 0.0966 mol
02

(b) Calculate moles of CO2 in the second mixture

Given, 5.66 g of CO2. To determine the number of moles, we will use the formula: moles = mass / molar mass moles = 5.66 g / 44.01 g/mol = 0.1286 mol
03

(c) Calculate moles of CO2 in the third mixture

Given, 6.48 g of CO2. To determine the number of moles, we will use the formula: moles = mass / molar mass moles = 6.48 g / 44.01 g/mol = 0.1472 mol ##Step 2: Compute the reaction quotient Qc for each mixture## Next, we will calculate the reaction quotient Qc for each mixture using the formula: Qc = [CO2]
04

(a) Calculate Qc for the first mixture

Qc = [0.0966 mol CO2 / 10 L] = 0.00966
05

(b) Calculate Qc for the second mixture

Qc = [0.1286 mol CO2 / 10 L] = 0.01286
06

(c) Calculate Qc for the third mixture

Qc = [0.1472 mol CO2 / 10 L] = 0.01472 ##Step 3: Compare Qc to Kc to determine the change in CaCO3## Finally, we will compare the Qc value of each mixture to the Kc value (0.0108) to determine if the CaCO3 will increase, decrease, or remain the same as the system approaches equilibrium:
07

(a) Determine the change in CaCO3 for the first mixture

Qc < Kc, therefore the reaction will proceed to the right to reach equilibrium. This means that the amount of CaCO3 will decrease.
08

(b) Determine the change in CaCO3 for the second mixture

Qc > Kc, therefore the reaction will proceed to the left to reach equilibrium. This means that the amount of CaCO3 will increase.
09

(c) Determine the change in CaCO3 for the third mixture

Qc > Kc, therefore the reaction will proceed to the left to reach equilibrium. This means that the amount of CaCO3 will increase.

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 \( Q_c \) for reactions in terms of concentrations, is a numerical value that provides insight into the current state of a chemical reaction. It is calculated in a similar fashion to the equilibrium constant \( K_c \), but for a system that hasn't reached equilibrium.
By comparing \( Q_c \) to \( K_c \), which is the equilibrium constant, we can predict the direction the reaction will need to shift to reach equilibrium.

For our given example, the reaction describes the decomposition of calcium carbonate (\(\text{CaCO}_3\)) into calcium oxide (\(\text{CaO}\)) and carbon dioxide (\(\text{CO}_2\)). Here, since \( \text{CaCO}_3 \) and \( \text{CaO} \) are solids, they don't appear in the \( Q_c \) expression. Only gaseous or aqueous species are included.
  • Calculation: For this reaction, the reaction quotient is \( Q_c = [\text{CO}_2] \).
  • If \( Q_c < K_c \), the reaction will proceed in the forward direction to increase \([\text{CO}_2]\), decreasing \(\text{CaCO}_3\).
  • If \( Q_c > K_c \), the reaction will shift in the reverse direction, increasing \(\text{CaCO}_3\).
Equilibrium Constant
The equilibrium constant, denoted as \( K_c \) for concentration, is a fundamental concept in chemical equilibrium. It quantifies the ratio of concentrations of products to reactants at equilibrium for a given reversible reaction at a specific temperature.
The value of \( K_c \) is unique for every reaction at a particular temperature and does not change regardless of the initial concentrations of reactants or products.
In our example reaction:\[ \text{CaCO}_3(s) \rightleftharpoons \text{CaO}(s) + \text{CO}_2(g) \] \( K_c \) is given as 0.0108. This signifies the concentration of \(\text{CO}_2\) at equilibrium when \(\text{CaCO}_3\) decomposes.
  • The equilibrium position tells us where the balance lies between reactants and products; whether reactants or products are favored in this balance.
  • A \(K_c\) much less than 1 suggests that at equilibrium, the reactants are favored over products, and vice versa for a \(K_c\) larger than 1.
Thus, knowing \(K_c\) helps us understand the stability and progression of a reaction under equilibrium conditions.
Le Chatelier's Principle
Le Chatelier's Principle is a foundational principle in chemistry that predicts how a system at equilibrium will respond to external changes. When an equilibrium system experiences a disturbance like changes in concentration, pressure, or temperature, the system will adjust itself to partially counteract the effect of the disturbance.
This principle can be applied to our reaction with the equilibrium constant \( K_c = 0.0108 \) at \( 900^{\circ} \text{C} \).
  • If we add more \(\text{CO}_2\), it will increase \( Q_c \) temporarily beyond \( K_c \), pushing the reaction toward the reverse direction to decrease the \(\text{CO}_2\) concentration. Consequently, more \(\text{CaCO}_3\) will be formed.
  • Conversely, if \(\text{CO}_2\) is removed, \( Q_c \) falls below \( K_c \), prompting the reaction to shift forward to produce \(\text{CO}_2\), thus decreasing \(\text{CaCO}_3\).
Le Chatelier provides a qualitative method for predicting the effect of changing conditions on equilibrium, offering an intuitive approach for evaluating how equilibrium can be altered by outside forces.

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

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 10.0 -L. vessel contains \(2.67 \times 10^{-2} \mathrm{g}\) of I(g), how many grams of \(\mathrm{I}_{2}\) are in the mixture?

An equilibrium mixture of \(\mathrm{H}_{2}, \mathrm{I}_{2},\) and \(\mathrm{HI}\) at \(458^{\circ} \mathrm{C}\) contains \(0.112 \mathrm{mol} \mathrm{H}_{2}, 0.112 \mathrm{mol} \mathrm{I}_{2},\) and 0.775 \(\mathrm{mol}\) HI in a 5.00 -L. vessel. What are the equilibrium partial pressures when equilibrium is reestablished following the addition of 0.200 \(\mathrm{mol}\) of \(\mathrm{HI}\) ?

Consider the following equilibrium for which \(\Delta H<0\) $$2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{SO}_{3}(g)$$ How will each of the following changes affect an equilibrium mixture of the three gases: (a) O \(_{2}(g)\) is added to the system; (b) the reaction mixture is heated; (c) the volume of the reaction vessel is doubled; (d) a catalyst is added to the mixture; (e) the total pressure of the system is increased by adding a noble gas; (f) \(\mathrm{SO}_{3}(g)\) is removed from the system?

At a temperature of 700 \(\mathrm{K}\) , the forward and reverse rate constants for the reaction \(2 \mathrm{HI}(g) \rightleftharpoons \mathrm{H}_{2}(g)+\mathrm{I}_{2}(g)\) are \(k_{f}=1.8 \times 10^{-3} \mathrm{M}^{-1} \mathrm{s}^{-1}\) and \(k_{r}=0.063 \mathrm{M}^{-1} \mathrm{s}^{-1}\) (a) What is the value of the equilibrium constant \(K_{c}\) at 700 \(\mathrm{K} ?\) (b) Is the forward reaction endothermic or exothermic if the rate constants for the same reaction have values of \(k_{f}=0.097 M^{-1} \mathrm{s}^{-1}\) and \(k_{r}=2.6 \mathrm{M}^{-1} \mathrm{s}^{-1}\) at 800 \(\mathrm{K} ?\)

Consider the hypothetical reaction \(\mathrm{A}(g)+2 \mathrm{B}(g) \rightleftharpoons\) \(2 \mathrm{C}(g),\) for which \(K_{c}=0.25\) at a certain temperature. A 1.00 -L reaction vessel is loaded with 1.00 mol of compound \(C,\) which is allowed to reach equilibrium. Let the variable \(x\) represent the number of mol/L of compound A present at equilibrium. (a) In terms of \(x,\) what are the equilibrium concentrations of compounds \(\mathrm{B}\) and \(\mathrm{C} ?\) (b) What limits must be placed on the value of \(x\) so that all concentrations are positive? (c) By putting the equilibrium concentrations (in terms of \(x\) ) into the equilibrium-constant expression, derive an equation that can be solved for \(x .(\mathbf{d})\) The equation from part (c) is a cubic equation (one that has the form \(a x^{3}+b x^{2}+c x+d=0 )\) . In general, cubic equations cannot be solved in closed form. However, you can estimate the solution by plotting the cubic equation in the allowed range of \(x\) that you specified in part (b). The point at which the cubic equation crosses the \(x\) -axis is the solution. (e) From the plot in part (d), estimate the equilibrium concentrations of \(A, B,\) and C. (Hint: You can check the accuracy of your answer by substituting these concentrations into the equilibrium expression.)

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.