
a.
Interpretation: The concentrations of all the species at equilibrium needs to be calculated for the given reaction when 2.0 moles of pure
Concept Introduction: The relationship between reactants and products of a reaction in equilibrium with respect to some unit is said to be equilibrium expression. It is the expression that gives ratio between products and reactants. The expression is:
a.

Answer to Problem 48E
Explanation of Solution
Given:
The reaction:
The concentration of
Substituting the values as:
The ICE table for the reaction is:
The expression for the equilibrium constant is:
Substituting the values:
Let the value of x be small so,
Solving for x:
Thus, the equilibrium concentration of all the species at equilibrium is:
Interpretation: The concentrations of all the species at equilibrium needs to be calculated for the given reaction when 2.0 moles of
Concept Introduction: The relationship between reactants and products of a reaction in equilibrium with respect to some unit is said to be equilibrium expression. It is the expression that gives ratio between products and reactants. The expression is:

Answer to Problem 48E
Explanation of Solution
The concentration of
Substituting the values as:
So, the amount of NO and Cl2 is 2.0 M and 1.0 M respectively.
As the value of K is very small so the reverse reaction can be assumed to proceed to completion and generation 2.0 M of NOCl. So,
The ICE table for the reaction is:
The expression for the equilibrium constant is:
Substituting the values:
Let the value of x be small so,
Solving for x:
Thus, the equilibrium concentration of all the species at equilibrium is:
b.
Interpretation: The concentrations of all the species at equilibrium needs to be calculated for the given reaction when 1.0 mole of
Concept Introduction: The relationship between reactants and products of a reaction in equilibrium with respect to some unit is said to be equilibrium expression. It is the expression that gives ratio between products and reactants. The expression is:
b.

Answer to Problem 48E
Explanation of Solution
The concentration of
Substituting the values as:
So, the amount of NOCl and NO is 1.0 M.
The ICE table for the reaction is:
The expression for the equilibrium constant is:
Substituting the values:
Let the value of x be small so,
Solving for x:
Thus, the equilibrium concentration of all the species at equilibrium is:
c.
Interpretation: The concentrations of all the species at equilibrium needs to be calculated for the given reaction when 3.0 moles of
Concept Introduction: The relationship between reactants and products of a reaction in equilibrium with respect to some unit is said to be equilibrium expression. It is the expression that gives ratio between products and reactants. The expression is:
c.

Answer to Problem 48E
Explanation of Solution
The concentration of
Substituting the values as:
So, the amount of NO and Cl2 is 3.0 M and 1.0 M respectively.
As the value of K is very small so the reverse reaction can be assumed to proceed to completion and generation of NOCl takes place.
Since, the ratio of NO:Cl2is 3:1 instead of 2:1 so, Cl2 will limit the production of NOCl that is Cl2 is the limiting reagent. So, the Cl2 will react completely and amount of NO reacted will be 3.0 − 2.0 = 1.0 M and amount of NOCl formed will be 2.0 M.
The ICE table for the reaction is:
The expression for the equilibrium constant is:
Substituting the values:
Let the value of x be small so,
Solving for x:
Thus, the equilibrium concentration of all the species at equilibrium is:
d.
Interpretation: The concentrations of all the species at equilibrium needs to be calculated for the given reaction when 2.0 moles of
Concept Introduction: The relationship between reactants and products of a reaction in equilibrium with respect to some unit is said to be equilibrium expression. It is the expression that gives ratio between products and reactants. The expression is:
d.

Answer to Problem 48E
Explanation of Solution
The concentration of
Substituting the values as:
As the value of K is very small so the reverse reaction can be assumed to proceed to completion and generation 4.0 M of NOCl. So,
The ICE table for the reaction is:
The expression for the equilibrium constant is:
Substituting the values:
Let the value of x be small so,
Solving for x:
Thus, the equilibrium concentration of all the species at equilibrium is:
e.
Interpretation: The concentrations of all the species at equilibrium needs to be calculated for the given reaction when the concentration of all the gases is 1.00 mol/L.
Concept Introduction: The relationship between reactants and products of a reaction in equilibrium with respect to some unit is said to be equilibrium expression. It is the expression that gives ratio between products and reactants. The expression is:
e.

Answer to Problem 48E
Explanation of Solution
As the value of K is very small so the reverse reaction can be assumed to proceed to completion and generation of NOCl takes place.
Since, the ratio of NO:Cl2is 1:1 instead of 2:1 so, NO will limit the production of NOCl that is NO is the limiting reagent. So, the NO will react completely and amount of Cl2 reacted will be 1.0 − 0.5 = 0.5 M and amount of NOCl formed will be 1.0 + 1.0 = 2.0 M.
The ICE table for the reaction is:
The expression for the equilibrium constant is:
Substituting the values:
Let the value of x be small so,
Solving for x:
Thus, the equilibrium concentration of all the species at equilibrium is:
Want to see more full solutions like this?
Chapter 6 Solutions
EBK CHEMICAL PRINCIPLES
- Curved arrows are used to illustrate the flow of electrons. Using the provided starting and product structures, draw the curved electrons-pushing arrows for the following reaction or mechanistic step(s).arrow_forwardCurved arrows are used to illustrate the flow of electrons. Using the provided starting and product structures, draw the curved electron-pushing arrows for the following reaction or mechanistic step(s). Be sure to account for all bond-breaking and bond-making steps. I I I H Select to Add Arrows HCI, CH3CH2OHarrow_forwardCurved arrows are used to illustrate the flow of electrons. Use the reaction conditions provided and the follow the arrows to draw the intermediate and product in this reaction or mechanistic step(s).arrow_forward
- Curved arrows are used to illustrate the flow of electrons. Use the reaction conditions provided and follow the curved arrows to draw the intermediates and product of the following reaction or mechanistic step(s).arrow_forwardCurved arrows are used to illustrate the flow of electrons. Use the reaction conditions provided and follow the arrows to draw the intermediate and the product in this reaction or mechanistic step(s).arrow_forwardLook at the following pairs of structures carefully to identify them as representing a) completely different compounds, b) compounds that are structural isomers of each other, c) compounds that are geometric isomers of each other, d) conformers of the same compound (part of structure rotated around a single bond) or e) the same structure.arrow_forward
- Given 10.0 g of NaOH, what volume of a 0.100 M solution of H2SO4 would be required to exactly react all the NaOH?arrow_forward3.50 g of Li are combined with 3.50 g of N2. What is the maximum mass of Li3N that can be produced? 6 Li + N2 ---> 2 Li3Narrow_forward3.50 g of Li are combined with 3.50 g of N2. What is the maximum mass of Li3N that can be produced? 6 Li + N2 ---> 2 Li3Narrow_forward
- Concentration Trial1 Concentration of iodide solution (mA) 255.8 Concentration of thiosulfate solution (mM) 47.0 Concentration of hydrogen peroxide solution (mM) 110.1 Temperature of iodide solution ('C) 25.0 Volume of iodide solution (1) used (mL) 10.0 Volume of thiosulfate solution (5:03) used (mL) Volume of DI water used (mL) Volume of hydrogen peroxide solution (H₂O₂) used (mL) 1.0 2.5 7.5 Time (s) 16.9 Dark blue Observations Initial concentration of iodide in reaction (mA) Initial concentration of thiosulfate in reaction (mA) Initial concentration of hydrogen peroxide in reaction (mA) Initial Rate (mA's)arrow_forwardDraw the condensed or line-angle structure for an alkene with the formula C5H10. Note: Avoid selecting cis-/trans- isomers in this exercise. Draw two additional condensed or line-angle structures for alkenes with the formula C5H10. Record the name of the isomers in Data Table 1. Repeat steps for 2 cyclic isomers of C5H10arrow_forwardExplain why the following names of the structures are incorrect. CH2CH3 CH3-C=CH-CH2-CH3 a. 2-ethyl-2-pentene CH3 | CH3-CH-CH2-CH=CH2 b. 2-methyl-4-pentenearrow_forward
- Chemistry: The Molecular ScienceChemistryISBN:9781285199047Author:John W. Moore, Conrad L. StanitskiPublisher:Cengage LearningChemistry & Chemical ReactivityChemistryISBN:9781337399074Author:John C. Kotz, Paul M. Treichel, John Townsend, David TreichelPublisher:Cengage LearningChemistry & Chemical ReactivityChemistryISBN:9781133949640Author:John C. Kotz, Paul M. Treichel, John Townsend, David TreichelPublisher:Cengage Learning
- General Chemistry - Standalone book (MindTap Cour...ChemistryISBN:9781305580343Author:Steven D. Gammon, Ebbing, Darrell Ebbing, Steven D., Darrell; Gammon, Darrell Ebbing; Steven D. Gammon, Darrell D.; Gammon, Ebbing; Steven D. Gammon; DarrellPublisher:Cengage LearningChemistryChemistryISBN:9781305957404Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCostePublisher:Cengage LearningChemistry: An Atoms First ApproachChemistryISBN:9781305079243Author:Steven S. Zumdahl, Susan A. ZumdahlPublisher:Cengage Learning





