(a)
Interpretation:
On increasing the total pressure, it has to be identified whether the reactant or products will be more favored.
Concept Introduction:
Le Chatelier's principle:
When a change is applied to a system at equilibrium, the equilibrium will shift against the change. The equilibrium of a system will be affected by the change in temperature, pressure, and concentration.
(b)
Interpretation:
On increasing the total pressure, it has to be identified whether the reactant or products will be more favored.
Concept Introduction:
Refer part (a).
(c)
Interpretation:
On increasing the total pressure, it has to be identified whether the reactant or products will be more favored.
Concept Introduction:
Refer part (a).
(d)
Interpretation:
On increasing the total pressure, it has to be identified whether the reactant or products will be more favored.
Concept Introduction:
Refer part (a).
(e)
Interpretation:
On increasing the total pressure, it has to be identified whether the reactant or products will be more favored.
Concept Introduction:
Refer part (a).
Want to see the full answer?
Check out a sample textbook solutionChapter 5 Solutions
Chemical Principles: The Quest for Insight
- At a certain temperature, K=0.29 for the decomposition of two moles of iodine trichloride, ICl3(s), to chlorine and iodine gases. The partial pressure of chlorine gas at equilibrium is three times that of iodine gas. What are the partial pressures of iodine and chlorine at equilibrium?arrow_forwardWrite an equation for an equilibrium system that would lead to the following expressions (ac) for K. (a) K=(Pco)2 (PH2)5(PC2H6)(PH2O)2 (b) K=(PNH3)4 (PO2)5(PNO)4 (PH2O)6 (c) K=[ ClO3 ]2 [ Mn2+ ]2(Pcl2)[ MNO4 ]2 [ H+ ]4 ; liquid water is a productarrow_forwardFor the system SO3(g)SO2(g)+12 O2(g)at 1000 K, K=0.45. Sulfur trioxide, originally at 1.00 atm pressure, partially dissociates to SO2 and O2 at 1000 K. What is its partial pressure at equilibrium?arrow_forward
- . What does it mean to say that a state of chemical or physical equilibrium is dynamic?arrow_forwardAfter a mixture of cis-2-butene and trans-2-butene has reached equilibrium at 600 K, where Kc = 1.47, half of the cis-2-butcne is suddenly removed. Answer these questions: (a) Is the new mixture at equilibrium? Explain why or why not. (b) In the new mixture, which rate is faster, cis trans or trans cis? Or are both rates the same? (c) In an equilibrium mixture, which concentration is larger, cis-2-butenc or trans-2-butene? (d) If the concentration of cis-2-butene at equilibrium is 0.10 mol/L, what will be the concentration of trans-2-butene?arrow_forwardSuppose a reaction has the equilibrium constant K = 1.3 108. What does the magnitude of this constant tell you about the relative concentrations of products and reactants that will be present once equilibrium is reached? Is this reaction likely to be a good source of the products?arrow_forward
- Chemistry by OpenStax (2015-05-04)ChemistryISBN:9781938168390Author:Klaus Theopold, Richard H Langley, Paul Flowers, William R. Robinson, Mark BlaserPublisher:OpenStaxChemistry: Principles and ReactionsChemistryISBN:9781305079373Author:William L. Masterton, Cecile N. HurleyPublisher:Cengage LearningChemistry: Principles and PracticeChemistryISBN:9780534420123Author:Daniel L. Reger, Scott R. Goode, David W. Ball, Edward MercerPublisher:Cengage Learning
- Chemistry: The Molecular ScienceChemistryISBN:9781285199047Author:John W. Moore, Conrad L. StanitskiPublisher:Cengage LearningPhysical ChemistryChemistryISBN:9781133958437Author:Ball, David W. (david Warren), BAER, TomasPublisher:Wadsworth Cengage Learning,