ACHIEVE/CHEMICAL PRINCIPLES ACCESS 1TERM
7th Edition
ISBN: 9781319399849
Author: ATKINS
Publisher: MAC HIGHER
expand_more
expand_more
format_list_bulleted
Question
Chapter 5, Problem 5.14E
(a)
Interpretation Introduction
Interpretation:
The numerical form of Clausius - Clapeyron equation for benzene has to be derived.
(b)
Interpretation Introduction
Interpretation:
The appropriate quantity that requires to give a straight line relation between vapor pressure and temperature should be given.
(c)
Interpretation Introduction
Interpretation:
When the external pressure is 0.655 atm, the boiling point of the benzene has be calculated.
(d)
Interpretation Introduction
Interpretation:
The standard Molar Entropy of benzene has be calculated.
Expert Solution & Answer
Want to see the full answer?
Check out a sample textbook solution
Students have asked these similar questions
(a) From the phase diagram for carbon dioxide, determine the state of co2 at 20 atm and 273 K.
(b) Explain why co2 can undergo sublimation under normal condition.
X(9) + 4Y(9) = 3Z(g)
K, = 3.12 x 10 2 bar 2 at a temperature of 287 °C.
Calculate the value of Ke.
Write the formula for hypoiodous acid.
formula: ||
SPECIAL
ΔΣΩ
λμπ
x"
Xo x (s)
(1) (aq)
(g)
()
[]
CLR
Chapter 5 Solutions
ACHIEVE/CHEMICAL PRINCIPLES ACCESS 1TERM
Ch. 5 - Prob. 5A.1ASTCh. 5 - Prob. 5A.1BSTCh. 5 - Prob. 5A.2ASTCh. 5 - Prob. 5A.2BSTCh. 5 - Prob. 5A.3ASTCh. 5 - Prob. 5A.3BSTCh. 5 - Prob. 5A.1ECh. 5 - Prob. 5A.2ECh. 5 - Prob. 5A.3ECh. 5 - Prob. 5A.4E
Ch. 5 - Prob. 5A.5ECh. 5 - Prob. 5A.6ECh. 5 - Prob. 5A.7ECh. 5 - Prob. 5A.8ECh. 5 - Prob. 5A.11ECh. 5 - Prob. 5B.1ASTCh. 5 - Prob. 5B.1BSTCh. 5 - Prob. 5B.2ASTCh. 5 - Prob. 5B.2BSTCh. 5 - Prob. 5B.3ASTCh. 5 - Prob. 5B.3BSTCh. 5 - Prob. 5B.1ECh. 5 - Prob. 5B.2ECh. 5 - Prob. 5B.3ECh. 5 - Prob. 5B.5ECh. 5 - Prob. 5B.7ECh. 5 - Prob. 5C.1ASTCh. 5 - Prob. 5C.1BSTCh. 5 - Prob. 5C.2ASTCh. 5 - Prob. 5C.2BSTCh. 5 - Prob. 5C.3ASTCh. 5 - Prob. 5C.3BSTCh. 5 - Prob. 5C.1ECh. 5 - Prob. 5C.3ECh. 5 - Prob. 5C.4ECh. 5 - Prob. 5C.5ECh. 5 - Prob. 5C.6ECh. 5 - Prob. 5C.7ECh. 5 - Prob. 5C.8ECh. 5 - Prob. 5C.9ECh. 5 - Prob. 5C.10ECh. 5 - Prob. 5C.11ECh. 5 - Prob. 5C.12ECh. 5 - Prob. 5C.15ECh. 5 - Prob. 5C.16ECh. 5 - Prob. 5D.1ASTCh. 5 - Prob. 5D.1BSTCh. 5 - Prob. 5D.1ECh. 5 - Prob. 5D.2ECh. 5 - Prob. 5D.3ECh. 5 - Prob. 5D.4ECh. 5 - Prob. 5D.5ECh. 5 - Prob. 5D.6ECh. 5 - Prob. 5D.7ECh. 5 - Prob. 5D.8ECh. 5 - Prob. 5D.9ECh. 5 - Prob. 5D.10ECh. 5 - Prob. 5D.11ECh. 5 - Prob. 5D.12ECh. 5 - Prob. 5D.13ECh. 5 - Prob. 5D.14ECh. 5 - Prob. 5D.15ECh. 5 - Prob. 5D.16ECh. 5 - Prob. 5D.18ECh. 5 - Prob. 5D.19ECh. 5 - Prob. 5D.20ECh. 5 - Prob. 5E.1ASTCh. 5 - Prob. 5E.1BSTCh. 5 - Prob. 5E.2ASTCh. 5 - Prob. 5E.2BSTCh. 5 - Prob. 5E.1ECh. 5 - Prob. 5E.2ECh. 5 - Prob. 5E.11ECh. 5 - Prob. 5E.12ECh. 5 - Prob. 5F.1ASTCh. 5 - Prob. 5F.1BSTCh. 5 - Prob. 5F.2ASTCh. 5 - Prob. 5F.2BSTCh. 5 - Prob. 5F.3ASTCh. 5 - Prob. 5F.3BSTCh. 5 - Prob. 5F.4ASTCh. 5 - Prob. 5F.4BSTCh. 5 - Prob. 5F.5ASTCh. 5 - Prob. 5F.5BSTCh. 5 - Prob. 5F.1ECh. 5 - Prob. 5F.2ECh. 5 - Prob. 5F.3ECh. 5 - Prob. 5F.5ECh. 5 - Prob. 5F.7ECh. 5 - Prob. 5F.9ECh. 5 - Prob. 5F.10ECh. 5 - Prob. 5F.11ECh. 5 - Prob. 5F.12ECh. 5 - Prob. 5F.13ECh. 5 - Prob. 5F.14ECh. 5 - Prob. 5F.15ECh. 5 - Prob. 5F.16ECh. 5 - Prob. 5G.1ASTCh. 5 - Prob. 5G.1BSTCh. 5 - Prob. 5G.2ASTCh. 5 - Prob. 5G.2BSTCh. 5 - Prob. 5G.3ASTCh. 5 - Prob. 5G.3BSTCh. 5 - Prob. 5G.4ASTCh. 5 - Prob. 5G.4BSTCh. 5 - Prob. 5G.5ASTCh. 5 - Prob. 5G.5BSTCh. 5 - Prob. 5G.1ECh. 5 - Prob. 5G.2ECh. 5 - Prob. 5G.3ECh. 5 - Prob. 5G.4ECh. 5 - Prob. 5G.7ECh. 5 - Prob. 5G.8ECh. 5 - Prob. 5G.9ECh. 5 - Prob. 5G.11ECh. 5 - Prob. 5G.12ECh. 5 - Prob. 5G.13ECh. 5 - Prob. 5G.14ECh. 5 - Prob. 5G.15ECh. 5 - Prob. 5G.16ECh. 5 - Prob. 5G.17ECh. 5 - Prob. 5G.19ECh. 5 - Prob. 5G.20ECh. 5 - Prob. 5G.21ECh. 5 - Prob. 5G.22ECh. 5 - Prob. 5H.1ASTCh. 5 - Prob. 5H.1BSTCh. 5 - Prob. 5H.2ASTCh. 5 - Prob. 5H.2BSTCh. 5 - Prob. 5H.1ECh. 5 - Prob. 5H.2ECh. 5 - Prob. 5H.3ECh. 5 - Prob. 5H.4ECh. 5 - Prob. 5H.5ECh. 5 - Prob. 5H.6ECh. 5 - Prob. 5I.1ASTCh. 5 - Prob. 5I.1BSTCh. 5 - Prob. 5I.2ASTCh. 5 - Prob. 5I.2BSTCh. 5 - Prob. 5I.3ASTCh. 5 - Prob. 5I.3BSTCh. 5 - Prob. 5I.4ASTCh. 5 - Prob. 5I.4BSTCh. 5 - Prob. 5I.1ECh. 5 - Prob. 5I.2ECh. 5 - Prob. 5I.3ECh. 5 - Prob. 5I.4ECh. 5 - Prob. 5I.5ECh. 5 - Prob. 5I.6ECh. 5 - Prob. 5I.7ECh. 5 - Prob. 5I.9ECh. 5 - Prob. 5I.10ECh. 5 - Prob. 5I.11ECh. 5 - Prob. 5I.12ECh. 5 - Prob. 5I.13ECh. 5 - Prob. 5I.14ECh. 5 - Prob. 5I.15ECh. 5 - Prob. 5I.16ECh. 5 - Prob. 5I.17ECh. 5 - Prob. 5I.18ECh. 5 - Prob. 5I.19ECh. 5 - Prob. 5I.20ECh. 5 - Prob. 5I.21ECh. 5 - Prob. 5I.22ECh. 5 - Prob. 5I.23ECh. 5 - Prob. 5I.24ECh. 5 - Prob. 5I.25ECh. 5 - Prob. 5I.26ECh. 5 - Prob. 5I.27ECh. 5 - Prob. 5I.28ECh. 5 - Prob. 5I.29ECh. 5 - Prob. 5I.30ECh. 5 - Prob. 5I.32ECh. 5 - Prob. 5I.33ECh. 5 - Prob. 5I.34ECh. 5 - Prob. 5I.35ECh. 5 - Prob. 5I.36ECh. 5 - Prob. 5J.1ASTCh. 5 - Prob. 5J.1BSTCh. 5 - Prob. 5J.3ASTCh. 5 - Prob. 5J.3BSTCh. 5 - Prob. 5J.4ASTCh. 5 - Prob. 5J.4BSTCh. 5 - Prob. 5J.5ASTCh. 5 - Prob. 5J.5BSTCh. 5 - Prob. 5J.1ECh. 5 - Prob. 5J.2ECh. 5 - Prob. 5J.3ECh. 5 - Prob. 5J.4ECh. 5 - Prob. 5J.5ECh. 5 - Prob. 5J.6ECh. 5 - Prob. 5J.9ECh. 5 - Prob. 5J.10ECh. 5 - Prob. 5J.11ECh. 5 - Prob. 5J.12ECh. 5 - Prob. 5J.13ECh. 5 - Prob. 5J.17ECh. 5 - Prob. 5.1ECh. 5 - Prob. 5.2ECh. 5 - Prob. 5.3ECh. 5 - Prob. 5.4ECh. 5 - Prob. 5.5ECh. 5 - Prob. 5.6ECh. 5 - Prob. 5.7ECh. 5 - Prob. 5.8ECh. 5 - Prob. 5.9ECh. 5 - Prob. 5.10ECh. 5 - Prob. 5.11ECh. 5 - Prob. 5.12ECh. 5 - Prob. 5.13ECh. 5 - Prob. 5.14ECh. 5 - Prob. 5.15ECh. 5 - Prob. 5.16ECh. 5 - Prob. 5.17ECh. 5 - Prob. 5.19ECh. 5 - Prob. 5.23ECh. 5 - Prob. 5.24ECh. 5 - Prob. 5.25ECh. 5 - Prob. 5.26ECh. 5 - Prob. 5.27ECh. 5 - Prob. 5.28ECh. 5 - Prob. 5.29ECh. 5 - Prob. 5.30ECh. 5 - Prob. 5.31ECh. 5 - Prob. 5.32ECh. 5 - Prob. 5.33ECh. 5 - Prob. 5.35ECh. 5 - Prob. 5.37ECh. 5 - Prob. 5.38ECh. 5 - Prob. 5.41ECh. 5 - Prob. 5.43ECh. 5 - Prob. 5.44ECh. 5 - Prob. 5.45ECh. 5 - Prob. 5.46ECh. 5 - Prob. 5.47ECh. 5 - Prob. 5.49ECh. 5 - Prob. 5.51ECh. 5 - Prob. 5.53ECh. 5 - Prob. 5.55ECh. 5 - Prob. 5.57ECh. 5 - Prob. 5.58ECh. 5 - Prob. 5.61ECh. 5 - Prob. 5.62E
Knowledge Booster
Similar questions
- Define the joule in terms of SI base units.arrow_forwardThe normal boiling point of a liquid is (a) the temperature at which the vapor pressure equals 1 atm. (b) the temperature at which the gas molecules have more kinetic energy than the molecules in the liquid. (c) the only temperature at which there can be equilibrium between liquid and gas. (d) the temperature at which the liquid will usually boil.arrow_forwardThe vapor pressure of water at 5 degrees Celsius is 0.008612 atm. Calculate the vapor pressure in kPa.arrow_forward
- Indicate whether each of the following statements is true orfalse. If it is false, correct it. (a) The feasibility of manufacturingNH3 from N2 and H2 depends entirely on the valueof ΔH for the process N2(g) + 3 H2(g)----->2 NH3(g).(b) The reaction of Na(s) with Cl2(g) to form NaCl(s) is aspontaneous process. (c) A spontaneous process can in principlebe conducted reversibly. (d) Spontaneous processes ingeneral require that work be done to force them to proceed.(e) Spontaneous processes are those that are exothermicand that lead to a higher degree of order in the system.arrow_forwardC6H12O6(aq) + 6O2(g) 6CO2(g) + 6H2O(l)ΔH = –2802.7 kJ mol –1a) Write an expression for the equilibrium constant for this reaction.b) At equilibrium, the concentration of the reactants and products are determined as [CO2] = 0.30 M, [O2] = 0.040 M and [C6H12O6] = 0.065 M. Determine the value of the equilibrium constant (Kc) and predict the whether the products or reactants will be favoured at equilibrium.c) Given that the concentrations of the reactants and products at a particular time are [CO2] = 0.65 M, [O2] = 0.020 M and [C6H12O6] = 0.055 M, determine the reaction quotient (Qc). Compare the Kc and Qc values and predict the favoured direction of the reaction.d) Explain the effect on equilibrium of:i) Increasing temperatureii) Increasing pressureiii) Decreasing the concentration of oxygeniv) Increasing the concentration of carbon dioxidev) Adding a catalystarrow_forwardIodine has an orthorhombic unit cell for which the a, b, and c lattice parameters are 0.479, 0.725, and 0.978 nm, respectively. (a) If the atomic packing factor and atomic radius are 0.547 and 0.177 nm, respectively, determine the number of atoms in each unit cell. (b) The atomic weight of iodine is 126.91 g/mol; compute its theoretical density.arrow_forward
- Examining the crystal structure of CsCl (Caesium Chloride), the Cs+ions form the 8 corners of a cube and the Cl−ion is a the center of the cube. From first-principles calculation, it was determined that the lattice constant of CsCl is 4.209 ̊A. (a) Calculate in detail the electrostatic force exerted by all the Cs+ atoms to the Cl−atom; (b) Assuming that 1 Cs+atom is missing in crystal structure (crystal is said to have a defect), calculate in detail what will be the net electrostatic force on the Cl−ion due to the remaining Cs+ions.arrow_forwardProvide the formal definition of enthalpy and explain all symbols used in it. Explain how the change in enthalpy is related to certain thermodynamic property (other than Gibbs energy).arrow_forwardBalanced chemical equation for conversion of Al(s) to KAl(SO4)2·12H2O(s) in aqueous solutionarrow_forward
- 6. When the ideal-gas reaction A+B=C+Dhas reached equilibrium, state whether or not each of the following relations must be true. Here n¡ is the number of moles of species i in equilibrium, P, is the partial pressure of i, and µ; is the chemical potential of i. Here a simple True or False answer is sufficient. (a) nc+np=nA +ng (b) Pc+Pp=PA+PB (c) na=ng (d) nc=na (e) If only A and B are present initially, then nc=np (f) Ha + HB= Hc+ Hp no matter what the initial composition. (g) If only A and B are present initially, then in equilibrium we must have nc # 0. (h) The equilibrium constant Kp(T)= PĄPB/(PcPp). (i) The value of –RT In Kp(T) = µE(T)+µ8(T)– H3(T) –- H§(T). ) The equilibrium constant is independent of the total pressure.arrow_forward(a) The equilibrium constant of the isomerization of cis-2-butene to trans-2-butene is 2.07 at 127.0 °C. Determine the standard molar Gibbs free energy, AGR º for the reaction. (b) For C6o, the standard molar Gibbs free energy (AG, º) of formation from the elemental carbon is 23.98 kJ/mol at 25.0 °C. Determine the equilibrium constant, Keq for the formation of C60.arrow_forwardThe phase diagram for SO2 is shown here. (a) What doesthis diagram tell you about the enthalpy change in thereaction SO2(l) ---->SO2(g)? (b) Calculate the equilibriumconstant for this reaction at 100 °C and at 0 °C.(c) Why is it not possible to calculate an equilibrium constantbetween the gas and liquid phases in the supercriticalregion? (d) At which of the three points marked in reddoes SO2(g) most closely approach ideal-gas behavior?(e) At which of the three red points does SO2(g) behaveleast ideally?arrow_forward
arrow_back_ios
SEE MORE QUESTIONS
arrow_forward_ios
Recommended textbooks for you
Chemistry: The Molecular ScienceChemistryISBN:9781285199047Author:John W. Moore, Conrad L. StanitskiPublisher: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 Learning
Chemistry: The Molecular Science
Chemistry
ISBN:9781285199047
Author:John W. Moore, Conrad L. Stanitski
Publisher:Cengage Learning
General Chemistry - Standalone book (MindTap Cour...
Chemistry
ISBN:9781305580343
Author:Steven D. Gammon, Ebbing, Darrell Ebbing, Steven D., Darrell; Gammon, Darrell Ebbing; Steven D. Gammon, Darrell D.; Gammon, Ebbing; Steven D. Gammon; Darrell
Publisher:Cengage Learning