Tungsten is usually produced by the reduction of WO3 with hydrogen:
Consider the following data:
WO3(s) | H2O(g) | ||
|
−839.9 | −241.8 | |
|
−763–1 | −228.6 |
- a It K > 1 or < 1 at 25°C? Explain your answer.
- b What is the value of ΔS° at 25°C?
- c What is the temperature at which ΔG° equals zero for this reaction at 1 atm pressure?
- d What is the driving force of this reaction?
(a)
Interpretation:
For the given reaction, the value of
Concept introduction:
Standard free energy change:
Standard free energy change is measured by subtracting the product of temperature and standard entropy change from the standard enthalpy change of a system.
Spontaneous process:
The chemical or physical change can takes place by itself without the help of surroundings are called as spontaneous process.
Answer to Problem 18.92QP
For the given reaction the sign of free energy change
Explanation of Solution
To calculate: The value of
Given reaction and information,
Calculate the value of
The sign of free energy change
(b)
Interpretation:
For the given reaction, the value of
Concept introduction:
Standard free energy change:
Standard free energy change is measured by subtracting the product of temperature and standard entropy change from the standard enthalpy change of a system.
Spontaneous process:
The chemical or physical change can takes place by itself without the help of surroundings are called as spontaneous process.
Answer to Problem 18.92QP
The value of entropy change
Explanation of Solution
To calculate: The value of
Calculate the value of
Calculate the value of
The value of entropy change
(c)
Interpretation:
For the given reaction, the value of
Concept introduction:
Standard free energy change:
Standard free energy change is measured by subtracting the product of temperature and standard entropy change from the standard enthalpy change of a system.
Spontaneous process:
The chemical or physical change can takes place by itself without the help of surroundings are called as spontaneous process.
Answer to Problem 18.92QP
At temperature
Explanation of Solution
To calculate: At which temperature the free energy change equals to zero
Consider free energy change
The value of
(d)
Interpretation:
For the given reaction, the value of
Concept introduction:
Standard free energy change:
Standard free energy change is measured by subtracting the product of temperature and standard entropy change from the standard enthalpy change of a system.
Spontaneous process:
The chemical or physical change can takes place by itself without the help of surroundings are called as spontaneous process.
Answer to Problem 18.92QP
The entropy change is the driving force. The value of
Explanation of Solution
To identify: The driving force for the given reaction
The entropy change is the driving force.
The value of
Want to see more full solutions like this?
Chapter 18 Solutions
General Chemistry - Standalone book (MindTap Course List)
- Actually, the carbon in CO2(g) is thermodynamically unstable with respect to the carbon in calcium carbonate(limestone). Verify this by determining the standardGibbs free energy change for the reaction of lime,CaO(s), with CO2(g) to make CaCO3(s).arrow_forwardConsider the reaction Fe2O3(s)+3H2(g)2Fe(s)+3H2O(g) a. Use Gf values in Appendix 4 to calculate G for this reaction. b. Is this reaction spontaneous under standard conditions at 298 K? c. The value of H for this reaction is 100. kJ. At what temperatures is this reaction spontaneous at standard conditions? Assume that H and S do not depend on temperature.arrow_forwardUse the standard free energy of formation data in Appendix G to determine the free energy change for each of the following reactions, which are run under standard state conditions and 25 C. Identify each as either spontaneous or nonspontaneous at these conditions. (a) MnO2(s)Mn(s)+O2(g) (b) H2(g)+Br2(l)2HBr(g) (c) Cu(s)+S(g)CuS(s) (d) 2LiOH(s)+CO2(g)Li2CO3(s)+H2O(g) (e) CH4(g)+O2(g)C(s,graphite)+2H2O(g) (f) CS2(g)+3Cl2(g)CCl4(g)+S2Cl2(g)arrow_forward
- Using values of fH and S, calculate the standard molar free energy of formation, fG, for each of the following: (a) Ca(OH)2(s) (b) Cl(g) (c) Na2CO3(s) Compare your calculated values of fG with those listed in Appendix L. Which of these formation reactions are predicted to be product-favored at equilibrium at 25 C?arrow_forwarda) Calculate the standard enthalpy of formation of NaHCO3(s) given the following information:Δf?°[CO2(g)] = −393.41 kJ mol-1; Δf?° [ NaOH(s) ] = −425.61 kJ mol-1; Δrxn?° [NaOH(s) + CO2(g)⟶NaHCO3(s) ] = −127.5 kJ mol-1 b) Construct a diagram depicting Δrxn?° of the above reaction and the standardenthalpies of formation of the associated products and reactants.arrow_forwardCalculate: C(graphite) + O2(g) → CO2(g) ΔH° = -393.5 kJH2(g) + ½O2(g) → H2O(l) ΔH° = -285.8 kJ CH3OH(l) + 3/2O2(g) → CO2(g) + 2H2O(l) ΔH° = -726.4 kJUsing the data above, calculate the enthalpy change for the reaction below.Reaction: C(graphite) + 2H2(g) + ½O2(g) → CH3OH(l)arrow_forward
- Calculate an equilibrium constant from a standard free energy at 25 ᴼC. The reaction is: ½ H2(g) + ½ I2(s) ⇆ HI(g). Take DG as -1.72 kJ/molarrow_forwardEstimate the difference between ΔHreaction and ΔUreaction (in kJ/mol) for the reaction: 6Li(s) + N2(g) →2Li3N(s) at 298 Karrow_forwardDetermine ΔH for the reaction of 50.0 g SO3 with MgO at 25°C. Values of ΔHf are −602 kJ/mol for MgO, −395 kJ/mol for SO3, and −1280 kJ/mol for MgSO4.arrow_forward
- What is the standard free energy change for the reaction below? Is the reaction expected to be spontaneous under standard conditions? FeS(s) + O2(g) → Fe(s) + SO2(g)arrow_forwardGiven: 2MNO(s) + O2(g) → 2MnO2(s) ArH = –269.6 kJ · mol-1arrow_forwardCalculate ΔH°25° for the reaction: Fe3O4(s)+2 C(s, graphite) = 3 Fe(s) + CO2(g) from the following data: 2 Fe(s)+3/2 O2(g) = Fe2O3(s) ΔH°25°c= -826 kJ Fe(s)+1/2 O2(g)= FeO(s) ΔH°25°c= -272 kJ FeO(s)+Fe2O3(s)= Fe3O4(s) ΔH°25°c= -1121 kJ C(s, graphite)+ CO2(g)= CO2(g) ΔH°25°c= -394 kJarrow_forward
- Chemistry & Chemical ReactivityChemistryISBN:9781337399074Author:John C. Kotz, Paul M. Treichel, John Townsend, David TreichelPublisher:Cengage LearningChemistry by OpenStax (2015-05-04)ChemistryISBN:9781938168390Author:Klaus Theopold, Richard H Langley, Paul Flowers, William R. Robinson, Mark BlaserPublisher:OpenStaxChemistry & Chemical ReactivityChemistryISBN:9781133949640Author:John C. Kotz, Paul M. Treichel, John Townsend, David TreichelPublisher:Cengage Learning
- Chemistry: An Atoms First ApproachChemistryISBN:9781305079243Author:Steven S. Zumdahl, Susan A. ZumdahlPublisher:Cengage LearningChemistryChemistryISBN:9781305957404Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCostePublisher:Cengage Learning