Use data from Appendix IIB to calculate the equilibriumconstants at 25 °C for each reaction.a. 2 CO(g) + O2(g) = 2 CO2(g)b. 2 H2S(g) = 2 H2(g) + S2(8)

The standard values at 250C: Temperature T= 250C =298K Gas constant R = 8.314×10-3KJ K-1 ∆G0f(CO2) =…

Answered: Use data from Appendix IIB to calculate…

Principles of Modern Chemistry

8th Edition

ISBN:9781305079113

Author:David W. Oxtoby, H. Pat Gillis, Laurie J. Butler

Publisher:David W. Oxtoby, H. Pat Gillis, Laurie J. Butler

Chapter14: Chemical Equilibrium

Section: Chapter Questions

Problem 16P

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Elemental boron, in the form of thin fibers, can be made by reducing a boron halide with H2. BCl3(g) + 3/2 H2(g) B(s) + 3HCl(g) Calculate H, S, and G at 25 C for this reaction. Is the reaction predicted to be product favored at equilibrium at 25 C? If so, is it enthalpy driven or entropy driven?
For each reaction, an equilibrium constant at 298 K is given. Calculate G for each reaction. (a) Br2()+ H2(g)2HBr(g) KP = 4.4 1018 (b) H2O()H2O(g) KP = 3.17 102 (c) N2(g) +3H2(g)2NH3(g) Kc = 3.5 108
Use the data in Appendix J to calculate rG andKPat 25 C for the reaction 2HBr(g)+Cl2(g)2HCl(g)+Br2() Comment on the connection between the sign of rG and the magnitude ofKP.
The decomposition of diamond to graphite [C(diamond) C(graphite)] is thermodynamically favored, but occurs slowly at room temperature. a. Use fG values from Appendix L to calculate rG and Keq for the reaction under standard conditions and 298.15 K. b. Use fH and S values from Appendix L to estimate rG and Keq for the reaction at 1000 K. Assume that enthalpy and entropy values are valid at these temperatures. Does heating shift the equilibrium toward the formation of diamond or graphite? c. Why is the formation of diamond favored at high pressures? d. The phase diagram shows that diamond is thermodynamically favored over graphite at 20,000 atmospheres pressure (about 2 GPa) at room temperature. Why is this conversion actually done at much higher temperatures and pressures?
For the reaction 2Cu(s)+S(s)Cu2S(s) H and G are negative and S is positive. a At equilibrium, will reactants or products predominate? Why? b Why must the reaction system be heated in order to produce copper(I) sulfide?
A crucial reaction for the production of synthetic fuels is the production of H2 by the reaction of coal with steam. The chemical reaction is C(s) + H2O(g) CO(g) + H2(g) (a) Calculate rG for this reaction at 25 C, assuming C(s) is graphite. (b) Calculate Kp for the reaction at 25 C. (c) Is the reaction predicted to be product-favored at equilibrium at 25 C? If not, at what temperature will it become so?
Elemental boron, in the form of thin fibers, can be made by reducing a boron halide with H2. BCl3(g) + 32 H2(g) B(s) + 3 HCl(g) Calculate rH, rS, and rG at 25 C for this reaction. Is the reaction predicted to be product-favored at equilibrium at 25 C? If so, is it enthalpy- or entropy-driven? [S for B(s) is 5.86 J/K mol.]
Consider the reaction CO(g)+H2O(g)CO2(g)+H2(g) Use the appropriate tables to calculate (a) G at 552C (b) K at 552C
Adenosine triphosphate, ATP, is used as a free-energy source by biological cells. (See the essay on page 624.) ATP hydrolyzes in the presence of enzymes to give ADP: ATP(aq)+H2O(l)ADP(aq)+H2PO4(aq);G=30.5kJ/molat25C Consider a hypothetical biochemical reaction of molecule A to give molecule B: A(aq)B(aq);G=+15.0kJ/molat25C Calculate the ratio [B]/[A] at 25C at equilibrium. Now consider this reaction coupled to the reaction for the hydrolysis of ATP: A(aq)+ATP(aq)+H2O(l)B(aq)+ADP(aq)+H2PO4(aq) If a cell maintains a high ratio of ATP to ADP and H2PO4 by continuously making ATP, the conversion of A to B can be made highly spontaneous. A characteristic value of this ratio is [ATP][ADP][H2PO4]=500 Calculate the ratio [B][A] in this case and compare it with the uncoupled reaction. Compared with the uncoupled reaction, how much larger is this ratio when coupled to the hydrolysis of ATP?
The standard free energies of formation and the standard enthalpies of formation at 298 K for difluoroacetylene (C2F2) and hexafluorobenzene (C6F6) are Gfo(KJ/mol) Hfo(KJ/mol) C2F2(g) 191.2 241.3 Hexane 78.2 132.8 For the following reaction: C6F6(g)3C2F2(g) a. calculate S at 298 K. b. calculate K at 298 K. c. estimate K at 3000. K, assuming H and S do not depend on temperature.
Consider the following reaction at 298 K: 2SO2(g)+O2(g)2SO3(g) An equilibrium mixture contains O2(g) and SO3(g) at partial pressures of 0.50 atm and 2.0 atm, respectively. Using data from Appendix 4, determine the equilibrium partial pressure of SO2 in the mixture. Will this reaction be most favored at a high or a low temperature, assuming standard conditions?
Titanium(IV) oxide is converted to titanium carbide with carbon at a high temperature. TiO2(s) + 3 C(s) 2 CO(g) + TiC(s) (a) Calculate rG and K at 727 C. (b) Is the reaction product-favored at equilibrium at this temperature? (c) How can the reactant or product concentrations be adjusted for the reaction to proceed at 727 C?

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Use data from Appendix IIB to calculate the equilibrium constants at 25 °C for each reaction. a. 2 CO(g) + O2(g) = 2 CO2(g) b. 2 H2S(g) = 2 H2(g) + S2(8)