(d) For the reaction: H3C. H3C, CC-OH Cc=CH2 H20 H3C H3C Reaction 2 Reaction 2 does not go to completion and the position of the reaction can be defined in terms of the equilibrium constant, K. (i) For the above reaction at 298K under specific conditions the following thermodynamic information were obtained, reaction Gibbs free energy (AreactG = -5.4 kJ mol-1), standard reaction enthalpy (AreactH® = -52.7 kJ mol-1) and standard reaction entropy (AreactS® = -0.16 kJ mol-1). From the above detail, calculate the equilibrium constant. (ii) Sketch a diagram of the Gibbs free energy versus reaction progress and on diagram label the reaction Gibbs free energy (A,react G) and standard reaction Gibbs free energy (AreactG®).

(d) For the reaction: H3C. H3C, CC-OH Cc=CH2 H20 H3C H3C Reaction 2 Reaction 2 does not go to completion and the position of the reaction can be defined in terms of the equilibrium constant, K. (i) For the above reaction at 298K under specific conditions the following thermodynamic information were obtained, reaction Gibbs free energy (AreactG = -5.4 kJ mol-1), standard reaction enthalpy (AreactH® = -52.7 kJ mol-1) and standard reaction entropy (AreactS® = -0.16 kJ mol-1). From the above detail, calculate the equilibrium constant. (ii) Sketch a diagram of the Gibbs free energy versus reaction progress and on diagram label the reaction Gibbs free energy (A,react G) and standard reaction Gibbs free energy (AreactG®).

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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the free energy for a reaction can be related to the equilibriumconstant through the formula below.K = e (-ΔG° / RT)Therefore if Kc for a reaction is known, Go can be determined, or vice versa. Furthermore, ifyou have the value for Go at two different temperatures, you can calculate H and S throughthe familiar equation for Gibbs energy below, since you have two unknowns but also twoequations.G = H – T SIn this lab you will be studying the solubility of borax (Na2B4O5(OH)4*8H2O), a slightly solublesodium salt, at two different temperatures. When solid borax is added to water, theequilibrium below is established.Na2B4O5(OH)4*8H2O (s) 2 Na+ (aq) + B4O5(OH)42- (aq) + 8 H2O(l)If you measure the concentrations for those substances that show up in the reaction quotient,then the Kc for the reaction at that temperature can be calculated. In this lab, theconcentration of borate ion (B4O5(OH)42-) in solution will be measured by titration with standardhydrochloric acid according to the…
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From these two reactions at 298 K,
For the reaction NH,C(aq)NH,(2) + HCI(aq) AH =86.4 kJ and AS° = 79.1 J/K %3D The equilibrium constant for this reaction at 286.0 K is Assume that AH° and AS° are independent of temperature.
Make a non-spontaneous process spontaneous A process that is nonspontaneous can be made spontaneous by coupling it with another process tha is highly spontaneous. The coupling of nonspontaneous reactions with highly spontaneous ones is important in biological system. The oxidization of glucose, for example, is highly spontaneous: C6H12O6(s) + 60₂(g) → 6 CO₂(g) + 6 H₂O(1) Spontaneous reactions such as this ultimately drive the nonspontaneous reactions necessary to sustain life. Can you use Gibbs free energy, enthalpy, and entropy to argue this reaction is highly spontaneous?
A chemical engineer is studying the two reactions shown in the table below. In each case, he fills a reaction vessel with some mixture of the reactants and products at a constant temperature of 136.0 °C and constant total pressure. Then, he measures the reaction enthalpy AH and reaction entropy AS of the first reaction, and the reaction enthalpy AH and reaction free energy AG of the second reaction. The results of his measurements are shown in the table. Complete the table. That is, calculate AG for the first reaction and AS for the second. (Round your answer to zero decimal places.) Then, decide whether, under the conditions the engineer has set up, the reaction is spontaneous, the reverse reaction is spontaneous, or neither forward nor reverse reaction is spontaneous because the system is at equilibrium. AH = -2220, kJ J AS = -5426. K C,H, (g) + 50,(8) → 3C0,g) + 4H,0() AG = kJ Which is spontaneous? O this reaction O the reverse reaction O neither ΔΗ 50. ky AS = 6CI, (g) + 2Fe,0.(s)…
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