Using enthalpies of formation, calculate AH° for the following reaction at 25°C. Also calculate AS° for this reaction from standard entropies at 25°C. Use these values to calculate AG for the reaction at this temperature. Species AH; (kJ/mol) S (J/mol-K) CH, OH() -238.7 126.8 02(9) 205.0 CO2(g) 393.5 213.7 H,0(1) -285.8 69.95 2CH, OH(1) + 302(9) → 2CO2 (9) + 4H,0(1) AH kJ AS = J/K AG° = kJ

Using enthalpies of formation, calculate AH° for the following reaction at 25°C. Also calculate AS° for this reaction from standard entropies at 25°C. Use these values to calculate AG for the reaction at this temperature. Species AH; (kJ/mol) S (J/mol-K) CH, OH() -238.7 126.8 02(9) 205.0 CO2(g) 393.5 213.7 H,0(1) -285.8 69.95 2CH, OH(1) + 302(9) → 2CO2 (9) + 4H,0(1) AH kJ AS = J/K AG° = kJ

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...

See similar textbooks

Similar questions

None
What is the standard free-energy change for the following reaction: 2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l) at 25 0C? Values of delta Gf of substances at standard conditions are as follows: Benzene = 124.5 kJ/mol, Carbon dioxide = -394.4 kJ/mol, Oxygen gas = 0.0 kJ/mol, and Liquid water =A. -237.2 kJ/mol. B. -64.05 kJC. +640.5 kJD. -6405 kJE, None of the above
For the reaction 2H₂S(g) + 302(g) → 2H₂O(1) + 2SO2 (9) AH° -1124.0 kJ and AS - 390.7 J/K The standard free energy change for the reaction of 1.74 moles of H₂S(g) at 300 K, 1 atm would be This reaction is favored under standard conditions at 300 K. Assume that AH and AS are independent of temperature. kJ.
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?
(Q70) Using the tabulated values from the text (located in Appendix G, beginning at p. 1211) calculate (and report in kJ) the standard Gibbs free energy (AG°rxn) for the net reaction that occurs during photosynthesis: 6 H2O (1) + 6 CO2 (g) --> C6H12O6 (s) + 6 O2 (g). The AG°; for glucose (C6H1206) is -1273.3 kJ/mol.
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)…
Explain pleaseeeeeeeeeee