9-10. At 298K and 100kPa, diamond and graphite's molar entropy and molar combustion heat are listed below. molar entropy S/ (J.K*mol) 2.45 A̟H„/(kJ · mol) molar combustion Diamond -395.40 Graphite 5.71 -393.51 m of the reaction graphite to diamond. 10) Which one is more stable, diamond or graphite? 9) Calculate the Atrs

9-10. At 298K and 100kPa, diamond and graphite's molar entropy and molar combustion heat are listed below. molar entropy S/ (J.K*mol) 2.45 A̟H„/(kJ · mol) molar combustion Diamond -395.40 Graphite 5.71 -393.51 m of the reaction graphite to diamond. 10) Which one is more stable, diamond or graphite? 9) Calculate the Atrs

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

Ll.153.
Use the information give below to calculate the value of the standard molar entropy, So m of a solid K2O ( in J/mole x K) 2 K2O (s) ---> 4 K (s) + O2 (g) DeltaSo m = +275 J/K So m [K(s)] = +65 J/Mole x K Som [O2 (g)] = + 205 j/mole x k
The standard enthalpy of formation of HCe (g) at 298 K is equal to -92.3 kJ mol-4. The constant-pressure molar heat capacity in this temperature range is given by: Cp,m (H2)/J.K-1.mol-1 = 27.28 + 3.25 x 10-3 T+ 0.50 x 105 T-2 Cp,m (Cl2)/J.K-1.mol-1 = 36.90 + 0.25 x 10-3 T - 2.84 x 105 T-2 %3D Cp.m (HCl2)/J.K-1.mol-1 = 26.53 + 4.60 x 10-3 T+ 1.09 x 105 Calculate the enthalpy formation of HCe (g), AH (in kJ.mol¯-), at 1000 K. [Give your final answer to 3 significant figures].
Using the standard entropies below, calculate ASºrx (in J/K) for the following reaction. Report your answer to two decimal places in standard notation (i.e. 123.45 J/K). 2Cu₂0(s) + C(s) → 4Cu(s) + CO₂(g) -> Compound so (J mol-¹ K-¹) Cu₂0 (s) 92.37 5.60 33.15 213.74 C (s) Cu (s) CO₂ (g) Submit Answer Tries 0/5
EMISTRY II N 19 What is the change in volume ratio A(V,V.) change required if the entropy of a gas (2 mol) is to be increased by 50 % at 298.15 K. (R = 8.314 J mol K ) swered it of 5 Select one: O a. A(V,V,) = 1.88 O b. A(V,V) = 1.83 O c. A(V,V,) = 2.47 O d. A(V,V) = 0.64 g С.
Gaseous SO2Cl2 reacts with liquid water to form hydrogen chloride gas and liquid sulfuric acid. Using data tabulated in Appendix G of your text, calculate the enthalpy change, in units of kJ/mol, for the formation of 2 mol of liquid sulfuric acid.
engine.html?ClassID31653724337# Calculate the standard molar entropy for Ca2+ using the equation and data given. CaSO4(s) Ca2+(aq) + SO;-(aq) AS° = -140 J/mol K S° (J/mol-K) Substance Caso, 107 so2- 20 Ca2+ S 2. = [?] J/mol K Enter either a + or - sign AND the magnitude in %3D your answer. Enter
Please answer 4-6.
O ENTROPY AND FREE ENERGY Calculating dG from dH and dS A chemical engineer is studying the two reactions shown in the table below. In each case, she fills a reaction vessel with some mixture of the reactants and products at a constant temperature of 57.0 °C and constant total pressure. Then, she measures the reaction enthalpy H 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 her measurements are shown in the table. 2CH₂OH(g) + 30₂(g) 2CO₂(g) + 4H₂O(g) 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. 3CO₂(g) + 4H₂O(1)→ C₂H₂(g) + 50₂ (g) ΔΗ – -1353. kJ AS-4109. AG - KJ Which is…
Please see the picture attached. Thanks!
Calculate ASor for the reaction 3H2(g) +N2(g) → 2NH3(g) at 744 K. Omit terms in the temperature-dependent heat capacities greater than T2/K2. CP,m(H2,g)=22.66+0.0438TK-1.0835x10-4T2K2J-K-1-mol-1 CP,m(N2,g)=30.81-0.01187TK+2.3968x10-5T2K2J-K-1.mol-1 CP,m(NH3,g)=29.29+0.01103TK+4.2446x10-5T2K2J-K-1.mol-1 H2(g) N2(g) NH3(g) So(J.mol-1 K-1) 130.7 191.6 192.8
M Preq 3 Consider the reaction 00 E D 4HCI(g) + O₂(g) 2H₂O(g) + 2Cl₂(8) for which AH° = -114.4 kJ and AS° = -128.9 J/K at 298.15 K. J/K (1) Calculate the entropy change of the UNIVERSE when 1.775 moles of HCI(g) react under standard conditions at 298.15 K. ASuniverse (2) Is this reaction reactant or product favored under standard conditions? Submit Answer (3) If the reaction is product favored, is it enthalpy favored, entropy favored, or favored by both enthalpy and entropy? If the reaction is reactant favored choose 'reactant favored'. 4 000 000 R F DO LO V % 5 [Review Topics] [References] Use the References to access important values if needed for this question. T Retry Entire Group 8 more group attempts remaining Cengage Learning Cengage Technical Support G 6 B Y H & 7 N U 1 * 8 J M 9 K बैं O < L P Previous Email Instructor Save and Exit { + 11 Next ? } ]