Compute AH °1000 of HCI(g) from enthalpies of formation data and these C, [J/(mol K)]´P,mexpressions, which hold from 298 K to 1500 K.27.14 + 0.009274 (T/K) – 1.381 (10°T°/K) + 7.645 (10¯°T°/K°)26.93 + 0.03384 (T/K) – 3.896 (10¯³T°/K°) + 15.47 (10°r°/K°)30.67 – 0.007201 (T/K) – 1.246 (10¯°Tʻ/K´) – 3.898 (10¯°T° /K`)-5-9for(g), Cl, (g), and HCI(g), in that order.

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Answered: Compute AH °1000 of HCI(g) from…

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 enthalpy of the following reaction at 1.00 atm and 298.15K is -56.9kJmol-1 H* (ag) + OH (ag) →H,Oµ, When 2mol of an aqueous acid (H* (aq) reacts with 2mol of an aqueous base (OH(a9) in a thermally isolated constant pressure calorimeter initially at 298.15K, the temperature was observed to increase to 304.35K. In the exact same calorimeter, the hydrolysis of 1.7mol of aqueous adenosine 5'- triphosphate (ATP) caused an increase in temperature from 298.15K to 300.95K. (a) What is the enthalpy of hydrolysis per mol of ATP? (b) The hydrolysis of ATP is the primary mechanism by which the energy required by the human body is released. Assuming that the human body can use all of the released energy from ATP hydrolysis, how many grams of ATP is necessary to undergo hydrolysis per day to provide the average human daily energy requirement of 2000kcals (Molar mass of ATP =507.181gmol-1), assuming no other source of energy.
The following are the heats of combustion at 25 °C and 1 atm pressure: C6H6 (1) AH = -3267.62 kJ mol-¹ C (graphite) AH = -393.51 kJ mol-¹ H₂ (g) AH = -285.83 kJ mol-¹ 1. Calculate the standard enthalpy of formation of liquid benzene. 2. If the enthalpy of vaporization of benzene at 25 °C is 33.93 kJ/mol, find the AH, of gaseous benzene. 3. Calculate AH of liquid benzene at 300 °C given the Cp.m. Cp.m (J mol-¹ K-¹) 136.1 C6H6 (1) C (graphite) H₂ (g) 16.86 +4.77 x 10-378.54 x 105 T² 27.38 +3.26 x 10-³T 0.50 x 105 T² -
shed by the 2. 22.5 g of Zn (Cp = 0.387 J/g°C) is heated to the boiling point of water, then added to 25 mL of water at thermodynamic standard temperature. What is the final temperature?
Is ther an error in the solution that ∆Grxn=−9.33 kJ/mol=9330 J/mol was not kept negative when converted tfrom kJ to J? Can steps 3-5 be checked for accuracy ?
What is A, H° for the following chemical reaction? CO(g) + NH3 (9)→HCN(g)+ H2O(g) You can use the following table of standard heats of formation (Aş H°) to calculate the enthalpy of the given reaction. Standard Heat of Standard Heat of Element/ Compound Formation Element/ Compound Formation (kJ mol 1) (kJ mol-1) H(g) 218 N(g) 473 На (9) O2 (9) NH3 (9) -45.90 O(g) 249 CO(g) -110.5 H2O(g) -241.8 C(g) 71 HCN(g) 130.5 C(s) HNO3 (ag) -206.6 Express the standard enthalpy of reaction to three significant figures and include the appropriate units.
14. Diamond and graphite are two crystalline forms of carbon. At 1 atm and 25oC, diamond, changes to graphite so slowly that the enthalpy change of the process must be obtained indirectly. Using equations from the numbered list below, determine delta H for C(diamond) -> C(graphite) (1) C(diamond) + O2(g) -> CO2(g) ΔH = -395.4 kJ (2) 2CO2(g) -> 2CO(g) + O2(g) ΔH = 566.0 kJ (3) C(graphite) + O2(g) -> CO2(g) ΔH = -393.5 kJ (4) 2CO(g) -> C(graphite) + CO2(g) ΔH = -172.5 kJ
Calculate AG (kJ/mol) for the following system using the formula AG=RT In K where R = 8.314 J/K-mol T = 25.2 °C K = 2.49 x 1054 The correct number of significant figures will be required for a correct answer.
Standard Enthalpy of Formations in kJ/mol at 298.15 K ΔHof ΔHof AlCl3(s) -704.2.0 Al2O3(s) -1676 H2O(l) -285.8 H2O(g) -241.8 NH4ClO3(s) -265.2 N2O(g) 81.6 Use data above to calculate the enthalpy change for the reaction: 6Al(s) + 6NH4ClO3(s) => 2Al2O3(s) + 2AlCl3(s) + 3 N2(g) + 12H2O(g) ΔH = ?
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At 25 °C, the equilibrium partial pressures for the reaction 3 A(g) + 4 B(g) 2C(g) + 2D(g) were found to be PA = 4.67 atm, På = 5.73 atm, Pc = 5.60 atm, and Pp = 5.84 atm. What is the standard change in Gibbs free energy of this reaction at 25 °C? kJ mol privacy policy contact us help terms of use 6:11 PM about us careers 3/25/202
· From ice to water vapor Data: Average values of the specific heat capacities at constant pressure: Cp (H2O, s) = 37.6 J K-' mol-' Cp (H2O, l) = 75.2 J K-' mol·l Standard molar enthalpy of melting of ice at 0 °C: AmeltH°(H2O) = 6.02 kJ mol-1 Standard molar enthalpy of vaporization of ice at 100 °C: AvapH° (H2O) = 40.59 kJ mol- One would like to convert 1 mole of ice at - 10 °C into its vapor at 100 °C under a 1 bar pressure. 'Write a cycle of transformations in order to calculate the heat Qp that is necessary for this conversion. 'Calculate the value of Qp.
4.00 g NaOH(s) (39.996 g mol–1) dissolves in 100.0 g of water in a constant–pressure calorimeter. The initial temperatures of both materials are the same at 20.0°C. The density of the solution is 1.00 g cm–3. The heat of dissolution is –44.51 kJ mol–1. The specific heat of the solution is 4.184 J g–1 °C–1. The specific heat of the calorimeter is 100 J °C–1. What is the final temperature of the solution? 30.23°C 25.94°C 28.32°C What is the energy lost to the calorimeter? 579 J 832 J 721 J

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