4) Using heat capacities from the table and formation enthalpy of HCl(g) A¢H° = -92.3 kJ molª (at 298K), determine standard enthalpies of reaction Cl2 (g) + H2(g) => 2HC1 (g)) Assume that the heat capacities are independent of temperature. at 298K and 498K? Cp, JK-' mol-1 Gas Cl2 34 На НСІ 29 29

4) Using heat capacities from the table and formation enthalpy of HCl(g) A¢H° = -92.3 kJ molª (at 298K), determine standard enthalpies of reaction Cl2 (g) + H2(g) => 2HC1 (g)) Assume that the heat capacities are independent of temperature. at 298K and 498K? Cp, JK-' mol-1 Gas Cl2 34 На НСІ 29 29

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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Please help me with the following question and explain it pls making sure its 100% correct. I keep getting this answer but it marks me wrong I am not sure whats wrong, pls help thank you
For the reaction P4010(s) + 6 H20(1) 4 H3PO4(aq) AG° = -448.9 kJ and AS° = -16.3 J/K at 261 K and 1 atm. This reaction is (reactant, product) favored under standard conditions at 261 K. The standard enthalpy change for the reaction of 2.15 moles of P4010(s) at this temperature would be kJ. Submit Answer Try Another Version 1 item attempt remaining rch a
3. Heats of reaction can also be calculated using heats of formation and the relationship - AHrxn = ΣnAHƒ.prod — ΣmAHƒ.reac Substance HCl(aq) MgCl2(aq) MgO(s) H₂O(/) H.(kJ/mol) -167.2 -796.8 -601.2 -285.8 0 a. Standard State Elements Calculate the theoretical AH° values for the following reaction Mg(s) + 2HCl(aq) → MgCl2(aq) + H2(g) MgO(s) + 2HCl(aq) → MgCl2(aq) + H2O(l) Mg(s) + O2(g) → MgO(s)
3. Use the appropriate data from the table to calculate the change in enthalpy for this reaction: Mg) + 2HClag) > M9CI2(ag) + H2(9) (aq) Species AH", , kJ/1mol HCl(g) -92.30 HCl(aq) -167.2 MgCl(s) -641.6 MgCla(aq) -796.9 H,O(1) -285.83 4. Using your answer for question #3, how much energy would be produced from the reaction of 0.680g of Mg with excess hydrochloric acid?
V:OY IA %YA N& 令 l PNG.9 → حداد H.W) calculate AçH° (500 K) for methane by using the data given on the temperature dependence of heat capacities. CH4(g)+2O2(g)→CO2(g)+2H2O(g) AÇH° (298 K) -891 kJ Cp,m(CO2) = 44.22 + 8.79 × 10-37 - 8.62 × 105 T2 Cp.m(H2O) = 75.29 + 0 + 0 Cp,m(O2) = 29.96 + 4.18 × 10-37 1.67 × 105 T2 Cp,m(CH4) = 35.31
A quantitity of 100mL of 0.500 MHCl was mixed with 50 mL of 0.500M NaOH in a constant pressure calorimeter of negligible heat capacity. The initial temperature of the HCl and NaOH solutions was the same, 20ºC, and the final temperature of the mixed solution was 22.3ºC. Calculate the heat change for neutralization reaction on a molar baisis.
(c) Make the molecule below using the Sonogashira and Suzuki-Miyaura couplings to make the new bolded C-C bonds. F HO
3) Calculate the reaction heat of formation of PbO from Pb and O2 at rxn temperatures of 500K and 800K seperately (using below reactions). Pb(s) + 1/2 O2(g) Pb(1) + 1/2 O2(g)- AH°298.PbO =-219400 J/mole Cp (PbO.s)-52.3 +8.66 x 10T- 8.2x10 T J/molK (298-1159 K) CP (Pb.s)-24.3 +8.7 x 10 T Cp (Pb.)= 32.5-3.1 x 10 T Cp(02g)-29.9 +4.18 x 10 T-1.7x10 T2 Pb(s)→ PbO(s) PbO(s) J/molK (298-600 K) J/molK (600-1300 K) J/molK (298-2500 K) Pb(1). AHmelting 4800 J/mole, Tmelting 600K
(d) The enthalpy change for the reaction (A,Hº) of ethyne (C,H,) is -2511.6 kJ. i. Write a balance equation for the combustion of ethyne. ii. Determine the enthalpy of formation A¡Hº of ethyne given the following: A¢H° CO2 = -393-5 kJ mol" A¡Hº H2O_ = -241.8 kJ mol" A¡H° O2 = 0.00 kJ mol-1 %3D
8) Kirchhoff's Law Calculate the reaction enthalpy for the following reaction (ARH° = - 282.98 kJ mol¹¹) at a temperature of 400 °C: CO+0.502 → CO₂ In your calculations, consider the temperature-dependency of the molar heat capacities as follows: Cp (CO) = (21.75 +0.064 T-4.1.10-5T2)Jmol-¹K-1 Cp(O2) = (27.96 +0.004 T-1.7 10-7T²)Jmol¯¹K−¹ Cp (CO2)=(27.63 +0.005 T)Jmol-1K-1
I got -1366.7 but I don't think that's right.