Calculate enthalpy changes for the following: a. 0.080 g of sulfur (rhombic) burns, forming SO₂(g) (AƒH° for SO₂ (g) = - 296.84 kJ/mol) Enthalpy change = kJ b. 0.33 mol of HgO(s) decomposes to Hg(l) and O₂(g) (AƒH° for HgO = -90.83 kJ/mol) Enthalpy change = c. 2.80 g of NH3(g) is formed from N₂(g) and excess H₂(g) (AƒH° for NH3 = -45.90 kJ/mol) Enthalpy change = KJ d. 1.36 × 10-2 mol of carbon is oxidized to CO₂(g) (AƒH° for CO₂ = -393.509 kJ/mol) Enthalpy change = kJ kJ

Calculate enthalpy changes for the following: a. 0.080 g of sulfur (rhombic) burns, forming SO₂(g) (AƒH° for SO₂ (g) = - 296.84 kJ/mol) Enthalpy change = kJ b. 0.33 mol of HgO(s) decomposes to Hg(l) and O₂(g) (AƒH° for HgO = -90.83 kJ/mol) Enthalpy change = c. 2.80 g of NH3(g) is formed from N₂(g) and excess H₂(g) (AƒH° for NH3 = -45.90 kJ/mol) Enthalpy change = KJ d. 1.36 × 10-2 mol of carbon is oxidized to CO₂(g) (AƒH° for CO₂ = -393.509 kJ/mol) Enthalpy change = kJ 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

In the recovery of iron from iron ore, the reduction of the ore is actually accomplished by reactions involving carbon monoxide. Use the following thermochemical equations, Fe,O3(s) + 3CO(g) 2Fe(s) + 3CO,(g) AH° = - 28 kJ 3Fc,O3() + CO(g) 2Fe3O4() + CO,(g) AH = -59 kJ FezO4(s) + CO(g) 3FcO(s) + CO,(g) AH° = +38 kJ to calculate AH° for the reaction FeO() + CO(g) Fe(s) + CO2(g) AH° = i kJ
Which of the following reactions is accompanied by an increase in enthalpy? Select one: O 2s03(g) – 2S02(g) + O2(g) A,H" = +197 kJ O 2NO(g) + O2(g) - 2NO2(g) + 113 kJ O 2H2(g) + O2(g) - 484 kJ – 2H20(g) 4HCI(g) + 02(g) - 2H20(g) + 2Cl2(g) A,H' = -111.4 kJ
Find the enthapy change for the reaction below in kJ . 4NH3(g) + 5O2(g) --> 4NO(g) + 6H2O(g) ΔH= ? Given : (NH3(g)) = -46.1, (NO(g)) = + 90.3 kJ/mol and (H2O(g)) = -241.8 kJ/mol.
Consider the following reaction. CH3OH(g) CO(g) + 2 H2(g) H = +90.7 kJ (a) Is the reaction exothermic or endothermic? endothermicexothermic (b) Calculate the amount of heat transferred when 40.0 g of CH3OH(g) are decomposed by this reaction at constant pressure.H = kJ(c) If the enthalpy change is 11.0 kJ, how many grams of hydrogen gas are produced? g(d) How many kilojoules of heat are released when 13.5 g of CO(g) reacts completely with H2(g) to form CH3OH(g) at constant pressure?H = kJ(e) Calculate E when 360.0 g of CH3OH(g) completely reacts at a constant temperature of 300 K and constant pressure of 0.95 atm. R = 8.314 J/mol*K and R = 0.08206 atm*L/mol*K kJ HopHelpCh5N1
(c) Consider the following chemical reaction: C;Hs(g) + H2(g) C2H6(g) + CH4(g) ΔΗ? %3D Calculate the enthalpy change for the reaction above using Hess's law. Thermochemical dat: H:(g) + ½O2(g) H2O(1) AH° =-285.8 kJ AH°combustion for CH4(g) AH° =-890.0 kJ/mol %3D CO:(g) C(s. graphite) + O:(g) AH° = +393.5 kJ AH°combustion for C2H6(g) AH = -1560.0 kJ/mol AH°, for C3H$(g) AH° =-103.8 kJ/mol
When ammonia reacts with dinitrogen oxide gas (ΔHf° = 82.05 kJ/mol), liquid water and nitrogen gas are formed. How much heat (in kJ) is liberated or absorbed by the reaction that produces 365 mL of nitrogen gas at 25°C and 743 mm Hg? ΔHf° (kJ/mol) NH3(g) –46.1 H2O(l) –285.8 N2(g) 0 N2O(g) 82.05
What is the reaction mechanism for the major product? Please draw.
Calculate the standard enthalpy for the below reaction (in kJ) using the standard enthalpies of formation provided: Reaction: 5 A + 3 B ⟶⟶ 7 C + 3 D Standard enthalpies of formation, ΔH°f (in kJ/mol) are: A: ‑15.7 B: ‑86.4 C: ‑52.7 D: ‑71.6
Study the following thermochemical data very carefully: Clelg) lag) ICI(g) Ia(s) 2C(g) 21(g) I(g) + CI(g) la(g) AH = 242.3 kJ AH = 151.0 kJ AH = 211.3 kJ AH = 62.8 kJ I2(s) reacts with Clag) to produce ICI(g). Write the chemical equation to produce ONLY one mole of ICI(g)
Why are the reactions of aldehydes and imines very similar? Be very specific.
Given the following two reactions:(1)Fe(s) + Cl2(g)->FeCl2(s) H(1) = -341.8 kJ(2)Hg(l) + Cl2(g)->HgCl2(s) H(2) = -224.3 kJcalculate the enthalpy change for the following reaction:(3) FeCl2(s) + Hg(l)-> Fe(s) + HgCl2(s) H(3) = kJ
20) The standard enthalpy change for the reaction below is -196.0 kJ. Use Table below to calculate the standard enthalpy of formation of H₂O (1). 2H₂O2 (1) ->>> 2H₂O (1) + O₂ (g) Substance H₂O2 (1) AH° (kj/mole) -187.8