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TC05ЕОЗ Use the enthalpies of formation given in the table below to calculate the standard molar...
TC05E04 Use the enthalpies of formation given in the table below to calculate the standard molar enthalpy change for this reaction at 298K: 2CuS(s) 302(g)2CuO(s)+ 2SO2(g) AfH / kJ mol at 298K, standard Substance state 1 bar. CuO(s) -157 SO2(g) -297 CuS(s) -53 Select one: a. -802 kJ mol1 b. +401 kJ mol1 c. +802 kJ mol1 d. -507 kJ mol1 e. -401 kJ mol1
The "roasting" of 48.7 g of ZnS at constant pressure gives off 220. kJ of heat. Calculate the DH for this reaction.2ZnS(s) + 3O2(g) ® 2ZnO(s) + 2SO2(g)A.) -110 kJ/mol rxnB.) -293 kJ/mol rxnC.) -440. kJ/mol rxnD.) -881 kJ/mol rxnE.) +440. kJ/mol rxn
4. Use a standard enthalpies of formation values below to determine the change in enthalpy for each of these reactions. Reaction: NaOH(s) + HCl(g) → NaCl(s) + H2O(g) Compound NaOH(s) HCl(g) NaCl(s) H2O(g) AHF (kJ/mol) -426.7 -92.3 -411.0 -241.8
Use a standard enthalpies of formation values below to determine the change in enthalpy for each of these reactions. SHOW WORK PLEASE. Reaction: NaOH(s) + HCl(g) →NaCl(s) + H2O(g) Compound ∆Hf (kJ/mol) NaOH(s) -426.7 HCl(g) -92.3 NaCl(s) -411.0 H2O(g) -241.8
Use Hess's law to calculate standard enthalpy change of formation at 298K for carbon (iv) sulphide compound, Cs2. Given that: Cs2(s) + 3O2(g) - Co2(g) + 2So2(g) ∆H°298= -1113KJ/mol C(s) + O2(g) - Co2(g). ∆H°298 = -407 KJ/mol S(s) + O2(g) - So2(g) ∆H°298 = -298 KJ/mol
TC04M01 Use the enthalpy changes given in the data below to calculate the enthalpy change for this reaction: Cus(s) + O2(g) → Cu(s) + SO2(9) Data: Reaction no. Reaction A.HⓇ/kJ morat 298K standard state = 1 bar. +314 2 2 CuO(s) – 2Cu(s) + O2(g) S(s) + O2(g) - S02(9) 2CuO(s) + 2S(s) - 2CuS(s) + O2(9) -297 +208 Select one: a. +225 kJ/mol b.-85 kJ/mol O C. -244 kJ/mol d. -191 kJ/mol e. -225 kJ/mol
Part A - Calculating an Enthalpy of Reaction from Enthalpies of Formation Calculate the enthalpy change for the reaction: 2 H2O2(l) → 2 H2O(l) + O2(g) using enthalpies of formation: ΔH∘f[H2O2]ΔH∘f[H2O]==−187.8 kJ/mol−285.8 kJ/mol Calculate the enthalpy change for the reaction: using enthalpies of formation: Multiple choice answers below: -98.0 kJ -196.0 kJ +98.0 kJ +196.0 kJ
c Given the following standard molar entropies of formation (S) and enthalpies of combustion to gaseous carbon dioxide and liquid water at 25 °c (AHe AH/kJ mol 393.5 -285.9 -1559.7 C(graphite) H2(g) C2Ho(g) 5.9 131.0 229.5 Calculate the enthalpy change (AH) and Gibbs energy change (AG) for the reaction 2C(graphite) +3H28)CH) datseatt Is this reaction thermodynamically possible? Give a reason for your answer. (10 marks) Explain why it is possible for endothermic processes to occur spontaneously. 15 marks]
Heat of Formation Calculations: 32) Use a standard enthalpies of formation (Ho) table to determine the change in enthalpy for each of these reactions Hrxn [n. Ho(products) - n. Ho(products)] CO (g): -110.5 kJ/mol; CO2 (g): -393.5 kJ/mol CH4 (g): -890.4 kJ/mol H2O (l): -285.8 kJ/mol; H2O (g): -241.8 kJ/mol H2S (g): -20.6 kJ/mol; NO: -90.2 kJ/mol NO2: +33.9 kJ/mol; HCl (g): -92.3 kJ/mol NaOH (s): -426.7 kJ/mol; SO2 (g): -296.8 kJ/mol a) CH4(g) + 2 O2(g) ---> CO2(g) +...
Use the References to access important values if needed for this question. Given the standard enthalpy changes for the following two reactions: (1) 2Fe(s) + O2(g) →2FeO(s) AH° = -544.0 kJ (2) 2Zn(s) + O2(g)—2ZnO(5) AH° = 696.6 kJ what is the standard enthalpy change for the reaction: (3) FeO(s) + Zn(s)—>Fe(s) + ZnO(s) AH° = ?