12) Use the following data to calculate the standard heat (enthalpy) of formation, ???, of manganese(IV)...
using Hess's law what is the standard enthalpy of formation of manganese(ll)oxide, MnO(s)? 2MnO2(s)-->2MnO(s)+O2(g) rxn= +272.0 kJ/mol MnO2(s)+Mn(s)-->2MnO(s) rxn=-248.9 kJ/mol A. -520.9 kJ/mol B. -396.5 kJ/mol C. -384.9 kJ/mol D. -147.6 kJ/mol E. 24 kJ/mol
Question 9 (1 point) Use the following equations to calculate the molar enthalpy of formation for manganese(IV) oxide, MnO2(s). MnO2(s) --> MnO(s) + 1/2O2(g) AH1 = 132 kJ/mol MnO2(s) + Mn(s) --> 2MnO(s) AH2 = -240 kJ/mol O +504 kJ/mol 0-504 kJ/mol 0-24.0 kJ/mol +24.0 kJ/mol 0-372 kJ/mol
Calculate the standard enthalpy of reaction for the following reaction using the appropriate standard enthalpies of formation ΔH°f [SiCl4 (g)] = -657.0 kJ/mol SiO2 (s) + 4 HCl (g) → SiCl4 (g) + 2 H2O (g)
Given the data in the table below, determine Hrxn for
the following reaction.
18. Given the data in the table below, determine AH for the following reaction: rxn SiCla (g) + 2 H20 (g) SiO2 (s)4 HCI (g) AH: (kJ/mol) Substance SiO2 (s) HCI (g) SiCla (g) H2O (g) -910.9 -92.3 -657.0 -241.8 A. -598.9 kJ B. +139.5 kJ C. -139.5 kJ D. +1106.7 kJ E. +598.9 kJ
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
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 given standard enthalpies of formation to determine the heat of reaction of the following reaction: 2 C5H10O(g) + 19 O2(g) => 10 CO2(g) + 10 H2O(g) ΔHof C5H10O(g) = -232.11 kJ/mol Δ Hof CO2(g) = -393.5 kJ/mol ΔHof H2O(g) = -241.8 kJ/mol
3) Calculate the AH,º for the following reactions based on the standard heat of enthalpy of each compound. AH° (kJ/mol): HCl(g) → -92.3, NaCl(s) →-411.0, NaOH(s) →-426.7. H2O(g) +-241.8 SO2(g) → -296.8, H2S(g) → -20.0, O2(g) → 0.0, H2O(l) →-285.8 a) NaOH(s) + HCl(g) ----> NaCl(s) + H2O(g) b) 2 H2S(g) + 3 O2(g) ---> 2 H2O(l) + 2 SO2(g)