Calculate ΔH° for the process ½N2(g) + ½O2(g) → NO(g)
from the following information
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c. Calculate ΔH° for the process Co3O4(s)→3Co(s) + 2O2(g) from the following information: Co(s) + 1/2O2(g)→CoO(s) ΔH° = -237.9kJ 3CoO(s) +1/2O2(g)→Co3O4(s) ΔH° = -177.5k d. Calculate the standard molar enthalpy of formation of NO(g) from the following data: N2(g) + 2O2 →2NO2(g) ΔH° = 66.4 k. 2NO(g) + O2 →2NO2(g) ΔH° = -114.1 kJ
Calculate ΔH⁰298 (in kJ) for the process Co3O4(s) → 3 Co(s) + 2 O2(g) from the following information. Co(s) + 1/2 O2(g) → CoO(s) ΔH⁰298 = −237.9 kJ 3 Co(s) + O2(g) → Co3O4(s) ΔH⁰298 = −177.5 kJ Please show detailed work and an explanation.
Calculate ΔH⁰298 (in kJ) for the process Co3O4(s) → 3 Co(s) + 2 O2(g) from the following information. Co(s) + 1/2 O2(g) → CoO(s) ΔH⁰298 = −237.9 kJ 3 CoO(s) + 1/2 O2(g) → Co3O4(s) ΔH⁰298 = −177.3 kJ Please show detailed work and an explanation.
Given the following reactions N2 (g) + O2 (g) → 2NO (g) ΔH = +180.7 kJ 2NO( g) + O2 (g) → 2NO2 (g) ΔH = -113.1 kJ the enthalpy for the decomposition of nitrogen dioxide into molecular nitrogen and oxygen 2NO2 (g) → N2 (g) + 2O2 (g) is ________ kJ.
Given the following reactions N2 (g) + O2 (g) → 2NO (g) ΔH = +180.7 kJ 2N2O (g) → O2 (g) + 2N2 (g) ΔH = -163.2 kJ the enthalpy of reaction for 2N2O (g) → 2NO (g) + N2 (g) is ________ kJ.
Using Hess’s Law to Calculate ΔH Calculate ΔH for 2 NO(g) + O2(g) → N2O4(g) using the following information: N2O4(g)2 NO(g) + O2(g)→→2 NO2(g)2 NO2(g)ΔHΔH==+57.9 kJ−114.1 kJ Calculate for using the following information: Select one 2.7 kJ -55.2 kJ -85.5 kJ -171.0 kJ +55.2 kJ
Calculate the enthalpy of the reaction 2NO(g)+O2(g)→2NO2(g) Given the following reactions and enthalpies of formation: N2(g) + 2O2(g)→ 2NO2(g), ΔH∘ = 66.4kJ N2(g)+ O2(g)→ 2 NO(g), ΔH∘=180.4 kJ Please explain the steps as well!
Use the ΔH°f and ΔH°rxn information provided to calculate ΔH°f for SO3(g): 2 SO2(g) + O2(g) → 2 SO3(g) ΔH°rxn = -198 kJ ΔH°f (kJ/mol) SO2(g) -297
ΔH is positive for the equilibrium N2 (g) + O2 (g) ↔ 2 NO (g). If we lower the temperature, (A) the equilibrium shifts to the left (more N2 + O2 gases formed) (B) the equilibrium shifts to the right (more NO gas formed) (C) Equilibrium does not shift because there are the same number of molecules on left side as on the right (D) Equilibrium does not shift because temperature does not affect the reaction rates why?
Consider the following data. 2 H2(g) + O2(g) 2 H2O(l) ΔH = -571.7 kJ N2O5(g) + H2O(l) 2 HNO3(l) ΔH = -92.0 kJ N2(g) + 3 O2(g) + H2(g) 2 HNO3O(l) ΔH = -348.2 kJ Use Hess's law to calculate ΔH for the reaction below. 2 N2O5(g) 2 N2(g) + 5 O2(g) ΔH = _____kJ