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
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Calculate ΔH⁰298 (in kJ) for the process Co3O4(s) → 3 Co(s) + 2 O2(g) from the...
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.
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 deltaH degree 298 for the process CO3O4(s) ---> 3Co(s) +2O2(g) From the following information: Co(s)+120^2(g) ---> CoC(s) deltaH degree 298 = -237.9kJ 3CoO(s)+120^2(g) ---> Co3O4(s) deltaH degree 298= -177.5kJ (The 120^2, is lower.. not upper. Don't have the sign on my phone.) Please show work, so I know how to do future problems. Thank you.
66. Calculate AHº for the process Co3O4(s)-3Co(s)+2O2(g) from the following information: Co(s)+12O2(g) →COO(s)AH°=-237.9kJ 3COO(s)+12O2(g) C0304(s)AH"=-177.5kJ
Given: C(s) + O2(g) ---> CO2(g) ΔH = −393.5 kJ/mol S(s) + O2(g) ---> SO2(g) ΔH = −296.8 kJ/mol C(s) + 2S(s) ---> CS2(ℓ) ΔH = +87.9 kJ/mol A) Calculate the standard enthalpy change for the following reaction CS2(ℓ) + 3O2(g) ---> CO2(g) + 2SO2(g) ΔH° rxn = -1075 kJ/mol B) Using the equation and standard enthalpy change for the reaction (from part A), calculate the amount of heat produced or consumed when 3.2 mol of CS2 reacts with excess...
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 ΔH° for the process ½N2(g) + ½O2(g) → NO(g) from the following information
You are given the following thermodynamic data. 2 Fe(s) + 3/2 O2(g) → Fe2O3(s) ΔH° = -823 kJ 3 Fe(s) + 2 O2(g) → Fe3O4(s) ΔH° = -1120. kJ Calculate the ΔH° for the following reaction. 3 Fe2O3(s) → 2 Fe3O4(s) + ½ O2(g)
The standard enthalpy change for the following reaction is -111 kJ at 298 K. C(s,graphite) + 1/2 O2(g) --> CO(g) ΔH° = -111 kJ What is the standard enthalpy change for the reaction at 298 K? 2 CO(g) --> 2 C(s,graphite) + O2(g) Answer in kJ
From the enthalpies of reaction 2C(s)+O2(g)→2CO(g)ΔH=−221.0kJ 2C(s)+O2(g)+4H2(g)→2CH3OH(g)ΔH=−402.4kJ calculate ΔH for the reaction CO(g)+2H2(g)→CH3OH(g)