please solve and explain your answer I) (15 points) Given the standard enthalpies of formation for...
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) +...
From the given enthalpies of formation, calculate the enthalpy change for the following reaction. Your label should be in kJ/mol. Again, leave a space between the answer and the label. N2H4(0) + 2H2O2(1) ► N2(g) + 4H2O(1) Thermochemical data: Substance AH(kJ/mol). H2O(1) -285.8 N Hall) 50.7 H2O2(1) -187.8 Answer:
Question 17 3 pts Using standard molar enthalpies of formation given in the table below, calculate AH/xnto one decimal place, for the combustion of ammonia: AHrxn° = E nAH (products) - E mAHt"reactants) 4 NH3(g) + 7 O2(g) → 4NO2(g) + 6H2O(1) molecule AHF (kJ/mol-rxn) NH3(g) -45.9 NO2(g) +33.1 H2O(1) -285.8 H2O(9) -241.8 - 1663.6 kJ/mol-rxn +30.24 kJ/mol-rxn -1398.8 kJ/mol-rxn -298.6 kJ/mol-rxn -206.9 kJ/mol-rxn Question 11 3 pts A gas absorbs 45 kJ of heat and does 29 kJ of...
58. Calculate the standard enthalpy of formation of PCI, (g), given that AH' = -136.0 ks for the reacti PCI, (g) + H2O(g) POCI, (g) + 2 HCI (8) and the following standard enthalpies of formation: H,0 (8), -241.8 kJ/mol; POCI, (), -592 kJ/mol; HCl (g), -92.5 kJ/mol.
Given the standard enthalpies of formation for the following substances, determine the reaction enthalpy in kJ/mol for the following reaction: 4A (g) + 2B (g) 2C (g) + 7D (g) Substance ΔHf in kJ/mol A (g) – 20.42 B (g) + 32.18 C (g) – 72.51 D (g) – 17.87
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
Standard Enthalpies of Formation, in kJ/mol N2(g) 0 NO2(g) +33.2 NH3(g) -45.9 H2O(l) -285.8 NO(g) +90.3 N2O(g) -82.1 H2O(g) -241.8 Use the data above to calculate ΔH for the reaction: 6 NO2(g) + 8 NH3(g) => 7 N2(g) + 12 H2O(g) ΔH = ?
Enthalpy of Formation Part 1: Part 2: Part 3: A scientist measures the standard enthalpy change for the following reaction to be 591.0 kJ : 2H2O(1)—>2H2(g) + O2(g) Based on this value and the standard enthalpies of formation for the other substances, the standard enthalpy of formation of H2O(l) is kJ/mol. A scientist measures the standard enthalpy change for the following reaction to be -2903.4 kJ: 2C2H6(g) + 7 O2(g)—>4CO2(g) + 6 H2O(g) Based on this value and the standard...
The following table lists some enthalpy of formation values for selected substances. Substance ΔfH∘ΔfH∘ (kJ mol−1)(kJ mol−1) CO2(g)CO2(g) −393.5−393.5 Ca(OH)2(s)Ca(OH)2(s) −986.1−986.1 H2O(l)H2O(l) −285.8−285.8 CaCO3(s)CaCO3(s) −1207−1207 H2O(g)H2O(g) −241.8−241.8 Part A: Determine the enthalpy for this reaction: Ca(OH)2(s)+CO2(g)→CaCO3(s)+H2O(l) C a ( O H ) 2 ( s ) + C O 2 ( g ) → C a C O 3 ( s ) + H 2 O ( l ) Express your answer in kJ mol−1 k J m o l...
A.Using standard heats of formation, calculate the standard enthalpy change for the following reaction. N2(g) + 3H2(g) = 2NH3(g) B.Using standard heats of formation, calculate the standard enthalpy change for the following reaction. CaCO3(s) = CaO(s) + CO2(g) C. A scientist measures the standard enthalpy change for the following reaction to be -2910.6 kJ: 2C2H6(g) + 7 O2(g) = 4CO2(g) + 6 H2O(g) Based on this value and the standard enthalpies of formation for the other substances, the standard enthalpy...