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 = ?
Standard Enthalpies of Formation, in kJ/mol N2(g) 0 NO2(g) +33.2 &nbs
Nitrogen AHºf (kJ/mol) AG°f (kJ/mol) Sº (J/mol K) N2 (g) 0 0 191.6 472.7 455.6 153.3 N(g) NH3 (g) -46.1 -16.5 192.5 NH3 (aq) -80.0 -27.0 111.0 NH4+ (aq) -132.0 -79.0 113.0 NO (g) 90.3 86.6 210.8 NOCI (g) 51.7 66.1 261.8 NO2 (g) 33.2 51.3 240.1 N20 (g) 82.1 104.2 219.9 N204 (8) 9.2 97.9 304.3 Consider the reaction N2(9) + 202(9)—2NO2(g) Using the standard thermodynamic data in the tables linked above, calculate Grxn for this reaction at 298.15K...
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...
1).From the standard enthalpies of formation, calculate ΔH°rxn for the reaction C6H12(l) + 9O2(g) → 6CO2(g) + 6H2O(l) For C6H12(l), ΔH°f = –151.9 kJ/mol (5 points) Substance ∆H°f , kJ/mol C6H12(l) –151.9 O2(g) 0 H2O(l) –285.8 CO2(g) –393.5 2).Determine the amount of heat (in kJ) given off when 1.26 × 104 g of ammonia are produced according to the equation N2(g) + 3H2(g) → 2NH3(g) ΔH°= –92.6 kJ/mol Assume that the reaction takes place under standard conditions at 25oC.
35. Calculate ΔrH° for the combustion of ammonia, 4 NH3(g) + 7 O2(g) → 4 NO2(g) + 6 H2O( ) using standard molar enthalpies of formation. molecule ΔfH° (kJ/mol-rxn) NH3(g) –45.9 NO2(g) +33.1 H2O( ) –285.8 a. +30.24 kJ/mol-rxn b. –206.9 kJ/mol-rxn c. –298.6 kJ/mol-rxn d. –1398.8 kJ/mol-rxn e. –1663.6 kJ/mol-rxn
Calculate the standard enthalpy of formation of NO(g) (in kJ/mol) from the following data. N2(g) + 2 O2(g) → 2 NO2(g) ΔH⁰298 = 66.4 kJ 2 NO(g) + O2(g) → 2 NO2(g) ΔH⁰298 = −116.2 kJ
The standard enthalpy of formation of NO(g) is +90.3 kJ·mol-1 and that of NO2(g) is +33.2 kJ·mol-1. Calculate the reaction enthalpy for 2NO(g) + O2(g) ® 2NO2(g) A. -47.8 kJ B. -57.1 kJ C. +123.5 kJ D. -23.9 kJ E. -114.2 kJ Please explain why/steps.
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) +...
Standard Enthalpies of Formation (ΔHof): C4H10(g) = -124.8 kJ/mol CO2(g) = -393.5 kJ/mol H2O(l) = -285.8 kJ/mol What is the total amount of heat required for the process? What is the molar heat of combustion for butane? What is the mass of butane needed for this process?
Hydrazine (N2H4) is a fuel used by some spacecraft. It is normally oxidized by N2O4according to this equation: N2H4(l)+N2O4(g)→2N2O(g)+2H2O(g) Standard Enthalpies of Formation at 25 ∘C Substance ΔH∘f(kJ/mol) N2H4(l) 50.6 N2H4(g) 95.4 N2O(g) 81.6 N2O4(l) −19.5 N2O4(g) 9.16 H2O(l) −285.8 H2O(g) −241.8 Calculate ΔH∘rxn for this reaction using standard enthalpies of formation. Express your answer using one decimal place.
Nitrogen Ahºf (kJ/mol) AG°f (kJ/mol) Sº (J/mol K) 191.6 N2(9) 0 0 N(g) 472.7 455.6 153.3 NH3(g) -46.1 -16.5 192.5 NH3(aq) -80.0 -27.0 111.0 NH4+ (aq) -132.0 -79.0 113.0 90.3 86.6 210.8 NO(9) NOCI(g) 51.7 66.1 261.8 NO2(g) 33.2 51.3 240.1 N20(9) 82.1 104.2 219.9 N204(9) 9.2 97.9 304.3 N204(0) -20.0 97.0 209.0 N205(s) -42.0 134.0 178.0 N2H4(0) 50.6 149.3 121.2 N2H3CH3 () 54.0 180.0 166.0 HNO3(aq) -207.4 -111.3 146.4 HNO3(1) -174.1 -80.7 155.6 HNO3(9) -135.1 -74.7 266.4 NH4ClO4(s) -295.0...