multiply first equation with 12
6N2 + 6O2 ------------> 12NO H = 1084.8 kJ
revers and multiply with 12 for second equation
12NO2 ------------> 6N2 + 12O2 H = - 405.6 kJ
multiply last equation with 3
12NH3 + 21O2 ------------> 12NO2 + 18H2O H = - 3978.6 KJ
add all 3 above equations
6N2 + 6O2 ------------> 12NO H = 1084.8 kJ
12NO2 ------------> 6N2 + 12O2 H = - 405.6 kJ
12NH3 + 21O2 ------------> 12NO2 + 18H2O H = - 3978.6 KJ
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12NH3 + 15O2 -----------> 12NO + 18H2O H = - 3.30 x 103 KJ
Calculate the standard enthalpy change, ΔΗ0rn, în k] for the following chemical equation, using o...
Calculate the standard enthalpy change, ΔH°rxn, in kJ for the following chemical equation, using only the thermochemical equations below: 2COBr2(g) + 4NO(g) → 4NOBr(g) + 2CO(g) Report your answer to three significant figures in scientific notation. Equations: ΔH°rxn (kJ) CO(g) + Br2(g) → COBr2(g) -3 1/2N2(g) + 1/2O2(g) → NO(g) 90.4 1/2N2(g) + 1/2O2(g) + 1/2Br2(g) → NOBr(g) 82.1
Calculate the standard enthalpy change, ΔH°rxn, in kJ for the following chemical equation, using only the thermochemical equations below: CaCO3(s) → CaO(s) + CO2(g) Report your answer to three significant figures in scientific notation. Equations: ΔH°rxn (kJ) Ca(s) + CO2(g) + 1/2O2(g) → CaCO3(s) -812.5 2Ca(s) + O2(g) → 2CaO(s) -1270.3
Using the enthalpy changes given below, calculate the enthalpy change for the following equation: N2(g) + 2O2(g) -----> 2 NO2 using the following equations: 2NO -----> N2 + O2 delta H + -180.0 kJ NO2 -----> NO + 1/2 O2 delta H= 112.0 kJ
Review Problem 6.086 Given the following thermochemical equations, 2Cu + S(s) - Cu2S(s) AH° = -79.5 kJ S(s) + O2(g) → SO2(9) AH = -297 kJ Cu2S(s) + 202(g) → 2CụO(s) + SO2(g) AH° = -527.5 kJ calculate the standard enthalpy of formation (in kilojoules per mole) of Cuo(s). AH°(Cuo(s)) - kJ/mol The number of significant digits is set to 3; the tolerance is +/-2% Show Work is REQUIRED for this question: Open Show Work Review Problem 6.087 Given the...
Using standard heats of formation, calculate the standard enthalpy change for the following reaction 2HBr(g) H2(g) + Br2() ANSWER kJ Submit Answer Retry Entire Group 2 more group attempts remaining Given the standard enthalpy changes for the following two reactions: (1) Zn(s)+ Cl2(g) ZnCl2(s) AH° = -415.0 kJ (2) Fe(s) + Cl2(g) FeCl2(s) AH° = -341.8 kJ what is the standard enthalpy change for the reaction: (3) Zn(s) + FeCl(s) AH° ? ZnCl2(s) + Fe(s) kJ Given the standard enthalpy...
Determine the enthalpy change for CS20 + 302(g) + CO2(g) + 2SO2(g); AH= ? using the following three thermochemical equations, which will be referred to as reactions (A), (B), and (C). (A) 2C(s) + 202(g) 2002(g); AH= -787.0 kJ (B) C(s) + 25(s) → CS2(); AH= 87.9 kJ (C) SO2(g) → S(s) + O2(g); AH= 296.8 kJ 1. The enthalpy change for the unknown reaction is 2. Describe the action performed on each thermochemical reaction. Reaction (A) was Reaction (B)...
Using Hess's Law to Calculate a Standard Enthalpy of Formation On the Solution Calorimetry Lab Report Form, vou will be asked to calculate a standard enthalpy of formation for magnesium oxide based on your experimental results. Below is an example of how to do this type of calculation. Calculate the standard enthalpy of formation of gaseous diborane (B2H6) using the following thermochemical equations: 4 B(s) + 3 O2(g)-> 2 B2O3(s) 2 H2(g) + O2(g) > 2 H20(0) B2H6(8) +3 O2(g)...
The standard enthalpy change for the following reaction is 66.4 kJ at 298 K. N2(g) + 2 O2(g) 2 NO2(g) AH° = 66.4 kJ What is the standard enthalpy change for this reaction at 298 K? 1/2 N2(g) + O2(g) — NO2(g) Submit Answer
1. Calculate the standard enthalpy of reaction for the reaction below, 3502(g) + 03(g) → 3503(g) Using only the following thermochemical data: 250(g) + O2(g) → 2502(g) AH° = -602.8 kJ 3503(g) → 350(g) + 203(9) AH° = +1485.0 kJ 202(g) → 03(9) AH° = +142.2 kJ
15. + -10.1 points 0/4 Submissions Used Calculate the standard enthalpy of formation of gaseous nitrogen monoxide (NO) using the following thermochemical information: N2(g) + 3 H2(g) = 2 NH3(9) 4 NO(g) + 6 H2O(l) = 4 NH3(g) + 5 O2(9) 2 H2O(1) = 2 H2(g) + O2(9) AH = -92.4 kJ AH = +1167.1 kJ AH = +571.7 kJ AH =