reverse and double the first equation
4NH3(g) -------------> 2N2(g) + 6H2(g) H0 = 198.44 kJ
double the second equation
2N2(g) + 2O2(g) ---------> 4NO(g) H0 = 361 KJ
triple the third equation
3H2(g) + 3O2(g) ------------> 6H2O(l) H0 = - 1714.8 KJ
add all the 3 equatio ns
4NH3(g) -------------> 2N2(g) + 6H2(g) H0 = 198.44 kJ
2N2(g) + 2O2(g) ---------> 4NO(g) H0 = 361 KJ
3H2(g) + 3O2(g) ------------> 6H2O(l) H0 = - 1714.8 KJ
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4NH3(g) + 5O2(g) -------------> 4NO(g) + 6H2O(l) H0 = - 1155.36 KJ
3. Determine the change in enthalpy for the oxidation of ammonia as follows 4 NH (g)5...
Use the following equations N2 (g) + 3 H2 (g) 2 NH3 (g) N2 (g) + O2 (g) → 2 NO (g) 2 H2 (g) + O2 (g) + 2 H20 (1) DH = -99.22 kJ DH = + 180.5 kJ DH = - 571.6 kJ to calculate the enthalpy change for the reaction 4 NH3 (g) + 5 O2 (g) → 4 NO (g) + 6 H20 (1) DH = ?
What is the enthalpy change (in kJ) to burn 28.6 g ammonia if 4 NH3(g) + 3 O2(g) → 2 N2(g) + 6 H2O(g) has ΔHrxn = -1267 kj?
1) Enthalpy changes for the following reactions can be determined: AH=-91.8k N2(g) + 3 H2(g) → 2 NH (B) 4 NH(B) + 5 O2(g) → 4 NO(g) + 6H2O () H2(g) + O2 (g) → H20 (8) AH = -906.2 kJ AH = -241.8 kJ What is the enthalpy change for the following reaction? % N2(g) + O2(g) → NO(g) 2) Consider the following thermochemical equation: 2CH.(g)+70:(g)4CO2(g)+6H,0(1) AH--3120 kJ If 2.4 L of ethane (C2H6) at STP are reacted with...
1. Calculate AH for the reaction C2H4 (8) + H2() → C2H6), from the following data. C2H4 (g) + 3 02 (®) → 2 CO2 (s) + 2 H20 (1) C2H6 (g) + 7/2 02(g) → 2 CO2(g) + 3 H20 (1) H2 + 1/2O2() → H20 (1) AH = -1411. kJ/mole AH = -1560. kJ/mole AH = -285.8 kJ/mole 2. Calculate AH for the reaction 4 NH3(g) +502 (g) → 4 NO(g) + 6 H20 (g), from the following...
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 =
Given the enthalpies of combustion of propane (C3Hg), carbon and hydrogen, C3H3(g) + 5 O2(g) + 3 CO2(g) + 4 H20(1) AH° = -2219.9 kJ C(s) + O2(g) + CO2(8) AH° = -393.5 kJ 2 H2(g) + O2 + 2 H2O(1) AH° = -571.6 kJ Calculate the enthalpy of formation of propane. The reaction is shown below. 3 C(s) + 4H2(g) → C3H3(g)
5. Given the following data: 2 H2(g) + O2(g) → 2 H20 (1) AH° = -571.6 kJ N,Os (g) + H20 (1) ► 2 HNO (1) AH = -76.6 kJ N2(g) + 3 O2 (g) + H2(g) → 2 HNO, (1) AH = -348.2 kJ a. Calculate the AHⓇ for the reaction: 2 N2 (g) + 5 O2(g) → 2 N2Os (g)
5. Given the following data: (2 H2 (g) + O2 (g) → 2 H2O (1) ro in each AH° = -571.6 kJ N20s (g) + H2O (1) 2 HNO3 (1) AH° = -76.6 kJ N2 (g) + 3 O2 (g) + H2 (g) → 2 HNO3 (1) AH° = -348.2 kJ a. Calculate the AHⓇ for the reaction: 2 N2 (g) + 5 O2 (g) → 2 N2O5 (g)
Calculate AH for the reaction N2H4(0) + O2(g) → N2(g) + 2 H2O(1) given the following data: Equation AH (kJ) 2 NH3(g) + 3 N2O(g) → 4 N2(g) + 3 H2O(l) -1010 N2O(g) + 3 H2(g) → N2H4(1) +H2O(1) -317 2 NH3(g) + 1202(g) → N2H4(1) + H20(1) -143 H2(g) + 1202(g) → H2O(H -286 AH=
6. Use the following thermochemical equation for the synthesis of ammonia N2(g) + 3 H2(g) → 2 NH3(g) AH = -91.8 kJ Rewrite the thermochemical equation to show 3 moles of hydrogen being produced? Rewrite the thermochemical equation to show 6 moles of hydrogen being produced? Rewrite the thermochemical equation to show the production of 6 moles of NH3 produced? 7. Use the thermochemical equation below to find the AH of the following: CH_(g) + H2O(g) → CO(g) + 3...