Determine ΔH for the following reaction: N2(g) + 2 H2(g) → N2H4(l) Given: N2H4(l) + O2(g)...
calculate enthalpy of H for the reaction N2H4(l) + 2H2O(l) -> N2(g) + 4H2)(l) Given the reactions N2H4(l) + O2(g) -> N2(g) + 2H2O(l) Enthalpy of H = -6.22.2 kJ H2(g) + (1/2)O2(g) -> H2O(l) enthalpy of H = -285.8 kJ/mol H2(g) + O2(g) -> H2O2(l) enthalpy of H = -187.8 kJ
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=
Determine delta H^0 for the reaction: N2H4(l) + O2(g) -----> N2(g) + 4H2O(l) From these data: N2H4(l) + 2H2O2(l) ----> N2(g) + 2H2O(l) delta H^0 = -622.2 KJ H2(g) + 1/2 O2(g) ----> H2O(l) delta H^0= -285.5KJ H2(g) + O2(g) -----> H2O2(l) delta H^0= -187.8KJ
72. Determine AHº for this reaction from the data below. N2H4(1) + 2 H2O2(1) -→ N2(g) + 4H2O(1) N2H4(1) + O2(g) →→ N2(g) + 2 H2O(1) A Hº = -622.2 kJ mol-1 H2(g) + + O2(g) →→ H2O(1) A Hº = -285.8 kJ mol-1 H2(g) + O2(g) —> H2O2(1) A,Hº = -187.8 kJ mol-1
Consider the following data. 2 H2(g) + O2(g) 2 H2O(l) ΔH = -571.7 kJ N2O5(g) + H2O(l) 2 HNO3(l) ΔH = -92.0 kJ N2(g) + 3 O2(g) + H2(g) 2 HNO3O(l) ΔH = -348.2 kJ Use Hess's law to calculate ΔH for the reaction below. 2 N2O5(g) 2 N2(g) + 5 O2(g) ΔH = _____kJ
Calculate the standard reaction enthalpy for the reaction N2H4(ℓ) + H2(g) → 2 NH3(g) given N2H4(ℓ) + O2(g) → N2(g) + 2H2O(g) ∆H ◦ = −543 kJ · mol−1 2 H2(g) + O2(g) → 2 H2O(g) ∆H◦ = −484 kJ · mol−1 N2(g) + 3 H2(g) → 2 NH3(g) ∆H◦ = −92.2 kJ · mol−1 1.) −243 kJ · mol−1 2.) −59 kJ · mol−1 3.) −935 kJ · mol−1 4.) −151 kJ · mol−1 5.) −1119 kJ · mol−1
(ii) Calculate AH° for the reaction N2H4(1) + O2(g) → N2(g) + 2H2O(1) using the data given below: 2NH3(g) + 3N2O(g) → 4N2(g) + 3H2O(1) N2O(g) + 3H2(g) → N2H4(l) + H2O(1) 2NH3(g) + O2(g) → N2H4(1) + H2O(1) H2(g) + 1/2O2(g) → H2O(1) AH° = -1010. kJ AH° = -317 kJ AH° = -143 kJ AH° = -286 kJ
The combustion reaction of ethane is as follows. C2H6(g) + 7/2 O2(g) → 2 CO2(g) + 3 H2O(l) Using Hess's law and the reaction enthalpies given below, find the change in enthalpy for this reaction. reaction (1): C(s) + O2(g) → CO2(g) ΔH = −393.5 kJ/mol reaction (2): H2(g) + 1/2 O2(g) → H2O(l) ΔH = −285.8 kJ/mol reaction (3): 2 C(s) + 3 H2(g) → C2H6(g) ΔH = −84.0 kJ/mol
Consider the reaction: H2(g) + (1/2)O2(g) -------> H2O(l) ΔH° = -286 kJ Which of the following is true? (Select all that apply) the reaction is endothermic heat is given off by the surroundings the reaction is exothermic heat is absorbed by the system the enthalpy of the products is less than the that of the reactants
2. Given the following data: H2O(l) → H2(g) + 1/2O2(g) ΔH° = 285.8 kJ 2HNO3(l) → N2O5(g) + H2O(l) ΔH° = 76.6 kJ 2N2(g) + 5O2(g) → 2N2O5(g) ΔH° = 28.4 kJ Calculate ΔH° for the reaction: 1/2N2(g) + 3/2O2(g) + 1/2H2(g) → HNO3(l) Note that you should be able to answer this one without needing to use any additional information from the thermo table. I've attempted this question multiple times. I am able to get to the simplified eqaution...