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 enthalpy of H for the reaction N2H4(l) + 2H2O(l) -> N2(g) + 4H2)(l)
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
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
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:
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=
Use the Data table to calculate ∆H for the reaction below:Reactions: Change in Enthalpy (∆H)(1) C (s) + O2 (g) -> CO2(g) ∆H1 = -393.5 kJ/ mol(2) H2 (g) + 1/2 O2 (g) -> H2O (l) ∆H2 = -285.8 kJ/mol(3) 2C2H6 (g) + 7O2 (g) -> 4 CO2 (g) + 6 H2O (l) ∆H3 = -283.0 kJ/molCalculate the enthalpy change for the reaction:2 C (s) + 3 H2 (g) -> C2H6(g) ∆H = ______________kJ/mol
Please explain step by step 12. Calculate the standard reaction enthalpy for the reaction: N2H4(4) + H2(g) → 2NH3(g) Given: N2H4(4) + O2(g) → N2(g) + 2H2O(g) AH° = 0543 kJ 2H2(g) + O2(g) → 2H2O(g) AH° = 1484 kJ N2(g) + 3H2(g) + 2NH3(g) AH° = 092.2 kJ A. - 1119 kJ B. - 33 kJ C. -151 kJ D. + 151 kJ E. + 1119 kJ
N2H4(l) + O2(g) --> N2(g) + 2H2O (l); LaTeX: \DeltaΔH= -285.8 kJ How many kJ of heat will be released when 14.4 g water is generated? FW: N = 14; H = 1; O = 16. The result should be positive. Keep one digit after decimal.
Use the example shown to calculate the reaction enthalpy, delta H, for the following reaction: CH4(g)+2O2(g)->CO2(g)2H2O(l). Use the series of reaction that follows: 1. C(s)+2H2(g)-> CH4(g), delta H= -74.8 kJ 2. C(s)+O2(g)->CO2(g), delta H= -393.5 kJ 3. 2H2(g)+O2(g)-> 2H2O(g), delta H= -484.0 kJ 4. H2O(l)->H2O(g), delta H= 44.0 kJ
a) Use Hess's law to calculate the enthalpy change for the reaction: 3C(s) + 4H2(g) + ½O2(g) → C3H8O(l) Given the following thermochemical equations: 2C3H8O(l) + 9O2(g) → 6CO2(g) + 8H2O(l) ΔH = -4042.6 kJ/mol C(s) + O2(g) → CO2(g) ΔH = -393.51 kJ/mol H2(g) + ½O2(g) → H2O(l) ΔH = -285.83 kJ/mol (in kJ/mol) A: -267.7 B: -302.6 C: -341.9 D: -386.3 E: -436.5 F: -493.3 G: -557.4 H: -629.9 b) Define if the following statement is an endothermic process or exothermic...