Lets number the reaction as 0, 1, 2, 3 from top to bottom
required reaction should be written in terms of other reaction
This is Hess Law
required reaction can be written as:
reaction 0 = +1 * (reaction 1) +2 * (reaction 2) -2 * (reaction 3)
So, ΔHo rxn for required reaction will be:
ΔHo rxn = +1 * ΔHo rxn(reaction 1) +2 * ΔHo rxn(reaction 2) -2 * ΔHo rxn(reaction 3)
= +1 * (-622.2) +2 * (-285.8) -2 * (-187.8)
= -818.2 KJ
Answer: -818.2 KJ
72. Determine AHº for this reaction from the data below. N2H4(1) + 2 H2O2(1) -→ N2(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
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 the value of AHº for the reaction 2 H2O2 (1) + 2 H20 (1) + O2 (9) given the following thermochemical equations: H2(g) + 1/2O2(g) → H2O (1) AH° = -286 kJ H2(g) + O2(g) → H2O2 (1) AH° = - 188 kJ - 196 kJ - 474 kJ + 196 kJ 98 kJ
Using the reactions below, determine the ∆Hº for 2 N2(g) + 5 O2(g) --> 2 N2O5(g) N2(g) + 3 O2(g) + H2(g) --> 2 HNO3(aq) ∆Hº = -415 kJ/mol N2O5(g) + H2O(l) --> 2 HNO3(aq) ∆Hº = -140 kJ/mol 2 H2(g) + O2(g) --> 2 H2O(l) ∆Hº = -572 kJ/mol
(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
Use Hess's law to determine A.Hº for the reaction C3H4(g) + 2 H2(g) -> C3H8(8), given that Hy(8) + O2(8) — H2O(1) A Hº = -285.8 kJ mol-1 C3H4(8) + 402(g) — 3 CO2(g) + 2 H2O(1) A Hº = -1937 kJ mol-1 C3H2(g) + 5O2(g) — 3CO2(g) + 4H2O(1) A Hº = -2219.1 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
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=
For the chemical reaction shown, 2H2O2(l)+N2H4(l)⟶4H2O(g)+N2(g) determine how many grams of N2 are produced from the reaction of 9.55g of H2O2 and 5.81 g of N2H4.
Calculate the enthalpy change for the following reaction: N2H4 (I)+ 2 H2O2 (I) → N2 (g) + 4 H2O (I); ΔH=? kJ Using the following thermochemical equations: