Lets number the reaction as 1, 2, 3, 4 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 4 = +2 * (reaction 1) +2 * (reaction 2) -3 * (reaction 3)
So, ΔHo rxn for required reaction will be:
ΔHo rxn = +2 * ΔHo rxn(reaction 1) +2 * ΔHo rxn(reaction 2) -3 * ΔHo rxn(reaction 3)
So,
ΔrHo = 2*ΔrHo1 + 2*ΔrHo2 - 3*ΔrHo3
70. Given the following information: N2(8) + H2(8) — NH3(8) AH NH3(g) + O2(8) —— NO(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=
5. Given the following chemical equilibria, N2(g) + O2(g) = 2 NO(g) N2(g) + 3 H2(g) = 2 NH3(g) H2(g) + 1/2 O2(g) =H2O(g) Determine the method used to calculate the equilibrium constant for the reaction below. 4 NH3(g) + 5 O2(g) = 4 NO(g) + 6 H2O(g) K
(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
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
6. Given H2(g) + 22 O2(g) → H2O(1), AH° = -286 kJ/mol, Determine the standard enthalpy change for the reaction 2H2O(l) → 2H2(g) + O2(g) (2 pt)
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...
Using the equations N2 (g) + 3 H2 (g) → 2 NH3 (g) AH° = -91.8 kJ/mol C(s) + 2 H2 (g) → CH4 (g) AH° = -74.9 kJ/ mol H2 (g) + 2 C(s) + N2 (g) → 2 HCN (g) AH° = 270.3 kJ/mol Determine the enthalpy for the reaction CH4 (g) + NH3 (g) → HCN (g) + 3 H2 (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)
3) N2(g) + 3 H2(g) → 2 NH3(g) AH = - 46.1kJ A) For the above reaction what is the sign of ASsys? (Is entropy of the reaction increasing or decreasing?) B) Calculate the value of Assurr using the information given at 25°C
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)