QUESTION 39 From the following enthalpy changes, calculate the value of AH for the reaction N2H4 N2 + 2 H2 AH = -95.4 kJ N2H4 + H2 2 NH3 AH = -187.6 kJ 3 H2 + N2 →>>> 2 NH3 AH = ? -283 kJ -92.2 kJ +92.2 kJ +283 kJ

Calculate the enthalpy change for the following reaction, _C₂H4 (9) + using the following standard heats of formation: AH(C₂H4) = +52.5 kJ/mol; AH(CF4) =-933.0 kJ/mol; AH (HF) =-271.6 kJ/mol _F₂(g) → _CF4(g) + _HF(g)

For the reaction below (which will be labeled reaction A), the reaction enthalpy is unknown:                                                   2 B    + 3 H2    B2H6    HA  = ? Use the following thermochemical data to calculate the reaction enthalpy ΔHA: Reaction 1:       4 B + 3 O2 → 2 B2O3                               ΔH1 = –2509 kJ Reaction 2:       2 H2 + O2 → 2 H2O                          ΔH2 = –572 kJ Reaction 3:       B2H6 + 3 O2 → B2O3 + 3 H2O   ΔH3 = –2,159 kJ Calculate the enthalpy of reaction A (in kJ) and enter the numerical value in the answer box. Do not write units in the answer box. Clearly show on a paper how each reaction can be combined to equal reaction A  (multiply or reverse  each of the above reactions).complete work must be uploaded

Calculate the enthalpy change in kJ for the following reaction given the data below.  Answer: -610.1 kJ      P4O10(g) + 6 PCl5(g) --> 10 Cl3PO(g)                A                1/4 P4(s) + 3/2 Cl2(g) --> PCl3(g)          Ho/kJ = -306.4  B                P4(s) + 5 O2(g) --> P4O10(g)               Ho/kJ = -2967.3            C                PCl3(g) + Cl2(g) --> PCl5(g)                    Ho/kJ = -84.2            D                PCl3(g) + 1/2 O2(g) --> Cl3PO(g)            Ho/kJ = -285.7