4) for calculating enthalpy we should get the desired equation by
adding on subtraction the equations from which we have to find
enthalpy
So consider al the given enthalpy equations as 1,2,3
equation 1+2
Then reverse equation 3 and then add to equation 1 and 2, we will add the desire equation I.e.
N2H4 + H2 –------ 2NH3
Now
Enthalpy can be calculated as
= (-37)+(-46)+65
=18 kj
3) Titanium dioxide is synthesized in the laboratory by hydrolysis, using the following reaction: TICIA() +...
3) Titanium dioxide is synthesized in the laboratory by hydrolysis, using the following reaction: н,--763 187.6 -945 -92.31 What change in internal energy would occur at 298K if 1 mol of TiC4) were added to H20(0) and all of the HCI evolved as gas? 4) Calculate the enthalpy of formation of acetylene: Given: 2C2H2lg)+502)4CO2)2H2 2599 kJ 2H2 O 2H20 393.5 k 285.8 kJ 5) The following question refers to the enthalpy of combustion of styrene monomer, CHa There should be...
(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
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
Exercise C For each of the reactions listed below. use the reactions after it to determine the enthalpy of the first reaction. Reaction 1 Find the AH for the reaction below. given the following reactions and subsequent AH values: 4 NH3(g) + 5O2(g) - 4NO(g) + 6H2O(g) N2(g) + O2(g) - 2NO(g) AH = +180.6 kJ N2(g) + 3H2(g) + 2NH3(g) AH = -91.8 kJ 2 H2(g) + O2(g) 2H2O(9) AH = -483.7 kJ Reaction 2 Find the AH for...
3. (3 points) Hydrazine (N2H4) can react with oxyger in the following chemical reaction: NaH.(1 +0:0) - N2(g) + 2H20 (1) Calculate AH, for the reaction above, given the following data: 2NH3(g) + 3N20(g) - 4260) + 3H2000 AH,º=-1010. kJ/mol N20(g) + 3H2(g) → N2H (1) + H2O(1 AH-=-317 k!/inol 2NH3(g) + 40:09) - NzH4O + H20(1 AH,"=-143 kJ/mol H2(g) +4202(9) - H20(1) AH"=-286 kJ/mol
(1) Find the H for the reaction below, given the following reactions and subsequent H values: 2CO2(g) + H2O(g) → C2H2(g) + 5/202(g) CH2(g) + 2H2(g) → CHg) H2O(g) - H2(g) + 1/20, (g) C2H6(g) + 7/202(g) → 2CO2(g) + 3H2O(g) answer - 235 kJ H =-94.5 kJ H =71.2 kJ H --283 kJ (2) Find the H for the reaction below, given the following reactions and subsequent H values: N2H4(I) + H2(g) + 2NH3(g) N2H4(1) + CH4O(1) - CH2O(g)...
15. The standard Gibbs free energy change (AG%) for the Haber reaction at 298K is -2.83 kJ/mol N2(g) + 3H2(g) + 2NH3(g) If the reaction begins with 2.55 atm N2(g) 0.115 atm Hz(g) and 0.488 atm NH3(9) at 298K what is the free energy change (AG) for the reaction under these conditions AND will the reaction be spontaneous under these conditions? (10 points) 16. For the following reactions, predict whether they will tend to be spontaneous at either high or...
PLEASE ANSWER THESE QUESTIONS What is the molar reaction enthalpy for the reaction below: NgH4(I) + CH4 (1) #CH2O(g) + Ng(g) + 3Hg(g) given the following thermodynamic data? 2NH3(g) + N2H4(1) + H2(g) A,Hm = 22.5 kJ mol-1 2NH3(g) + N2(g) + 3H2(9) A,Hm = 57.5 kJ mol-1 CH2O(g) + H2(g) + CHAO(0) A,Hm = 81.2 kJ mol-1 0 -11.5 kJ mol-1 0 -32.8 kJ mol-1 0 -46.2 kJ mol-1 O-59.7 kJ mol? A particular reaction has a standard molar...
12. Use Hess's Law to find AH for this reaction. (5) NO2(g) +2H2(g) ---> 2H2O(e) + NHs(g) Given the following two equations: 2NH3(g) -->N2(g) + 3H2(g) AHo +92 kJ 2N2(g)+ 2H2O(l) ---> NO2(g) + 2H2(g) AH° = +170 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=