calculate ΔH for the following reaction?

Question

2N2(g) + 6H2O(g) -> 3O2(g) + 4NH3(g)

you are given the following data:
NH3(g) -> 1/2N2(g) + 3/2H2(g) ΔH= 46KJ
2H2(g) + O2(g)-> 2H2O(g) ΔH= -484 KJ

Answer 1

2N2(g) + 6H2O(g) --> 3O2(g) + 4NH3(g)

Turn the first reaction around, and multiply it by 4. Multiply the dH by 4, and change the sign on it.

2N2(g) + 6H2(g) --> 4NH3(g) ...dH = - 184 kJ

Turn the 2nd reaction around and multiply by 3. Mulitply the dH by 3, and change the sign on it.

6H2O (g) --> 6H2(g) + 3O2(g)....dH = + 1452 kJ

Add the two reactions, cancelling out the 6H2O since they show up on both sides, and add the dH's.

2N2(g) + 6H2O(g) --> 3O2(g) + 4NH3(g)...dH = + 1268 kJ

Answer 2

You need to combine the reactions for which you have information to give you the one you are interested in.

You want to make 4 NH3 and 3O2. So try the reverse of 4 x the first eq and 3 x the second. Write them down, add them up, cancel out anything on both sides.

Now for the heats, assuming that works: you want 4 x - 46 kJ + 3 x - (-484 kJ); multiplying by number of moles of reaction and reversing signs because reversing direction.

Principle: DeltaH doesn't depend on the route taken, however contorted, only on start and finish. A bit like difference in height above sea level.

IMPORTANT NOTE: Heat data refer to 1 mole reaction as written, NOT 1 mole any particular reagent. So in previous problem, data referred to 4 moles Fe or 2moles Fe2O3. Other answere got that wrong.

Answer 3

u in basic terms multiply 2d one with -a million, and upload them mutually.. that is an removing technique.. via multiplying a reaction with -a million, it potential u make it grow to be D-->E+A so that is -208.1kJ