Question

Calculate Delta H rxn for the following reaction (watch the significant figures)

2 NOCl (g) --> N2(g) + 02(g) + Cl2(g)

Given the following set of reactions;

1/2 N2(g) + 1/2 02(g) ---> NO(g) Delta H= 90.3KJ
NO(g) + 1/2 Cl2(g) ---> NOCl(g) Delta H = - 38.6 KJ

Answer 1

We know that our required reaction contains N2 and 2 NOCl while given equations contain 1/2 N2 and NOCl so we must multiply both of these equations by 2. Furthermore, our required formulas for example N2 and NOCl are on the opposite sides, so it is a good idea to invert these equations. This means interchanging reactants with the products and this also changes the sign of delta H.
(1/2 N2(g) + 1/2 02(g) ---> NO(g) Delta H= 90.3KJ)x2

N2(g) + 02(g) ---> 2NO(g) Delta H= 180.6KJ

and, invert: 2NO(g) ---> N2(g) + 02(g) Delta H= -181KJ
Do the same with the second equation:
(NO(g) + 1/2 Cl2(g) ---> NOCl(g) Delta H = - 38.6 KJ)x2

2NO(g) + Cl2(g) ---> 2 NOCl(g) Delta H = - 77.2 KJ and invert:

2NOCl(g) ---> 2NO(g) + Cl2(g) Delta H = + 77.2 KJ

Finally, we have to add both our resultant equations:

2NO(g) ---> N2(g) + 02(g) Delta H= -181KJ
2NOCl(g) ---> 2NO(g) + Cl2(g) Delta H = + 77.2 KJ, Watch that 2 NO will be cancelled, we get our required equation:
2 NOCl (g) --> N2(g) + 02(g) + Cl2(g) delta H = -103.8KJ. rounding to three figures: -104 KJ

I hope this helps.