A couple of tricky Energy/Bonding/Kinetics Questions..?

Question

2 questions im stuck on, any help would be appreciated, may thanks in advance!
Question 2 of 15
Predict the shapes of
(i) phosphine, PH3

(ii) sulphur trioxide, SO3

(iii) the sulphite ion, SO32-

(iii) the amide ion, NH2-

Question 4 of 15
(a) Given the following data, construct a Hess's Law cycle and calculate the standard enthalpy of formation of ethane, Hf [C2H6].
2C(s) + 3H2(g) → C2H6 (g)

delta Hc carbon = -394 kJ mol-1
delta Hc hydrogen = -286 kJ mol-1
delta Hc ethane = -1560 kJ mol-1

Answer 1

It's been a couple of years since A level chemistry and I can;t remember how to predict shapes of molecules..however it's something to do with the bonds 'wanting' to be equally distributed around the centre of the molecule to prevent stearic hindrance I think...
Hess' Law states that the overall enthalpy change accompanying a chemical reaction is independent of the route taken in going from reactants to products. So basically, you can use the energy changes you have been given, in this case combustion, and from that calculate the total energy change.
Firstly, use delta Hc for C and H to write the equation to totally burn all the atoms in pure oxygen:
2C(s) ---> 2CO2(g) delta Hc: 2 x -394 kJ mol-1
= -788 kJ mol-1
3H2(g) ----> 3H2O(l) delta Hc: 3 x -286 kJ mol-1
= -858 kJ mol-1
So your total energy change for this first part is -858 + (-788)
= 1646 kJ mol-1
This uses 5 atoms of oxygen (2 1/2 molecules). Next, you want to re-combine the C and H to form ethane, which is basically the reverse of the combustion reaction for ethane which you have been given the energy change for. So, because it is reversed, you reverse the sign to give +1560 kJ mol-1:
2CO2(g) + 3H2O(l) ----> C2H6 + 2 1/2 O2

The total energy change for the reaction ie. the enthalpy of formation is the sum of all the individual energy changes, ie. (-788)+(-858)+1560 (remember to keep the sign reversed) = -86 kJmol-1

The whole thing should be written with the equation you want to find delta Hf for (2C(s) + 3H2(g) → C2H6 (g)) across the top, the intermediates that you are using based on the energy values given (2CO2(g) + 3H2O(l)) at the bottom in the middle, and two addition arrows, one from the reactants to the intermediates and another from the intermediates to the products like this:

reactants -------> products



intermediates

So the whole thing forms a cycle.
Remember to put "+ xO2" by the first arrow to indicate that oxygen is used in combustion and "- xO2" by the second arrow to indicate that the same amount of oxygen is then lost (in this case x = 2 1/2). Also remember to put the state symbols in brackets all the way through; I seem to remember they mark you down if you don't.
Good luck!