Question

1) determine the heat of reaction from the bond enthalpies for the reaction given

H2      +     Br2   >>    2 HBr

436.4         192.5            366.1

2) calculate the formal charges in the phosphate ion.

3) what elements can have "expanded octets" and how many electrons can they carry?

Answer 1

Solution :-

1) Reaction equation

H2 + Br2 --------- > 2 HBr

formula to calculate the enthalpy change using the bond eneregies is as follows

Delta H rxn = sum of bond energies of reactant - sum of bond energies of product

                  = [(H2*1)+(Br2*1)] - [HBr*2]

                  = [(436.4*1)+(192.5*1)] - [ 366.1*2]

                   = -103.3 kJ

so the heat of reaction is -103.3 kJ

2) formula to calculate the formal charge is

FC= number of valence electrons - (number of bonds + non bonding electrons )

FC on P = 5 - (5+0) = 0

FC on oxygen labeled a = 6 -(2+4) = 0

FC on oxygen labeled b = 6 -(1+6) = -1

FC on oxygen labeled c = 6 -(1+6) = -1

FC on oxygen labeled d = 6 -(1+6) = -1

Lewis structure is as follows

3) Elements which have 3d orbitals where the extra electrons can reside can form the expanded octer compounds

elements like S, P, Cl , I, Os can form expanded octets

they can have 10 or 12 electrons

PCl5 in this molecule P have 10 electrons

SF4 in this molecule S have 12 electrons.