enthalpy change...please help

Question

please help....

1.Consider the reaction:
Ch4(g)+Cl2(g)  CH3Cl(g) +HCl(g)
a.
what bonds are broken and what bonds are formed?
b.
Calculate:
i. the energy required to break these bonds
ii. the energy released on forming the bonds
E(C---H)=415kJmol-1; E(Cl---Cl) = 242 kJmol-1; E(C---Cl) = 338 kJmol-1;
E(H---Cl)=431kJmol-1

c.
Hence calculate △H for the reaction
d.
Explain why the following reaction would have a similar value of △H:
C2H6(g) +Cl2(g)  C2H5Cl(g) + HCl(g)
e.
Explain why the experimental values for △H for the chlorination of methane and ethane are not identical in practice.

20 marks will be given to the best solution!!

Answer 1

1a.
CH3-H + Cl-Cl → CH3-Cl + H-Cl
Bonds broken : a C-H bond, and a Cl-Cl bond
Bonds formed : a C-Cl bond, and an H-Cl bond

1.b.i.
Energy needed for bond breaking
= 415 + 242
= 657 kJ mol-1

1.b.ii.
Energy released in bond formation
= 338 + 431
= 769 kJ mol-1

1.c.
ΔH = 657 - 769 = -112 kJ mol-1

1.d.
CH3-H + Cl-Cl → CH3-Cl + H-Cl
CH3CH2-H + Cl-Cl → CH3CH2-Cl + H-Cl
In either reaction, a C-H bond and a Cl-Cl are broken, while a C-Cl bond and an H-Cl bond are formed.

1.e.
The exact bond enthalpy of a particular chemical bond depends on the molecular environment in which the bond exists.
Therefore, the bond enthalpy of C-H bond in CH4 is not exactly the same as that in CH3CH3, and the bond enthalpy of C-Cl bond in CH3Cl is also not exactly the same as that in CH3CH2Cl.