05-09-2018, 03:45 PM

The dilemma I am having with the following question is that I have the right answer for both part (a) and (b), but I don’t know how to calculate part (a). Even though part (b) can’t be answered without knowing part (a), I was able to calculate it as I was told by a friend what part (a) is. The question is as follows:

Q4. (a) Given that the enthalpy of formation of CH3Cl (g) is -80.83 kJ mol-1 estimate the average bond dissociation energy of a C-Cl bond. (b) Estimate the enthalpy change for the reaction:

H2C=CH2 + HCl ------------- CH3CH2Cl

Bond Ave. Bond Dissociation energy / kJ mol-1 Element H°f / kJ mol-1

C-H 414 C (g) 716.7

C-C 343 H (g) 218.0

C=C 611 Cl (g) 121.3

H-Cl 431

I will illustrate below my attempt for part (a):

First I wrote the word equation to the reaction which is:

H2C=CH2 + HCl ------------- CH3CH2Cl

Then the values given for enthalpy of formation were used to calculate the change in enthalpy:

H2C=CH2 = 2(716.7) = 1433.4

HCl = 218.0 + 121.3 = 339.3

CH3CH2Cl = -80.83 + 716.7 = 635.17

Enthalpy of formation for H2 equals 0 so it was excluded from the calculations above.

Enthalpy of change = 635.17 – 1433.4 – 339.3

= -1377.53 KJ mol ^-1 (energy to break C=C and Cl-H bonds).

Then I tried working out the dissociation energy of (C-Cl) using the following method:

-1377.53 = D_have(C=C) + D_have(Cl-H)

-1377.53 - D_have(C=C) = D_have(Cl-H)

-1377.53 - 611 = D_have(Cl-H)

-1377.53 – 611 = -1988.53

The above was my attempt at working out the bond dissociation energy of (C-Cl), I was totally off as the real bond dissociation energy is 330.8 kj mol^-1 and not what I calculated which was -1988.53.

Where did I go wrong? How do I calculate this correctly? I know how to calculate the part (b) but I just can’t seem to get my head around on how to calculate the bond dissociation energy of (C-Cl).

Any help would be greatly appreciated, Thank You.

Q4. (a) Given that the enthalpy of formation of CH3Cl (g) is -80.83 kJ mol-1 estimate the average bond dissociation energy of a C-Cl bond. (b) Estimate the enthalpy change for the reaction:

H2C=CH2 + HCl ------------- CH3CH2Cl

Bond Ave. Bond Dissociation energy / kJ mol-1 Element H°f / kJ mol-1

C-H 414 C (g) 716.7

C-C 343 H (g) 218.0

C=C 611 Cl (g) 121.3

H-Cl 431

I will illustrate below my attempt for part (a):

First I wrote the word equation to the reaction which is:

H2C=CH2 + HCl ------------- CH3CH2Cl

Then the values given for enthalpy of formation were used to calculate the change in enthalpy:

H2C=CH2 = 2(716.7) = 1433.4

HCl = 218.0 + 121.3 = 339.3

CH3CH2Cl = -80.83 + 716.7 = 635.17

Enthalpy of formation for H2 equals 0 so it was excluded from the calculations above.

Enthalpy of change = 635.17 – 1433.4 – 339.3

= -1377.53 KJ mol ^-1 (energy to break C=C and Cl-H bonds).

Then I tried working out the dissociation energy of (C-Cl) using the following method:

-1377.53 = D_have(C=C) + D_have(Cl-H)

-1377.53 - D_have(C=C) = D_have(Cl-H)

-1377.53 - 611 = D_have(Cl-H)

-1377.53 – 611 = -1988.53

The above was my attempt at working out the bond dissociation energy of (C-Cl), I was totally off as the real bond dissociation energy is 330.8 kj mol^-1 and not what I calculated which was -1988.53.

Where did I go wrong? How do I calculate this correctly? I know how to calculate the part (b) but I just can’t seem to get my head around on how to calculate the bond dissociation energy of (C-Cl).

Any help would be greatly appreciated, Thank You.