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enthalpy change?

I'm a little bit confused as im trying to work out the enthalpy change for the reaction between methane and chlorine free radical -
CH4 + CL. ---> CH3CL + H.

I worked it out to be 74 kjmol-1 but this cant be right :frown:
Help!
Reply 1
Avatar for Vanny17
Vanny17
19
Surely we need to know the mass of water and heat change of the reaction as enthaply change is, q= m x c x delta T
(edited 13 years ago)
Reply 2
Avatar for sc0307
sc0307
8
what she said ^^
which method are you using?
also I don't think you're far off with your answer.
Reply 4
Avatar for gemstar!
gemstar!
OP
1
Original post by Vanny17
Surely we need to know the mass of water and heat change of the reaction as enthaply chnage is, q= m x c x delta T


We dont need to do anything that complicated thank goodnesss! :P
I just did it by working out the bond enthaply of the reactants and products then added them together to find out the energy change ?
I just assumed it had to be a negetive answer..
Reply 5
Avatar for Tateco
Tateco
14
Original post by gemstar!
I'm a little bit confused as im trying to work out the enthalpy change for the reaction between methane and chlorine free radical -
CH4 + CL. ---> CH3CL + H.

I worked it out to be 74 kjmol-1 but this cant be right :frown:
Help!


Isn't the original equation supposed to be CH4 + Cl. --> CH3. + HCL?

Anyway if you're using bond enthalpy to work it out then add up all the bonds on the left hand side 4(C-H) and take away all the bonds on the right side (3(C-H) + (H-Cl)) then you should have your answer
Reply 6
Avatar for Vanny17
Vanny17
19
Original post by gemstar!
We dont need to do anything that complicated thank goodnesss! :P
I just did it by working out the bond enthaply of the reactants and products then added them together to find out the energy change ?
I just assumed it had to be a negetive answer..


:smile:. I have never worked out enthalpy change like that though :s-smilie:. It would be a negative answer if energy is given out e.g combustion or burning meanwhile heat will be taken in and would therefore be positive for an endothermic reaction e.g photosynthesis.
Reply 7
Avatar for gemstar!
gemstar!
OP
1
Original post by tateco
Isn't the original equation supposed to be CH4 + Cl. --> CH3. + HCL?

Anyway if you're using bond enthalpy to work it out then add up all the bonds on the left hand side 4(C-H) and take away all the bonds on the right side (3(C-H) + (H-Cl)) then you should have your answer


Thats exactly what I'm doing but im ending up with a postive enthalpy change for an exothermic reaction :s
Anyway for this reastion - N2H4 + 02 ----> N2 + 2H20
I got the enthalpy change to be 38 kjmol-1???
Whats the correct answer, i cant see where im going wrong :frown:
Energy is taken in when bonds are broken and then energy is given out when the bonds are made so:

Enthaply Change = (Bond Enthalpies of Reactants) - (Bond enthalpies of Products)

I know it sounds silly but did you do it the right way round? :p:
Reply 9
Avatar for gemstar!
gemstar!
OP
1
Original post by soutioirsim
Energy is taken in when bonds are broken and then energy is given out when the bonds are made so:

Enthaply Change = (Bond Enthalpies of Reactants) - (Bond enthalpies of Products)

I know it sounds silly but did you do it the right way round? :p:


hmm now its coming out as 4058kjmol-1 which is still positive...grrrr!
Reply 10
Avatar for Tateco
Tateco
14
Original post by gemstar!
Thats exactly what I'm doing but im ending up with a postive enthalpy change for an exothermic reaction :s
Anyway for this reastion - N2H4 + 02 ----> N2 + 2H20
I got the enthalpy change to be 38 kjmol-1???
Whats the correct answer, i cant see where im going wrong :frown:


Unfortunately I haven't had the time to memorise all the average bond enthalpies so don't know the correct answer, if you're doing it the way I said then it's right
Original post by gemstar!
hmm now its coming out as 4058kjmol-1 which is still positive...grrrr!


How do you know it's supposed to be exothermic?
C-H 412
C-Cl 339

energy change = 412-339 = +73kJ

These values may vary slightly according to source.

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