# Calculate the standard enthalapy change? Help please :/?

HI i have asked for help with questions via this and thankfully the answer community has helped me :)
I need more help... just to show me how to get the answer... i have searched the net for the solution to this but cant seem to find it..
a) Calculate the standard enthalapy change for the addition of hydrogen...
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HI i have asked for help with questions via this and thankfully the answer community has helped me :)

I need more help... just to show me how to get the answer... i have searched the net for the solution to this but cant seem to find it..

a) Calculate the standard enthalapy change for the addition of hydrogen to pent-1-ene.

CH2 = CHCH2CH2CH3 + H2 -------------> CH3CH2CH2CH2CH3

Given the following bond dissociation data : C - H 413kJmol-1, C - C 346kJmol-1, C = C 610 kJmol-1, H - H 435kJmol-1?

The answer is 127Kjmol-1.... i need help with how you get the answer :/

Then there is part 2 :/.... as follows

b) Calculate the enthalapy of formation of hexane (C6H14) , from the following enthalapy of combustion data?

i) C (s) + O2 (g) -----------> CO2 (g) = -393.5 kJmol-1

ii) H2 (g) + 12 O (g) -----------> H20 = -285.9kjmol-1

iii) C6H14 (l) + 912O2 (g) -------------> 6CO2 (g) + 7H2O (l) = -4163kJmol-1

The answers are as follows but again i need the method of how to get the answers...

(1) x 6

(2) x 7

(3) R

Correct answer and calculation is (-199.3kJmol-1)

PS please help any help in showing me how to get the answer and what working out has to be done... i would be so grateful

I need more help... just to show me how to get the answer... i have searched the net for the solution to this but cant seem to find it..

a) Calculate the standard enthalapy change for the addition of hydrogen to pent-1-ene.

CH2 = CHCH2CH2CH3 + H2 -------------> CH3CH2CH2CH2CH3

Given the following bond dissociation data : C - H 413kJmol-1, C - C 346kJmol-1, C = C 610 kJmol-1, H - H 435kJmol-1?

The answer is 127Kjmol-1.... i need help with how you get the answer :/

Then there is part 2 :/.... as follows

b) Calculate the enthalapy of formation of hexane (C6H14) , from the following enthalapy of combustion data?

i) C (s) + O2 (g) -----------> CO2 (g) = -393.5 kJmol-1

ii) H2 (g) + 12 O (g) -----------> H20 = -285.9kjmol-1

iii) C6H14 (l) + 912O2 (g) -------------> 6CO2 (g) + 7H2O (l) = -4163kJmol-1

The answers are as follows but again i need the method of how to get the answers...

(1) x 6

(2) x 7

(3) R

Correct answer and calculation is (-199.3kJmol-1)

PS please help any help in showing me how to get the answer and what working out has to be done... i would be so grateful

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