# AS ENERGETICS help

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#1
Given the following data
C(s) + 2H2(g) → CH4(g) ∆H = −75 kJ mol−1
**H2(g) → 2H(g) ∆H = +436 kJ mol−1
which one of the following is the enthalpy change, in kJ mol−1, of the reaction below?
CH4(g) → C(s) + 4H(g)
A−947
B+511
C+797
D+947

Please explain why. Thank you guys!!!!!
0
5 years ago
#2
If you reverse and add the equations until you get the desired equation and do the same to the enthalpies, you'll get the answer.

So, reverse the methane formation equation, to give CH4 -> 2H2 + C with ∆H=75

Double the hydrogen equation to get 2H2 -> 4H with ∆H=872

Add both to give CH4 + 2H2 -> 4H + C + 2H2

The 2H2's are on both sides of the equation, so cancel them to give CH4 -> 4H + C

Then do the same to the enthalpies as you did to the equations (doubling one, and adding both) to give 872 + 75
1
#3
(Original post by h3rmit)
if you reverse and add the equations until you get the desired equation and do the same to the enthalpies, you'll get the answer.

So, reverse the methane formation equation, to give ch4 -> 2h2 + c with ∆h=75

double the hydrogen equation to get 2h2 -> 4h with ∆h=872

add both to give ch4 + 2h2 -> 4h + c + 2h2

the 2h2's are on both sides of the equation, so cancel them to give ch4 -> 4h + c

then do the same to the enthalpies as you did to the equations (doubling one, and adding both) to give 872 + 75
life saver!!!!thank you!!!
1
#4
Another Question .... Anyone for help.

Substance
ethanamidecarbon dioxidewater
Enthalpy of formation ([img]file:///C:/Users/HUIHUI~1/AppData/Local/Temp/ksohtml/wps478D.tmp.jpg[/img]) / kJ mol−1
−320−394−286

Using the data above, which one of the following is a correct value for the enthalpy of combustion of ethanamide?

A −1823 kJ mol−1
B −1183 kJ mol−1
C −1000 kJ mol−1
D −360 kJ mo1−1

Please explain why. Thank you guys!!!!!
0
1 year ago
#5
why do you have to reverse it?
0
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