# Enthelpy chage for the production of ethanol

• Chemistry
DottZakapa
Homework Statement:
Calculate the standard enthalpy change related to the production of 2.00L of ethanol(##CH_3CH_2OH##, ##d=0,79\frac {g}{cm^3} ##)through the reaction
##CH_{2(g)}+H_2O_{l}->CH_3CH_2OH_{(l)}##
Relevant Equations:
thermochemistry
from table the standard enthalpy of ethanol is ##\Delta H°=-277 \frac{KJ}{mol}##

the grams of ethanol are

##0,002 cm^3*0,79\frac {g}{cm^3}=1,58*10^{-3}g##

and the moles are

##\frac{1,58*10^{-3}g}{(24,02+16,00+6,06)\frac g {mol}}= 3,43*10^{-5}##mol

By proportionality, having already the standard enthalpy for 1 mol of ethanol, the enthalpy change wil be
## 3,43*10^{-5}*-277 \frac{KJ}{mol}##=

is this procedure correct? result doesn't match book expected result. Any help?

Mentor
Check volume conversion.

DottZakapa
DottZakapa
Check volume conversion.
corrected but still doesn't match -1,5*10^3 KJ
from table the standard enthalpy of ethanol is ##\Delta H°=-277 \frac{KJ}{mol}##

the grams of ethanol are

##2000 cm^3*0,79\frac {g}{cm^3}=1580g##

and the moles are

##\frac{1580g}{(24,02+16,00+6,06)\frac g {mol}}= 34,3##mol

By proportionality, having already the standard enthalpy for 1 mol of ethanol, the enthalpy change wil be
## 34,3*-277 \frac{KJ}{mol}##=-9501.1= -9,5*10^3 KJ

still far away from expected -1,5*10^3 KJ

Mentor
How is standard enthalpy of formation defined?

DottZakapa
corrected but still doesn't match -1,5*10^3 KJ
from table the standard enthalpy of ethanol is ##\Delta H°=-277 \frac{KJ}{mol}##

the grams of ethanol are

##2000 cm^3*0,79\frac {g}{cm^3}=1580g##

and the moles are

##\frac{1580g}{(24,02+16,00+6,06)\frac g {mol}}= 34,3##mol

By proportionality, having already the standard enthalpy for 1 mol of ethanol, the enthalpy change wil be
## 34,3*-277 \frac{KJ}{mol}##=-9501.1= -9,5*10^3 KJ

still far away from expected -1,5*10^3 KJ
How is standard enthalpy of formation defined?
I've also tried like this
##\Delta H°_{ethanol}=-277\frac{KJ}{mol}##
##\Delta H°_{H_2O_{(l)}}=-285,8 \frac{KJ}{mol}##
##\Delta H°_R=\sum \Delta H°_{products}-\sum \Delta H°_{reagents}##=
##\Delta H°_R= (-277\frac{KJ}{mol})-(-285,8 \frac{KJ}{mol})= 8,8##
but i think, being positive si not the correct result.

for what concerns ## CH_2 ## I have no values on my table, i thought then, it does not contribute, but a this point i am making some mistake somewhere, i don't knw how to deal with the reagents, expecially why is written ## CH_2= CH_2 + H_2O_{(l)} -> ##

Gold Member
Maybe, the following reaction and its standard enthalpy of reaction are meant:

##C_2H_4(g)+H_2O(l)\to CH_3CH_2OH(l)##

DottZakapa
Maybe, the following reaction and its standard enthalpy of reaction are meant:

##C_2H_4(g)+H_2O(l)\to CH_3CH_2OH(l)##
thank you now it makes more sense, all works perfectly.
Are we sure is like that ? ##CH_2= CH_2 + H_2O_{(l)} -> ## so this is a way of writing that or is just a typo? never seen before such writing

Lord Jestocost
Homework Helper
The = indicates a double bond between the two carbon atoms of C2H4. It is not an equals sign.

Lord Jestocost and DottZakapa
DottZakapa
The = indicates a double bond between the two carbon atoms of C2H4. It is not an equals sign.
o ok, thank you, didn't know it. Thanks

Mentor
Ah, I though CH2 is just a typo and the question really meant starting from CO2/H2O.

DottZakapa
Ah, I though CH2 is just a typo and the question really meant starting from CO2/H2O.

🤷‍♂️ results coincide, so i guess he is right

Mentor
Yes, once you wrote it as CH2=CH2 the meaning is unambiguous and clear.

DottZakapa