Enthelpy chage for the production of ethanol

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  • #1
DottZakapa
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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?
 
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  • #2
Check volume conversion.
 
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  • #3
Borek said:
Check volume conversion.
corrected but still doesn't match -1,5*10^3 KJ :confused:
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
 
  • #4
How is standard enthalpy of formation defined?
 
  • #5
DottZakapa said:
corrected but still doesn't match -1,5*10^3 KJ :confused:
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
Borek said:
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)} -> ## :oldconfused:
 
  • #6
Maybe, the following reaction and its standard enthalpy of reaction are meant:

##C_2H_4(g)+H_2O(l)\to CH_3CH_2OH(l)##
 
  • #7
Lord Jestocost said:
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
 
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  • #8
The = indicates a double bond between the two carbon atoms of C2H4. It is not an equals sign.
 
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  • #9
mjc123 said:
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
 
  • #10
Ah, I though CH2 is just a typo and the question really meant starting from CO2/H2O.
 
  • #11
Borek said:
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
 
  • #12
Yes, once you wrote it as CH2=CH2 the meaning is unambiguous and clear.
 
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