Chemistry Enthelpy chage for the production of ethanol

[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?

 

[Borek]

Check volume conversion.

 

[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

 

[Borek]

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 $CH_2= CH_2 + H_2O_{(l)} ->$

 

[Lord Jestocost]

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

 

[mjc123]

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

 

[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

 

[Borek]

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

 

[Borek]

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