Enthelpy chage for the production of ethanol

• Chemistry
• DottZakapa
In summary, the conversation discusses the calculation of the enthalpy change for the formation of ethanol using the standard enthalpy of formation. After converting the volume to grams and moles, the enthalpy change is calculated using proportionality. However, the result does not match the expected value and further discussion reveals that the correct reaction should be C2H4(g)+H2O(l)->CH3CH2OH(l). After correcting for this, the calculations are successful and the results match. The use of the = symbol in CH2=CH2 is not a typo, but rather indicates a double bond between the two carbon atoms.
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?

Check volume conversion.

DottZakapa
Borek said:
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

How is standard enthalpy of formation defined?

DottZakapa said:
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 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)} -> ##

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

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

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

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

Lord Jestocost and DottZakapa
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

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

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

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

DottZakapa

1. What is enthalpy change?

Enthalpy change is a measure of the amount of energy absorbed or released during a chemical reaction or process.

2. How is enthalpy change related to the production of ethanol?

Enthalpy change is important in the production of ethanol because it is a key factor in determining the efficiency and cost-effectiveness of the process. The enthalpy change for the production of ethanol refers to the amount of energy that is absorbed or released during the conversion of raw materials, such as corn or sugar, into ethanol.

3. What factors affect the enthalpy change for the production of ethanol?

The enthalpy change for the production of ethanol can be affected by various factors, including the type of raw materials used, the efficiency of the conversion process, and the conditions under which the reaction takes place (e.g. temperature, pressure, catalysts).

4. Why is it important to consider enthalpy change in ethanol production?

Considering enthalpy change in ethanol production is important because it can impact the overall cost and environmental impact of the process. By optimizing the enthalpy change, we can reduce the energy consumption and emissions associated with ethanol production, making it more sustainable and economically viable.

5. How is enthalpy change calculated in the production of ethanol?

The enthalpy change for the production of ethanol can be calculated using the difference in the enthalpy values of the reactants and products. This can be determined experimentally or through thermodynamic calculations using known enthalpy values and reaction stoichiometry.

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