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Find H for the combustion of ethanol (C2H6O) to carbon dioxide and liquid water from the following data. The heat capacity of the bomb calorimeter is 34.65 kJ/K, and the combustion of 1.765 g of ethanol raises the temperature of the calorimeter from 294.33 K to 295.84 K.
$-1365.69 \quad k J / m 01$
Chemistry 102
Chemistry 101
Chapter 9
Thermochemistry
Thermodynamics
Chemical reactions and Stoichiometry
Carleton College
Brown University
University of Toronto
Lectures
00:42
In thermodynamics, the zer…
01:47
A spontaneous process is o…
04:03
Find $\Delta H$ for the co…
03:46
A bomb calorimetric experi…
01:54
The combustion of one mole…
05:08
The enthalpy of formation …
05:25
Write a complete thermoche…
01:21
The complete combustion of…
04:26
The following substances u…
01:12
The heat capacity of a bom…
04:14
Determine the standard ent…
02:05
Ethanol, $\mathrm{C}_{2} \…
06:23
The heat of combustion of …
01:22
Ethanol $\left(\mathrm{C}_…
03:59
02:47
Using standard enthalpies …
04:32
Methanol, ethanol, and $n$…
05:26
A bomb calorimetry experim…
01:19
The combustion of liquid e…
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09:59
A calorimeter that measure…
05:11
So here we have a bomb calorie meter problem, and we want to determine the delta h of combustion of ethanol. And so we know that our temperature ends at two. Ninety five point eight four Calvin and begins at two. Ninety four point three three Calvin and therefore a delta T is one point five one Calvin. And so we also know that the heat capacity of the calorie meter is thirty four point six five Killer Jules per Calvin. And so we multiply that by our one point five one Calvin and get that we have fifty two point three two one five. Kill it, Jules. So now we can determine the number of moles of ethanol and just take Killa Jules over moles and that'LL give us our adult h of combustion. And so we have one point seven six five grams of ethanol. We divide that by forty six point zero seven grams per mole, which is the Mueller massive ethanol. And that gives us zero point zero three eight three moles of ethanol. And so now that we have that, we just take our killer Jules, divided by our moles. So we take our fifty two point three two one five Killer Jules, divided by zero point zero three eight three moles. And that gives us Ah one three six, five point six nine killer Jules Permal and that will be our delta h of combustion. However, we know this is negative because combustion is always extra thermic and so will always have a negative delta H and therefore that is our final answer.
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