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Topic: Standard Enthalpy of Reaction Question.  (Read 11060 times)

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gameniac85

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Standard Enthalpy of Reaction Question.
« on: March 25, 2006, 12:45:44 AM »
I have been racking my brain for some time with this problem, I see what I have to do and I really think I get an answer but it’s continually not right. The problem is as follows:

Propane, C3H8(g), is used in many instances to produce heat by burning: C3H8(g) + 5O2(g) -->3CO2(g) + 4H2O(g) The standard enthalpy of reaction is -2,044 kJ. How many grams of oxygen would be needed to produce 605.0 kJ of heat? The answer needs to be in scientific notation.

I have tried various stoichiometry setups and I get nothing. I am not even sure what units go with what to be honest. I am trying to find grams of oxygen, but I don't even know what I have to start with. I have tried the following:

605.0 kJ x 1 mol x 16.00 g O2 / 2044 kJ x 5 mol O2
605.0 kJ x 1 mol x 32.00 g O2 / 2044 kJ x 5 mol O2

I have tried starting with the 2044 kJ and still nothing. I have also tried:

605.0 kJ x 1 mol x 16.00 / 1 kJ x 5 mol
605.9 kJ x 1 mol x 32.00 / 1 kJ x  5 mol

I have tried other various things with this as well trying to include the C3H8 in there to see what happens including using its formula weight of 44.01, but it seems nothing works.

I am missing something, I know I am. It is right there and I can't see it. I was able to get the other problems for my homework without any problems, even some of the harder Hess' Law problems, but this is just baffling me. Perhaps it’s because its 12:40 A.M. Any guidance or direction anyone could give me would be appreciated.

Offline green-goblin

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Re:Standard Enthalpy of Reaction Question.
« Reply #1 on: March 25, 2006, 06:06:39 AM »
Well one mole of propane reacts with 5 moles of oxygn to give -2044kjmol. you only need 605kj which is..(wait well he grabs his calculator)..0,296 % of 2044 so you only need 0.296 moles of porpane, times that by five to get your no of moles of O and times that by 16 to get the amount of oxygen in grams
« Last Edit: March 25, 2006, 06:07:45 AM by green-goblin »

gameniac85

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Re:Standard Enthalpy of Reaction Question.
« Reply #2 on: March 25, 2006, 10:59:01 AM »
I see now. I was kind of close, but totally off all at the same time. I tried what you told me and I still did not get the answer, but then I realized it is O2 and I needed to multiply by 32.00 g rather then 16.00 g and that led me to the right answer. Anyway, thank you very much for your assistance.

Offline green-goblin

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Re:Standard Enthalpy of Reaction Question.
« Reply #3 on: March 25, 2006, 12:08:15 PM »
lol, I forgot to times 16 by two as well  :-\ opps. Glad youv'e got it solved

gameniac85

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Re:Standard Enthalpy of Reaction Question.
« Reply #4 on: March 25, 2006, 01:11:34 PM »
Ok, now that I have the figured out part A, I can't figure out part B of that problem. It’s really weird, the rest of this homework was relatively easy, but this question is really hard. Part B is as follows:

A swimming pool 38.5 meters long, 15.8 meters deep, and 4.2 meters wide is filled with water. The temperature of the water is 25.5 deg C, but the owner of the pool would like the temperature to be 33.3 deg C. How many grams of propane would have to be consumed if all of the heat from the combustion was to be absorbed by the water? Refer to the previous problem. Enter your answer in scientific notation.

Now I determined that the volume of the water is 2554.86 m3 and that you need to convert it to liters. I used meters cubed to decimeters cubed and then to liters and I got 2.55 x 109 L. Now the density of water is 1.0 g/mL so I converted the liters to mL to get 2.55 x 1012 mL. Then I multiplied that by 1.0 g/mL, the mLs cancel and I get 2.55 x 1012 grams of water. Now I plug that into q=mc(deltaT) .

So I multiplied 2.55 x 1012 g H2O x 4.18 J/(g x degC) x 7.8 degC.

That got me 8.31 x 1013 J of H2O.

This is where I get lost. I am not sure where to go from here. I am not even sure if what I have done so far is right. I am also not sure what I need from part A to help me with part B. So, once again I am stuck with this question. Any further assistance anyone could give would be once again appreciated.

Offline Borek

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Re:Standard Enthalpy of Reaction Question.
« Reply #5 on: March 25, 2006, 01:41:53 PM »
That got me 8.31 x 1013 J of H2O.

Why 'of H2O'? Just J, you were calculating energy amount.

Quote
Propane, C3H8(g), is used in many instances to produce heat by burning: C3H8(g) + 5O2(g) -->3CO2(g) + 4H2O(g) The standard enthalpy of reaction is -2,044 kJ.

Quote
How many grams of propane would have to be consumed if all of the heat from the combustion was to be absorbed by the water?

Assuming 8.31 x 1013 J is correct - what I have left is all you need to find out the answer.
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Offline green-goblin

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Re:Standard Enthalpy of Reaction Question.
« Reply #6 on: March 25, 2006, 01:43:39 PM »
You have just worked out the amount of energy you require. 8x31x1013j, according to you calculations.

I think all you should need to do now is devide that number by the amount of joules given out by the combustion of one mole of propane. This will tell you how many moles of propane you need, then times that by it's RMM.

Offline green-goblin

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Re:Standard Enthalpy of Reaction Question.
« Reply #7 on: March 25, 2006, 01:45:36 PM »
sorry borek i was still writing my reply so didn't see yours go up.
« Last Edit: March 25, 2006, 01:45:50 PM by green-goblin »

gameniac85

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Re:Standard Enthalpy of Reaction Question.
« Reply #8 on: March 25, 2006, 02:06:17 PM »
Ok so this is what I have done.

8.31 x 1013 J x 1000 kJ x 1 mol x 44.11 g / 2044 kJ

That gets me 1.79 x 1015 g of propane, which sadly is incorrect.

I also tried it as just (8.31 x 1013 J / 2044 kJ) x 44.11 g, which gets me 1.79 x 1012, also incorrect.

I have actually already come across those numbers before in my other failed attempts. So, perhaps the 8.31 x 1013 is incorrect?

Offline Borek

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Re:Standard Enthalpy of Reaction Question.
« Reply #9 on: March 25, 2006, 03:18:54 PM »
I also tried it as just (8.31 x 1013 J / 2044 kJ) x 44.11 g, which gets me 1.79 x 1012, also incorrect.

That's correct approach, although it looks like you forgot what k in kJ means.

Besides, getting back:

Quote
Now I determined that the volume of the water is 2554.86 m3

Correct.

Quote
and that you need to convert it to liters. I used meters cubed to decimeters cubed and then to liters and I got 2.55 x 109 L.

Wrong. Seems like you have a problem with unit conversions. Decimeter cubed IS already 1L, so there is no further conversion needed.

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gameniac85

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Re:Standard Enthalpy of Reaction Question.
« Reply #10 on: March 25, 2006, 03:37:58 PM »
I want to make sure this is right.

Ok so we know 2554.86 m3 is right. So I multiplied that by 1000 to get me 2.55 x 106 L. I then multiplied that by 1000 to get that into mL. So now I have I have 2.55 x 109 mL. I then plugged that into the density for water and got the grams. I then plugged that into q=mc(deltaT) and got 8.31 x 1010 J.

I then did (8.31 x 1010 J x 1000 kJ x 1 mol x 44.11 g) / 2044 kJ

That gets me 1.79 x 1012 g. So, I am still screwing something up.

Offline green-goblin

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Re:Standard Enthalpy of Reaction Question.
« Reply #11 on: March 25, 2006, 04:04:08 PM »
like borek said, it's your conversions. a litre is 1000cm3 or 1m3
« Last Edit: March 25, 2006, 04:05:54 PM by green-goblin »

gameniac85

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Re:Standard Enthalpy of Reaction Question.
« Reply #12 on: March 25, 2006, 04:15:23 PM »
So it's basically there’s no need to convert anything, and its just 2554.86 L? From there I would multiply that by 1000 to get it to mL and then that becomes grams using the density of water. I then plug 2.55 x 106 in to q=m(deltaT) and get 8.31 x 107 J. So now I multiply that to 1000 to get it into kJ, divide by 2044 kJ and then multiply that by 44.11 g to get 1.79 x 109 g of propane.

I googled 2544.86 cubic meters to liters and it said this:

2 544.86 (cubic meters) = 2 544 860 liters

So, now if I use that I get the same answer as before of 1.79 x 1012.

So now I am just totally confused. Thanks for all the help so far.
« Last Edit: March 25, 2006, 04:33:44 PM by gameniac85 »

Offline green-goblin

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Re:Standard Enthalpy of Reaction Question.
« Reply #13 on: March 25, 2006, 04:39:24 PM »
2554.86m3 is 2445.86 litres, try it with that
« Last Edit: March 25, 2006, 04:40:28 PM by green-goblin »

Offline green-goblin

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Re:Standard Enthalpy of Reaction Question.
« Reply #14 on: March 25, 2006, 04:49:32 PM »
I get 1793.121 grams of propane. Do you know what the answer should be?

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