Chemistry (Enthalpy) Problem - Printable Version

Chemistry (Enthalpy) Problem - Printable Version

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Chemistry (Enthalpy) Problem - Panacea - 2011-04-02

Problem:

Spoiler

What I've done:

I know the ice has to be broken down into three equations. (q = mC(T1-T2))

I've gotten the heat, q, for the change in temperature from -9.3 -> 0 C. But, from here, I have no idea. I'm assuming I have to do something with the enthalpy fusion number in the second equation, get some answer, then use that answer in the third equation and equate it to the equation for the water.

(q I got for the first ice equation is 391.6 Joules).

So, yeah. Any suggestions as to what I should do?

Chemistry (Enthalpy) Problem - Shidoshi - 2011-04-03

Since they already tell you the ice will be melt at then end, all you have to do is calculate how much energy you need to go from:
-9,3ºC to 0ºC as ice (ice heat capacity 37.5 J/K.mol times the temperature difference and total ammount of ice in mols)
Energy1 = (0 - (-9.3)) * (37.5) * (mols of ice)

-0ºC ice to 0ºC water (entalpy of fusion, 6.01*10³ J/mol times the ammount of ice in mols).
Energy2 = (mols of ice) * (6.01*10³)

Then you take these two ammounts of energy you used to heat the ice and you use it to calculate how much the water has cooled:
(Energy1 + Energy2) = (water heat capacity) * (Nº of mols of water) * (95.3º - NewTemperature)

After that, to the final temperature you just need to do a weighted average of the two bodies of water, the one at 0ºC and the one at the temperature you just calculated.
(mass of water at ºC) * (0ºC) + (mass of water at NewTemperature) * (NewTemperature) = (total mass of water) * (Final Temperature)

Also, take care for when the values of entalpy/heat capacity are in kJ or J. And be sure to use mols when the aforementioned are per mol.

Chemistry (Enthalpy) Problem - Panacea - 2011-04-03

Got it! Thanks.

Chemistry (Enthalpy) Problem - Panacea - 2011-04-11

Spoiler

This problem's making me mad, because I'm doing exactly as someone else says and getting the wrong answer, supposedly. Got three heat equations, get delta H for the combustion of methane, then divide delta H by Qtotal, and multiplied by the molar mass of methane. Which is wrong.

Haven't slept and about to have to leave for work. Sorry I couldn't go into more detail about what I did, but if someone could help me I'd really appreciate it.

Chemistry (Enthalpy) Problem - OB3LISK - 2011-04-12

Okay I looked up some constants. I hope they're right.

Q = mst

T is the change in temperature. 113 - 30 = 83 Celsius.

Specific heat of water is 4.186.

Mass is 377 grams.

Q = (83) (377) (4.186)
Q = 130984 Joules = 130.98 kilo Joules

Now we need to find a constant that methane burns at. I'm not too sure about this one. I looked at the back of my book and got the heat of FORMATION to be -74.8. But I also googled it and got that it could be 55. Both in kilojoules.

So we do 130.98/74.8 (or 130.98/55) to get the number of moles of methane needed.

I get 1.75 moles (or 2.38 moles).

Then to get grams. Carbon weighs 12 grams, 4 x Hydrogen weighs 1 gram. 16 grams total.

16 Molecular mass x 1.75 moles = 28 grams. (Or 2.38 x 16 = 38.08 grams)

So 28 or 38. Maybe. I'm not sure, this question gives little information.

Chemistry (Enthalpy) Problem - Panacea - 2011-04-12

I know for sure that you have to separate the heat for the water into three different equations.

One for when the water is heated from 30 -> 100 degrees C. (C = 4.18 J / (g * C))

Two for the energy needed to break bonds within the 377g of H20 for it to evaporate. (H-Vap for water is 40.65 kJ/mol)

And third, the energy needed to heat the rest of the way (13 degrees C). (C = 2.03 J / (g*C)).

It's after here I think I mess up, where I need to set up the combustion equation for CH4: After balancing the equation (1 2 2 1 moles, respectively for each element), in order to acquire the DeltaH I have to subtract the heats of formations of the products minus the reactants.

Sorry for the late reply. I fell asleep.

EDIT: Son of a peach, finally got it.

q1 = (377g/18g = mol H20) * (75.291 J * mol ^ -1 * C ^ -1) * (70 C)

q2 = (377g/18g) * (40.65 kJ/mol H20)

q3 = (377g/18g) * (33.577 J * mol ^ -1 * C ^ -1) * (13 C)

q_tot = q1 + q2 + q3

DeltaH for methane combustion = Formation energy of productions minus reactants =

[(2*-241.8 kJ/mol)(H20 (g)) - 393.509 kJ/mol) (CO2)] - [-74.81 kJ/mol (Methane)] = DeltaH

n * DeltaH = -q_tot -> n = -q_tot/DeltaH

Mass methane (g) = n * 16g = 19.4 g

Chemistry (Enthalpy) Problem - OB3LISK - 2011-04-13

Never did something like that lol but nice anyway.