[lmvent]

When ice at 0.0oC melts to liquid water at 0.0oC, it absorbs, 0.334 kJ of heat per gram. Suppose the heat needed to melt 32.0 g of ice is absorbed from the water contained in a glass. If this water has a mass of 0.510 kg and a temperature of 26.0oC, what is the final temperature of the water (in C)? (Note that you will also have 32.0 g of water at 0oC from the ice). Do not enter unit.

Every time I do this problem I get either 21.8 degrees or 20.986 degrees. I have been doing m(H2O)*4.18*(26-T) = 0.334*32 and solving for T.  Anybody know what I'm doing wrong?




A 33.0-mL sample of 1.05 M KOH is mixed with 17.0 mL of 1.07 M HBr in a coffee-cup calorimeter (see Section 6.6 of your text for a description of a coffee-cup calorimeter). The enthalpy of the reaction, written with the lowest whole-number coefficients, is -55.8 kJ. Both solutions are 21.2oC prior to mixing and reacting. What is the final temperature (in C) of the reaction mixture? When solving this problem, assume that no heat is lost from the calorimeter to the surroundings, the density of all solutions is 1.00 g/mL, and volumes are additive. Do not enter unit.

For this one I started by writing the balanced equation KOH + HBr ---> KBr + H2O but because there are a different number of moles of KOH and HBr, I don't really know how to find what deltaH would be for this reaction.




A 10.0-g sample of a mixture of CH4 and C2H4 reacts with oxygen at 25.0oC and 1.00 atm to produce CO2(g) and H2O(l). If the reaction produces 548 kJ of heat, what is the mass percentage of CH4 in the mixture? Do not enter unit.

For this one, I was using the standard enthalpies of formation, but I can't really balance the chemical equation without knowing how much of the mixture is CH4 or C2H4 and that is what the question asks.





If anyone could give me some help I would really appreciate it.  The questions aren't woth much, I just really want to understand how to do them. Thanks

[sdekivit]

you carry on in the process. When the ice is melted, the water at 0,0 degrees also takes up heat from the water.

Suppose we have 32 g ice at 0,0 degrees C then this takes up 334 * 32 = 10688 J heat from the water.

But:

When the water releases 10688 J heat, the temperature of the water will be:

10688 = 4180 * 0,510 * delta T
--> delta T = 5,01360 degrees
--> T(water) = 26 - 5,01360 = 20,98640 degrees C

Thus: the water will still give heat to the water at 0,0 degrees C till there is equilibrium:

c * m * delta T(water at 0,0 C) = c * m * delta T (water at 21 C)

[lmvent]

Thanks, that was very helpful and I got it right. Any ideas for the other two??

[sdekivit]

KOH + HBr --> KBr + H2O --> delta H = -55,8 kJ

then we have 34,65 mmol KOH and 18,19 mmol HBr

thus we know from our reaction that 55,8 kJ of heat is produced per mol HBr.

--> calculate the amount of heat produced due to the amount of HBr that has reacted.

Then use Q = c * m * delta T and you know that m = density * volume (volumes were additive)

assume a specific heat equal to that of water.

The reaction releases het thus delta T must be added to the prior temperature.

Try the last one on your own now

[lmvent]

Thanks, I get that one too now.  The last one I tried a few times and I keep getting the wrong answer.  Do you think you could give me a hint of where to start maybe?

[sdekivit]

get 2 equations with x = mol Ch4 and y = mol C2H4 for the mass and the energy and then solve for x and y.

[lmvent]

Okay, here's what I did...

x = moles CH4 and y = moles C2H4

(x/16) = grams CH4 and (y/28) = grams C2H4

CH4 + 2O2 --> CO2 + 2H2O  delta H = -890.23 kJ
C2H4 + 3O2 --> 2CO2 + 2H2O  delta H = -1141.07 kJ

(-890.23 kJ/mol)*x + (-1141.07 kJ/mol)*y = -548 kJ
                             and
(x/16) + (y/28) = 10 g

I get x = 288.216, which isn't even reasonable.  What am I doing wrong??

[sdekivit]

you must do x * 16 = g CH4 and y * 28 = g C2H4

--> thus: 16x + 28y = 10

Then we take combustion energies, which tell us the enthalpychange per mol compound. Combustion energy for CH4 = -8,9 x 10^5 Jmol^(-1) and for C2H4: -14,1 x 10^5 Jmol^(-1)

--> thus: 8,9 x 10^5 x + 14,1 x 10^5 y = 548 x 10^3

Now solve for x and y.

[lmvent]

when I do that I get 84% CH4 and that is incorrect.

[lmvent]

I also can't find the correct answer, because I used all of my tries, so I have no way to check my answers anymore until after the homework is due.

[sdekivit]

when I do that I get 84% CH4 and that is incorrect.

when i use 12,01 and 1,008 as molar masses of C and H i get 87 %

[lmvent]

I see...thanks!