[NewtoAtoms]

Hello Physical Chemists,

I have this really stumping question, in which I have tried to work out myself.  Would someone be so kind as to review my work and tell me if I am at all in the correct ball park??!?!?  This stuff is making me sweat!

Q:  The heat capacity of a bomb calorimeter is 87.5 kJ/K (this value is for the total heat capacity including that of the water jacket around the reaction chamber). If 67.2 g of methane, CH4 (g), is combusted under such reaction conditions, what will be the increase in temperature of the calorimeter?  ΔEcombustion for CH4 (g) is -885.4 kJ/mol

A:  I have done the leg work, but I could be totally wrong, and I would be so grateful if you could review it and help me back onto the correct path again. 

GIVEN:

C_cal = 87.5 kJ/K
67.2 g CH₄ added
Calculate Δt
ΔE_comb = -885.4 kJ/mol (IS THIS SAME AS ΔE_rxn)

WORK:

STEP #1

ΔE_rxn= q_rxn / mol CH₄
-885.4 kJ/mol = q_rxn / 4.19 mol
-211.3 kJ = q_rxn

STEP #2

q_rxn + q_cal = 0
q_cal = -q_rxn
q_cal = -(-211.3 kJ)
q_cal = 211.3 kJ

Step #3

Δt = q_cal / C_cal
Δt = 211.3 kJ / 87.5 kJ/ K
Δt = 2.41 K

In my mind I try to rational the answer.  The original ΔE= -885.4 kJ/mol which means that this reaction is exothermic, and heat is given off.  According to my calculations above there is a change in temperature of 2.41 K, which means that there is definately heat given off from the exothermic reaction and the temperature rises to support my findings.

Is this at all correct, or am I totally lost?

Someone would you please be so kind to help.
Thank you for your time

Newtoatoms

[Borek]

ΔE_rxn= q_rxn / mol CH₄

You mean - the more moles of methane will be burnt, the less energy will be produced?

[NewtoAtoms]

yes Borek,  but did you understand my calculations?

Am I at all in the correct field?

I deeply appreciate your help Borek, as your chemistry knowledge far outways mine.

Have a great day.

[Borek]

yes Borek,  but did you understand my calculations?

Did you understand that the equation must be wrong, if it leads to absurd conclusions?

[NewtoAtoms]

I am sorry Borek, I absolutely don't understand.  The equation ΔErxn = qrxn/mol CH4 is what is given in my text, and therefore I referred it to this question. I was under the impression that the ΔErxn is calculated using qrxn, per moles of CH4 in 67.2 g.  Therefore now I am more perplexed than when I initially posted the problem.

Can anyone offer a bit more help and direction, I would so deeply appreciate it. 

I am new to chemistry, so what might seem very obvious to you, is sometimes a bit more difficult for others, especially me.  I am however so grateful for everyone's time and assistance.

Thank you,

Newtoatoms

[ARGOS++]

Dear NewtoAtome;

Why you divide in step 1.?:   Think - That’s not the  ∆E_RxN !!
(Release 4.19 moles less energy than 1.0 mole?)

Your calculation does otherwise the right thing at the right moment.
If you know calculate the ∆T you will get a much different/higher one!

I hope I have been of help to you.
Good Luck!
                    ARGOS⁺⁺

[NewtoAtoms]

Thank you for your reply Argos and the explanations!

If I remove moles from step #1 such as:

STEP #1

ΔE_rxn= q_rxn
-885.4 kJ/mol = q_rxn
-885.4 kJ/mol = q_rxn

Then where in the calculations should I take into account that there are 4.19 moles of CH₄ in the original reaction and not just 1???

If I don't put this into step #1, where should I calculate it in??

[ARGOS++]

Dear NewtoAtome;

NO!,  - that’s the wrong action!  You have to multiply!
 
For Step 1:
   ∆E_Total = ∆E_RxN = 4.19 moles _* -885.4 kJ/mole = ?

Can you now continue?

Good Luck!
                    ARGOS⁺⁺

[NewtoAtoms]

Hello Argos++,

Of course we have to multiply, that way the moles cancel out leaving only kJ. 
My lightbulb went on, thank you for your patience!

Therefore:


STEP #1

-885.4 kJ/mol= q_rxn / 4.19 moles CH₄
qrxn =  -885.4 kJ/mol x 4.19 mol
-3709.8 kJ = qrxn

STEP #2

qrxn + qcal = 0
qcal = -qrxn
qcal = -(-3709.8 kJ)
qcal = 3709.8 kJ

Step #3

Δt = qcal / Ccal
Δt = 3709.8 kJ / 87.5 kJ/ K
Δt = 42.40 K

And just so that I understand:  This means that when 67.2 g of CH4 is added to a calorimeter, the temperature of the calorimeter will increase by 42.40 k.
My only issue with this is that 42.40 K is -230.6^oC, what does that mean??
The original reaction was exothermic (ΔErxn = -885.4 kJ/mol) which means that through the reaction heat will be released by the system to the surroundings - water in the calorimeter in this case.  However according to my calculations there will be a Δt = 42.40 K which is a change of -230.5^oC... what on earth does that mean?

THANK YOU for helping me walk through this question. I am so grateful!!!

[ARGOS++]

Dear NewtoAtome;

I got exactly the same value!

You 're welcome!   ─   Soon again.

Good Luck!
                    ARGOS⁺⁺

[ARGOS++]

Dear NewtoAtome;

Can you tell us about the start temperature, because you never told anything about?
I can’t believe a calorimeter at a start temperature of ~ 190°C!

Good Luck!
                    ARGOS⁺⁺

P.S.: Can you next time use an other/additional posting and not edit the former one?