PLEASE HELP! A 0.900 sample of toluene, C7H8, was completely burned a bomb calorimeter containing 4560.g of water which increased in temperature from 23.800C to 25.718C. What is Delta E for the reaction in kJ/mole C7H8? The heat capacity of the calorimeter was 780. J/C. The specific heat of water is 4.184 J/gC. The answer to this question is -3900 kJ/mol. If anyone can help me know how to solve this problem it could be greatly appreciated. You would be my lifesaver! Please Help! Thank you. :)

Question

PLEASE HELP! A 0.900 sample of toluene, C7H8, was completely burned a bomb calorimeter containing 4560.g of water which increased in temperature from 23.800C to 25.718C. What is Delta E for the reaction in kJ/mole C7H8? The heat capacity of the calorimeter was 780. J/C. The specific heat of water is 4.184 J/gC.  The answer to this question is -3900 kJ/mol. If anyone can help me know how to solve this problem it could be greatly appreciated. You would be my lifesaver! Please Help! Thank you. :)

aaronq · Answer

You're finding the heat evolved by combustion by monitoring the water

$\sf \large \Delta H_{rxn}=m_{water}*C_p*(T_f-T_i)$ 

After, you need to scale it by taking into account the number of moles (n) of toluene used, to find the standard enthalpy of combustion (with units specified - note you will have to convert J to kJ at the end).

$\sf \large \Delta H_{rxn}=H^o*n_{toluene}$

anonymous · Answer

Thank you so much! I asked my lab teacher, the tutor, and my chemistry I teacher to help me solve the problem, and none of them could help me solve the Chem II problem.

aaronq · Answer

You should try it out, if you can't with what i gave you, i'll do it for you

anonymous · Answer

Can you tell me what the mwater and the Cp stand for?

aaronq · Answer

$m_{water}$=mass of water
$C_p$ is the specific heat capacity

anonymous · Answer

So would the Cp be the heat capacity that =780 J/C?

aaronq · Answer

Oh i didn't read that, you need to account the heat absorbed by the calorimeter, so you need this extended formula:

$\Delta H_{rxn}=m_{water}*C_p*(T_f-T_i)+C_p*(T_f-T_i)$
                              water                         calorimeter

anonymous · Answer

Is the temperature of the calorimeter the same temperature of the water?

aaronq · Answer

yeah, they're in thermal equilibrium

anonymous · Answer

If you possibly can like take me step by step through it. I think I plugged in the right numbers. It should be Delta Hrzn = 49.57 mol (I went ahead and converted to moles) * 4.184*1.918*780*1.918 = 547094.7066. But now I don't know how to use the other equation.

aaronq · Answer

$\Delta H_{rxn}$ should give you a value in terms of energy, not moles.

aaronq · Answer

I have to go, try it again and i'll check it when i'm on again

anonymous · Answer

Ok thank you.

aaronq · Answer

so it's like this:

$\sf  \Delta H_{rxn}=(4560~g)* ( 4.184~ J/gC)*(25.718-23.800)C+780 J/C*(25.718-23.800)C$

(Note how the units cancel out and you're left with Joules)

$\sf \Delta H_{rxn}=38089.63872~J=38.09~ kJ$

Because the water and the calorimeter absorbed the heat (endothermic) this is a positive value.We need to change it to a negative value because the reaction is exothermic.

---------------------------------------------------------
Then use the other formula to scale it up to 1 mole of toluene

$\sf-\Delta H_{rxn}=H^o_{combustion}*n_{toluene}$

$\sf H^o_{combustionn}=\dfrac{-\Delta H_{rxn}}{n_{toluene}}=\dfrac{-38.09~ kJ}{0.9~mol}=-42.22 ~kJ/mol$

This number is a little different than the one you posted, but i'm 99.99% sure this is right.

anonymous · Answer

I'm sorry that I couldn't get back to you sooner, but thank you so much! You really helped, and you were right. The teacher said there was a mistake in the answer choices and the answer key on that question. Thank you again! :)

aaronq · Answer

Great, i'm glad you got it all figured out! no problem!