ATAR Notes: Forum

VCE Stuff => VCE Science => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Chemistry => Topic started by: dfgjgddjidfg on November 09, 2012, 09:24:40 pm

Title: Calorimetry help
Post by: dfgjgddjidfg on November 09, 2012, 09:24:40 pm

how do i do this?

Title: Re: Calorimetry help
Post by: stephanieteddy on November 09, 2012, 10:01:35 pm

Do you have the answer? I have worked it out, but don't want to mislead you by posting incorrect solutions!

Title: Re: Calorimetry help
Post by: dfgjgddjidfg on November 09, 2012, 10:11:40 pm

yep its 24.8 degrees

Title: Re: Calorimetry help
Post by: stephanieteddy on November 09, 2012, 10:20:28 pm

This is what I did, hopefully no mistakes along the way!

Okay so, we have:

2HCl + Ba(OH)2 ----> BaCl2 + 2H20

If we remove the spectator ions, we end up with: H+(aq) + OH-(aq) ---> H20(l)

Now if we find the mole we have of each, we can see that n(OH-)= 2*0.04*0.431= 0.03448mol (which is limiting as it is less than the n(H+)
Now since 1mol of reaction = 56.2kJ of energy
                 0.03448mol        = x               
Cross multiply and you get a total of 1.9378kJ of energy released. Since we are dealing with a calibration factor with the units J/degree Celsius we have to convert this to J, so multiply by 1000 = 1937.8J of energy.

Pop this into out calorimeter rule thingo

C.F = E/Change in temp   (CT)                   ------------>                453 = (1937. 8 )/(CT) 

So then we solve and see that change in temp is 4.278 degrees celsius. Add this to our initial temperature and the final temperature is hopefully 24.8 degrees celsius.

Title: Re: Calorimetry help
Post by: dfgjgddjidfg on November 09, 2012, 10:24:38 pm

thanks. Whats the point of removing the spectator ions from the equation, did we even need to use that equation?

Title: Re: Calorimetry help
Post by: stephanieteddy on November 09, 2012, 10:28:10 pm

No, I just didn't want to confuse you, sorry :)