Look at this calculation: The accepted value for the enthalpy of this reaction i

Question

Look at this calculation:

The accepted value for the enthalpy of this reaction is -56.2 kJ/mol, so the experimental determination was 5.56% high.

a) Redo this calculation, assuming a “bad” Ccal of 175 J/ oC rather than the 35.6 J/ oC. (Show work!) By what percentage is your calculated delta Hrxn too high or low, compared to the accepted value of -56.2 kJ/mol?

b) In the original calculation, note that the size of the calorimeter term, 348.88 J, is about 8.5% of the dominant term, qsolution, 4100.32 J. This is typical; the qcal term is not the dominant factor in these determinations. Therefore, if there is a problem in your Ccal determination, it is recommended that the estimate Ccal ~= 0 is used (with proper explanation, and only as an estimation), in the determination of smetal and delta Hneutralization. Redo the calculation of delta Hneutralization, only this time assume Ccal ~= 0 rather than 35.6 J/ oC. (Show work!) By what percentage is your calculated delta Hrxn too high or low, compared to the accepted value of -56.2 kJ/mol?

solution calorimeter t q E 0 4.184 J/(g°C)10 100. ml)(1.00 g/ml) (31.9 -22.1) C) (35.6 J/ oC)(31.9-22.1) C) (AHxn)(1.50 mol/1000 ml0050.0 ml)) 0 4100.32 J 348.88 J (AHron (0.0750 mol) 0 -4449.20 J (AHron )(0.0750 mol) qrxn 59.322.7 J/mol AHDn 59.3 mo AHran

Explanation / Answer

a) Redo this calculation, assuming a “bad” Ccal of 175 J/ oC rather than the 35.6 J/ oC. (Show work!) By what percentage is your calculated delta Hrxn too high or low, compared to the accepted value of -56.2 kJ/mol?

Simply change CCal --< 175 for 35.6

so..

4.184*100*1*(31.9-22.1) + (175)(31.9-22.1) + HRxn*0.075 = 0

solve normally for HRxn

HRxn = -5815.32/0.075

HRxn = -77537.6 J/mol

HRxn= -77.53 kJ/mol

b)

Assume Ccal = 0

so..

4.184*100*1*(31.9-22.1) + (0)(31.9-22.1) + HRxn*0.075 = 0

solve normally

HRxn = -4100.32 /0.075

HRxn = -54670.9333 J/mol

HRxn = -54.67 kJ

the error isnot that high compared to -56.2 kJ/mol

so this is fine to assume Ccal = 0

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