Posted by Sindy on Friday, October 9, 2009 at 12:29pm.
First you need to address the data. I see 26.318 kJ/g for delta H in the table but 26.38 in the question. Be sure to take care of that first. To calculate the calorimeter constant, that is
mass x specific heat x delta T.
q = delta H, 26,380 J (check that number) = 1.045 x Cp x 5.985. Solve for Cp.
For the octanol,
[mass octanol x specific heat octanol x delta T] + [Cp x delta T] = 0
Solve for specific heat octanol which will be in units of J/g*C.
Then change that to kJ/mol.
Check my thinking. Check my work.
Okie i get Cp= 42178.8J/g delta C
and for the octanol
im using q= m*Cs*T
q= 0.876*Cs*8.158
q= 7.1395g degreeC * Cs
In your work i don't understand thisequation:
[mass octanol x specific heat octanol x delta T] + [Cp x delta T] = 0
Okie I get 36913.6J/g* degree C but how do I change that into kJ/mol?
36913.6 J/g x (grams octanol/mol) = ??
In other words, look up the molar mass of octanol.
36913.6 J/g * 114g/mol= -4208150.4J/mol and to convert that to kJ= -4208150400 kJ/mol.
the correct answer is -5116 kJ/mol
this is so confusing =(
You need to clarify.
Is benzoic 26.38 or 26.318?
Is the temperature rise 8.158, 8.518, or 8.15?
Benzoic is 26.38 and the temp rises 8.518
I did is again:
Benzoic acid: q= m*Cs*T
Cs = 4.217kJ/g°C
q1 + q2 = 0
[m*Cs*T] + [26.38kJ] = 0
[.876g*Cs*8.518°C]+ [26.38kJ] = 0
7.462g°C * Cs = -26.38kJ
Cs= -3.535kJ/g°C
to get the molar enthalpy:
-3.535kJ/g°C * 114g/mol= 403.03kJ/mol* °C
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