Posted by Emma on Monday, November 12, 2012 at 11:05pm.
The process is the same as the NaOH +HCl problem we did below. The only difference is Ccal is given and you calculate dH in kJ/mol.
I don't know the heat capacity of NH4Cl or the density of it, can you please help me figure those out.
I think you are supposed to assume the heat capacity of the NH4Cl solution is the same as that of water which is 4.184 J/g*C.
I think you are to assume that the density of the NH4Cl solution is the same as that of water or 1.00 g/mL which is the usual quoted value.
I was surprised that the NaOH/HCl problem used 3.91 for the solution instead of 4.184. Most of these problems assume the solutions to be the same as water and the density to be the same as water.
Thankyou Dr. Bob and can you please help me on one more question if this reaction goes to completion what would the final concentration of NH4Cl in the calorimeter. I know how to figure it out if its not in the calorimeter but I am wondering if the calculations are different when its inside a calorimeter.
I believe you are over thinking this. mols NH4Cl formed is the same whether we do it in a calorimeter or a bucket. mols mol HCl = M x L = 2.00M x 0.050L = 0100 mol. HCl is the limiting reagent (and it will go to completion) so this reaction produces 0.100 mol NH4Cl.
How much heat is released.
q, that letter again, = [mass H2O x specific heat H2O x delta T] + [Ccal x delta T] = 0
q = [100 x 4.184 x 9.28] + [28.9 x 9.28]= 0
Calculate q which of course is delta H for the reaction. This is the heat produced by the reaction.
You want delta H in kJ/mol.
Take your number which is in joules, convert to kJ, then dH/0.1 = dH in kJ/mol.
kik