What will be the final temp. of the solution in a coffee cip calorimeter if a 50.00mL sample of 0.250 M HCL is added to a 50.00 mL sample of 0.250. The initial temp. is 19.50 °C and the Delta H is -57.2 Kj/mol NaOH

Question

assume density of solution is 1.00 g/mL and specific heat of the solution is 4.18 J/g°c

MY TEACHER DID NOT GIVE US AN EQUATION
I KNOW
Cs is 4.18 j/g°c
Mass is 100g
q idk
I do not know T2 but the T1 is 19.50

Answer 1

You didn't tell us 0.250 WHAT.

Answer 2

First of all, since it's 0.250 M and 50 mL for both reactants, you don't really need to know the equation since both are limiting reactants, but you put that delta H is 57.2 Kj/mol of NaOH. So I'm guessing that NaOH is the other reactant, anyways HCl plus NaOH is a favorite reaction among chem teachers.

First you need to convert 57.2 kJ/mol NaOH to just joules. So since you have 50 g of NaOH*(1mol/40g)*(57.2KJ/1mol NaOH)*(1000J/1kJ) = 71500 J.

Use Q=CmdeltaT, C=4.18, m=100, so then 71500J/(4.18*100)=171.05, which is the change in T. We know that Ti is 19.5 so Tf-19.5=171.05, and that means Tf=190.55 degrees Celsius.

Answer 3

Also, I just noticed, the delta H for NaOH is for the standard enthalpy of formation, which is different from the enthalpy change of the reaction.

If 57.2 was meant to be the enthalpy change of the reaction, then just convert to J, and substitute 57200J instead of 71500J.

**BTW if you have the answer to the question, that would help me understand how to do the question better.

Answer 4

You have how many mols? That's M x L = 0.25 x 0.05 - 0.0125
The reaction releases 57,200 J/mol so in this reaction it releases 57,200 x 0.0125 = 715 J. Now, as Victor has said, substitute that into Q = mcdT
715 = 100 x 4.18 x (Tfinal - 19.5) and solve for Tfinal. I obtained about 21 C but that's just a close estimate. You need to work through it.