I need to find the final temurature of a solution. Starting solutions are combined in a perfect calorimeter. The initial temperature is given and ΔH is given.

Question

The density of each solution is given as 1.00 g/m

The enthalpy of neutralization for the reaction of HBr with NaOH is given below.

HBr (aq) + NaOH (aq) → H₂O (l) + NaBr (aq)           ΔH = -57.3 kJ

100.5 ml of 0.420 M HBr is combined with 100.5 ml of 0.42 M NaOH in a perfect calorimeter. Both of the starting solutions were initially at a temperature of 30.44 °C. The density of each solution is 1.00 g/ml.

 

I am supposed to find the final temperature of the solution in the calorimeter in celcius and honestly I don't even know where to start.

Answer 1

The given delta H (-57.3 kJ) is for 1 mole HBr + 1 mole NaOH

Calculate moles HBr (and NaOH) and then use that value and the value given for 1 mole to determine heat liberated (q). Then to find final temperature use q = mCdeltaT where q is the heat you calculated above, m is the mass which will be 100.5 g + 100.5 g (201 g ignoring mass of NaBr and H2O formed), and C = 4.184 J/g/deg. Solve for deltaT and add that value to 30.44 to get final temp.