The two were mixed together at exactly 400 min in

The two were mixed together at exactly 4.00 min in calorimeter and the temperature was obtained and
recorded for a duration of 4.50 min with at least 8 readings. After collecting the readings, a graph was
created to obtain the final temperature (Tf) of the reaction mixture.
To obtain the final temperature of the reaction, the graph was used. The linearly declining temperature
points after the maximum temp was used to capture the equation of the line for the selected data. The
equation, Temp=mt+b was used to calculate what the temperature should have been at the time of
mixing (4.00 min).
The heat of neutralization per mole of water formed- ∆ H(rxn/dissolution)=5.7KJ/mol
For trial #2 ,
HN O
3
(
aq
)
+ NaOH (aq) → NaNO3 (aq) + H2O (l) (Heat of Neutralization)
Repeat the procedure for trials #1 using 1.0 M HNO3 instead of HCL.
The heat of neutralization per mole of water formed- ∆ H(rxn/dissolution)=(-66.96 kj/mol)

3
For trial #3, NaOH(s) → NaOH(aq) (Heat of Solution)
50.0 ml of deionized water was carefully measured and put into the calorimeter to obtain initial
temperature readings at 30 -second intervals for 3 minutes. After turning on the stirrer bar, at the 4rth
minute the carefully measured 2g of solid NaOH was added and continued to collect the temperatures of
the reactants at 9 minutes.
A graph was used just like trail #1 and #2 to obtain the final temperature of the solution from the
equation. The heat of solution per mole of NaOH was calculated to be (-0.04936 kj/mol)