This document describes an experiment to measure the energy changes of neutralization reactions using a calorimeter. The experiment involves adding sodium hydroxide and hydrochloric acid to water in a calorimeter and measuring the temperature changes before and after the neutralization reaction using a thermometer. The temperature changes are then used to calculate the energy released by the neutralization of sodium hydroxide and hydrochloric acid to form sodium chloride and water.

1. INSTRUCTROR
DR.MD:ANWARUL KARIM
FACULTY OF CHM 116

2. NO NAME ID
2
1 Masum Billah
(Group Leader)
12103096
2 Md.Kyum Ahmed 12103098
3 Mariam jamil 12103080

4.  To measure, using a calorimeter, the energy
changes accompanying neutralization reactions.
 Learn the Kelvin temperature scale.
 Define heat capacity.

5. Calorimeter
Thermometer
HCL
NaOH
Funnel
 Pipette
 Buret
 Stand
5

6.  Obtain a calorimeter
and add it 100 ml of
distilled water.
• Secure a thermometer
to stand up in the
calorimeter, using ring
stand.
• Measure the initial
temperature (T1) of
water.

7.  Putting 4 gm NaOH and
storing them carefully.
 Measure the temperature
(T2) of this solution.
• Then have to 1 drop
phenolphthalein as an
indicator.

8.  Then putting HCl drop by
drop and stirring carefully
up to change the color of the
solution.
 After changing the color the
reaction will be finished.
•Then take the final
temperature (T3) of solution
and finally take the total
volume of the solution by
using measuring flask.

9. Acid Base Salt Water
Neutralisation is a reaction between acid and base to
produce salt and water.

10. HClaq + NaOHaq  NaClaq + H2O
acid + base  salt + water

11. 11
Result
Energy change = mc ΔT
In which,
m = the mass of the aqueous
reaction mixture
c = the specific heat capacity of
the aqueous reaction mixture
= the change in temperature
Qsolution = mc ΔT
= 91*4.184*(-13)
= -4.949 × 10^3 j
Here,
M=91 g
C=4.184j/g
=28-41
= -13
0c
0c
0c
0c
ΔT
ΔT

12. Result
0c
0c
Qcalorimeter = mc ΔT
=91*4.184*(-7)
= -2.665×10^3 j
Here,
M =91 g
C =4.184j/g
=21-28
= -70c
ΔT
- ΔT neutaralization = Qcalorimeter +Qsolution
= (- 2.665×10^3 j) + (-4.949 × 10^3j)
= -7.614×10^3 j