heat of neutralization

13
INSTRUCTROR DR.MD:ANWARUL KARIM FACULTY OF CHM 116

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Page 1: HEAT OF NEUTRALIZATION

INSTRUCTROR

DR.MD:ANWARUL KARIM

FACULTY OF CHM 116

Page 2: HEAT OF NEUTRALIZATION

2

NO NAME ID

1 Masum Billah

(Group Leader)

12103096

2 Md.Kyum Ahmed 12103098

3 Mariam jamil 12103080

Page 3: HEAT OF NEUTRALIZATION
Page 4: HEAT OF NEUTRALIZATION

To measure, using a calorimeter, the energy

changes accompanying neutralization reactions.

Learn the Kelvin temperature scale.

Define heat capacity.

Page 5: HEAT OF NEUTRALIZATION

Calorimeter

Thermometer

HCL

NaOH

Funnel

Pipette

Buret

Stand

5

Page 6: HEAT OF NEUTRALIZATION

Obtain a calorimeter

and add it 100 ml of

distilled water.

• Measure the initial

temperature (T1) ofwater.

• Secure a thermometer

to stand up in the

calorimeter, using ring

stand.

Page 7: HEAT OF NEUTRALIZATION

Putting 4 gm NaOH and

storing them carefully.

Measure the temperature

(T2) of this solution.

• Then have to 1 drop

phenolphthalein as an

indicator.

Page 8: HEAT OF NEUTRALIZATION

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.

Page 9: HEAT OF NEUTRALIZATION

Neutralisation is a reaction between acid and base to

produce salt and water.

Acid Base Salt Water

Page 10: HEAT OF NEUTRALIZATION

HClaq + NaOHaq NaClaq + H2O

acid + base salt + water

Page 11: HEAT OF NEUTRALIZATION

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

0c0c

0c

ΔT

ΔT

Page 12: HEAT OF NEUTRALIZATION

Result

0c0c

Qcalorimeter = mc ΔT

=91*4.184*(-7)

= -2.665×10^3 j

Here,

M =91 g

C =4.184j/g

=21-28

= -7 0c

- ΔT neutaralization = Qcalorimeter + Qsolution

= (- 2.665×10^3 j) + (-4.949 × 10^3j)

= -7.614×10^3 j

ΔT

Page 13: HEAT OF NEUTRALIZATION