APPARATUS REQUIRED:
1. Thermometer
2. Coffee cup
3. Beaker
CHEMICAL REQUIRED:
1. 1N NAOH
2. 1N HCI
3. Hotwater and cold water
(70).webp)
THEORY
To determine water equivalent of colorimeter (coffee cup) is the quantity of heat required to rise the temperature of cup by x; it is measured based on the principle.
Mn•S(dt)n= Mc S(dt)c + Mcc•Scc (dt)cc
Let, [Mcc•Scc= w]
where,
W = mcc- Scc = mass of coffeecup and xspecific heat capacity.
S = Specific heat capacity of water.
Therefore, W = Mn•S(dt)n - Mc•S(dt)c ÷ (dt)cc
To Determine the heat of neutralization of strong acid and strong base,
Zonic thermochemical aqueous solution for a neutralization revn H+ +OH- = H2O,ΔH = -Q
tempt of aqueous solution and calorimeter. Therefore,
the heat of neutralization is determined by
finding out the quantity of heat gained by
the aqueous solution and calorimeter (C.C.)
containing by 1gm equ. of acid and 1 gm equ. of base.
PROCEDURE
To determine water equivalent.
2.50 ml of hot water at about 60° to 65° is teken in Coffee cup
3. The temps of both cold and hot water were
recorded just before their mixing.
4. Hot water was added quickly and carefully
in the Coffee Cup. containing cold water. The mixture was stirred well and the final temp is recorded.
5. water equivalent of calorimeter was then calculated.
To determine heat of neutralization.
2. Somey 1N HCl (Aq) was taken in beaker and
the temp the solution also noted. If the tempt. of acid and bace before minive were found to be slightly different, the mean these two tempt. was noted.
3. 50 ml of HCL was poured quickly into the beaker.Containing NAOH solution and a thermometer.The rise in temp was noted at every 5 to lose until the constant tempt was reached. Then the time temp curve was plotted in a graph and the final tempt of the rest mixture was determine
4. The water eq. of calorimeter was taken and the heat of neutralization was then calculated.
OBSERVATION
Mass of hot water (Mn) = 50gMass of cold water (Mc) = 50g
Specific heat of water (S) = 4.2 Jlg°c
Temp of hot water (Tn) = 60°c
Tempt. of cold water (tc) = 27°c
After mixing, final temp" of mix. (tp): 44°c
Decrease of temp of hot water (dt):tn-tt(60-44)=16°C
Increase the temp of cold water (dt):
tt-tc = (14-27) = 17°c
To determine heat of neutralization,
Tempt. of NaOH solution (tb) = 27°C
Temp of HCL solution (ta): 27°c
Temperature of mix solution (ts): 32°c
Mass of NaOH solution (mb): 50g
Mass of HCL Solution (ma): 50g
Mass of mix Solution (Ms): 100g
CALCULATION
To determine water equivalent
Heat loss by hot water = Heat gain by cold water + Heat gain by coffee cup
Mn s (dt)h = Mcc Scc(dt)cc
W = Mn S(dt)h - Mc S(dt)c ÷ (dt)cc
= 50×4.2×16 - 50×4.2×17 ÷ (44-27)
= 12.35
To determine heat neutralization.
Heat produced by neutralization (Q) =
(Heat gained by solution) + Heat gained by coffee cup
Q =Ms S(dt)s + Mcc Scc (dt)cc
= Ms S(dt)s + W(dt)cc
= 100×4.2×(32-27) + 12.35X17
= -2309.95 Cal
Again,
100 ml of 0.5N solution = -2309.95
1000 ml of 0.5N solution = -2309.95x10
1000 ml of 1N solution = -2309.95x10x2
= -46,199 Cal.
= -46.19 kcal
Result
The heat of neutralization of HCL And NAOH is - 46.19 kcal and water equivalent of coffee cup is 12.35
(70).webp)
.jpeg)
(70).webp&description=TO DETERMINE THE HEAT OF NEUTRALIZATION OF STRONG ACID- STRONG BASE REACTION){kind=link}
0 Comments