A student determined the calorimeter constant for a calorimeter and found from a

Question

A student determined the calorimeter constant for a calorimeter and found from a plot of the time-temperature data that Tinitial of the 100.0 g sample of hot water was 30.10oC and of the 100.0 g sample of cold water was 20.72oC and the Tfinal of the mixture was 25.05oC.

Using the data provided in the excel file, show all of your work for the following calculations:

a.) heat transferred from hot water (J)

b.) heat transferred to cool water (J)

c.) calorimeter constant (J/K or J/oC)

A student reacted 100.0 mL of 0.9800 M HCl with 100.0 mL of 0.9900 M NH3. The density of the reaction mixture was 1.02 g/mL and the heat capacity was 4.016 J/g K. Calculate the enthalpy of neutralization by plotting and using the data shown below.

Time(min) Temp(oC)

0.0 23.25

0.5 23.27

1.0 23.28

1.5 23.30

2.0 23.30

3.0 23.35

4.0 23.44

4.5 23.47

mix ---------

5.5 28.75

6.0 28.50

7.0 28.55

8.0 28.48

9.0 28.32

10.0 28.25

11.0 28.20

12.0 28.05

13.0 27.96

14.0 27.80

15.0 27.75

Using the data provided in the excel file, show all of your work for the following calculations:

a.) mean temperature of unmixed reagents (oC)

b.) T from graph (oC)

c.) q absorbed by reaction mixture (J)

d.) q absorbed by calorimeter, stirrer, and thermometer (J)

e.) q total absorbed (J)

f.) q total released (J)

g.) calculation to show limiting reagent

h.) deltaH neutralization for the reaction (kJ/mole of acid)

Explanation / Answer

Experiment : Mixing hot and cold water

a) heat transferred from hot water = mCpdT

                                                       = 100 x 4.18 x (30.10 - 25.05) = 2110.9 J

b) heat transferred to cold water = mCpdT

                                                    = 100 x 4.18 x (25.05 - 20.72) = 1809.94 J

c) calorimeter constant = (2110.9 -1809.94)/(25.05 - 20.72) = 69.50 J/oC

Enthalpy of neutralization

a) mean temperature of unmixed reagent = 23.33 oC

b) dT = 28.75 - 23.33 = 5.42 oC

c) Total mass of solution = 200 ml x 1.02 g/ml = 204 g

q absorbed by reaction mixture = 204 x 4.18 x 5.42 = 4621.74 J

d) q absorbed by calorimeter = 4.016 x 5.42 = 21.77 J

e) total q absorbed = 4643.51 J

g) limiting reagent,

moles of NH3 = 0.99 M x 100 ml = 99 mmol

moles HCl = 0.98 x 100 ml = 98 mmol

since moles of HCl is lower, this is the limiting reagent

h) dH neutralization = 4643.51/0.098 mol = 46.90 kJ/mol

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