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EXPERIMENT 8 HEATS OF REACTION 4. In a neutralization reaction in a well-insulated calorimeter. 112.9 g of 1.2915 M NaOH were
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Time (s) Temperature (degree celcius)
0 20
5 24
10 27.5
15 27.2
20 26.9
25 26.6
30 26.2
35 25.9
40 25.6
45 25.3
50 25
55 24.6
60 24.3

Tmax = Maximum temperature = 27.5 degree celcius Temperature Temperature (degree celcius) 0 10 20 30 40 Time (second) 50 60 7ANSWER Q, a ) :

ANSWER Q, b ) :

Given:

1. Tmax = 27.5 °C

2. T initial = 24.3 °C

3. ΔT=T2T1 = (27.5 – 24.3) °C = 3.2 °C

4. The equation for the reaction is

NaOH + HCl → NaCl + HO

5. 112.9 g of 1.2915 M NaOH

6. 129.1 g of 1.1294 M HCl

Moles of HCl = n = 112.9 g HCl × 1.2915 mole HCl/1 kg HCl = 0.1458 mol HCl

7. Mass of solution = m = (112.9 + 129.1) g = 242 g

8. C = Negligible or zero

The quantity of heat involved in reaction (Exothermic) are:

qreaction + qsolution + qinstruments used = 0

∆H rxn = qreaction / moles of limiting reactant (here in this case HCl)

heat from neutralization + heat to warm solution + heat to warm calorimeter = 0

                q1                     +           q2                              +                  q3                   = 0

             nΔH   +   mcΔT                    +              CΔT = 0

0.1458 mol HCl × ΔH + 0.1458 g × 4.184 J·g¹°C¹ × 3.2 °C + (C = Negligible or zero)   = 0

0.1458 mol × ΔH + 1952.08= 0

0.1458 mol × ΔH = -1952.08J

∆H rxn = qreaction / moles of limiting reactant

ΔHrxn =−1952.08 /0.1458 mol = -13388 J/mol = -13.39 kJ/mol

ΔHrxn = -13388 J/mol = -13.39 kJ/mol of HCl

ANSWER Q, c ) :

Given:

1. Tmax = 27.5 °C

2. T initial = 24.3 °C

3. ΔT=T2T1 = (27.5 – 24.3) °C = 3.2 °C

4. The equation for the reaction is

NaOH + HCl → NaCl + HO

5. 112.9 g of 1.2915 M NaOH

6. 129.1 g of 1.1294 M HCl

Moles of NaOH = n = 129.1 g NaOH × 1.1294 mole NaOH /1 kg NaOH = 0.1458 mol HCl

7. Mass of solution = m = (112.9 + 129.1) g = 242 g

ΔHrxn =−1952.08 /0.1458 mol of NaOH = -13388 J/mol = -13.39 kJ/mol of NaOH

ΔHrxn = -13388 J/mol = -13.39 kJ/mol of NaOH

ANSWER Q, d ) :

Given:

1. Tmax = 27.5 °C

2. T initial = 24.3 °C

3. ΔT=T2T1 = (27.5 – 24.3) °C = 3.2 °C

4. The equation for the reaction is

NaOH + HCl → NaCl + HO

5. 112.9 g of 1.2915 M NaOH

6. 129.1 g of 1.1294 M HCl

Moles of NaOH = n = 129.1 g NaOH × 1.1294 mole NaOH /1 kg NaOH = 0.1458 mol HCl

7. Mass of solution = m = (112.9 + 129.1) g = 242 g

8. heat to equipments used = heat to warm calorimeter + heat to warm thermometer + heat to warm alumminium stirrer

9. heat to equipments used = (s = 1.131 J/g °C x 30 g) + (s = 0.840 J/g °C x 20 g) = (s = 0.900 J/g °C x 10 g) = 59.73 J/°C

heat from neutralization + heat to warm solution + heat to warm calorimeter + heat to warm thermometer + heat to warm alumminium stirrer = 0

heat from neutralization + heat to warm solution + heat to warm calorimeter = 0

                q1     +           q2   +                  q3                           = 0

             nΔH +      mcΔT   +              CΔT                    = 0

0.1458 mol HCl × ΔH + 0.1458 g × 4.184 J·g¹°C¹ × 3.2 °C + 59.73 J°C¹ × 3.2 °C = 0

0.1458 mol × ΔH + 373.11= 0

0.1458 mol × ΔH = -373.11J

∆H rxn = qreaction / moles of limiting reactant

ΔHrxn =−373.11/0.1458 mol of NaOH = -2559.05 J/mol = -2.256 kJ/mol of NaOH

ΔHrxn = -2559.05 J/mol = -2.256 kJ/mol of NaOH

ANSWER Q, e ) :

ΔHrxn = literature value of Neutralization reaction is -55.9 kJ/mol which is far away from

ΔHrxn = -13388 J/mol = -13.39 kJ/mol of NaOH . When we consider value for heat of instruments (Calorimeter, Thermometer & Stirrer) large difference is there as they absorbed all raction heat and changed ΔHrxn value. Although system is closed but evolved heat is transfered into closed environment components till equilibrium.

ΔHrxn = -2559.05 J/mol = -2.256 kJ/mol of NaOH

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