Question

EXPERIMENT 8 HEATS OF REACTION 4. In a neutralization reaction in a well-insulated calorimeter. 112.9 g of 1.2915 M NaOH were added to 129.1 g of 1.1294 M HCI. The following data were obtained. Assume the density of each solution is 1.00 g/mL. Time (8) 0 5 10 15 20 25 30 Temperature (°C) 20.00 24.00 27.50 27.20 26.90 26.60 26.20 25.90 25.60 25.30 25.00 24.60 24.30 35 40 45 SO 55 60 a. Using the graphing technique outlined in the discussion (see Figure 8.2), determine the maximum temperature graphically on graph paper. Clearly mark the maximum temperature on the graph. b. Assuming the calorimeter, thermometer, and stirrer have a negligible impact on the AHans calculate the AH (in kJ) for this reaction. C. Calculate the AHxn per mole of NaOH reacted. d. If the Styrofoam calorimeter (s = 1.131 J/g °C) weighs 30.0 g, the glass thermometer (s = 0.840 J/g °C) weighs 20.0 g, and the aluminum stirrer (s = 0.900 J/g °C) weighs 10.0 g. calculate the AH per mole of NaOH reacted. The literature value of this reaction is -55.9 kJ/mole. Qualitatively compare the AH.... from (c) and (d) and comment on the effects the calorimeter, thermometer, and stirrer have on the AHex and if they can be considered negligible in the calculations.

Answer 1

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

ANSWER Q, a ) :

Tmax = Maximum temperature = 27.5 degree celcius Temperature Temperature (degree celcius) 0 10 20 30 40 Time (second) 50 60 70

ANSWER Q, b ) :

Given:

1. Tmax = 27.5 °C

2. T initial = 24.3 °C

3. ΔT=T2–T1 = (27.5 – 24.3) °C = 3.2 °C

4. The equation for the reaction is

NaOH + HCl → NaCl + H₂O

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·g⁻¹°C⁻¹ × 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=T2–T1 = (27.5 – 24.3) °C = 3.2 °C

4. The equation for the reaction is

NaOH + HCl → NaCl + H₂O

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=T2–T1 = (27.5 – 24.3) °C = 3.2 °C

4. The equation for the reaction is

NaOH + HCl → NaCl + H₂O

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·g⁻¹°C⁻¹ × 3.2 °C + 59.73 J°C⁻¹ × 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