1. Consider the following data from a Heat of Vaporization experiment: ( the vol

Question

1.         Consider the following data from a Heat of Vaporization experiment: (the volume correction factor has not been applied – this is the raw data)

Temp              Volume                       Barometric Pressure = 758.3 torr

72.2°C             9.02 mL

65.4°C             7.78 mL

59.8°C             6.98 mL

55.1°C             6.54 mL

50.5°C             6.22 mL

2.5°C             4.65 mL (Data Point to calculate the moles of air)

a.   Calculate the moles of air trapped in the cylinder. (Use the one low temperature point and assume Barometric pressure equals the pressure of only air since PH2O will be negligible.)

b.   Calculate the vapor pressure of water, Pwater, at 55.1°C. (First calculate Pair at 55.1°C using the moles of air. Then PH2O = Barometric Pressure – Pair)

c.   Assuming the slope of the line resulting from a Claussius-Clapeyron plot (LnV.P. vs 1/T) of the data to be -5285 K, what is the value of DHvap in kJ/mol.

2.         The following is data from a heat of fusion experiment:

Mass of calorimeter = 4.25 g

Mass of calorimeter + water at 35.7°C = 78.52 g

Mass of calorimeter and water at 8.3°C = 103.03 g

Heat Capacity of Calorimeter Apparatus = 18.5 J/°C

a.   Calculate the qcal in Joules. (should be negative since T is negative)

b.   Calculate the qwarm water in Joules. (should be negative for the approx. 75 grams cooling)

c.   Calculate the qcold water in Joules. (should be positive for the approx. 25 grams warming)

d.   Calculate the qfusion in Joules. (should be positive)

e.   Calculate the moles of ice that melted.

f.    Calculate the DHfusion in kJ/mol.

Explanation / Answer

1.a T = 2.5 C = 275.5 k

P = 758.3 torr = 0.9977 atm

V = 4.65 ml

R = 0.0821 l.atm.K-1.mol-1

n = No of moles of gas = 0.9977*0.00465 /0.0821*275.5

   = 0.0002 mol

b.

Pair = nRT/v = 0.0002*0.0821*328.1/0.00654
    
      = 0.824 atm

PH2O = Barometric Pressure – Pair
  
     = 0.9977 - 0.824 = 0.1737 atm

c. slope = -DHvap/R = -5285

DHvap = 5285*8.314

      = 43.94 kj/mol

2.

a. qcal = Cp*DT = 18.5*(8.3-35.7) = -506.9 joule.

b. qwarmwater = m*s*DT = (78.52-4.5)*4.18*(8.3-35.7)

       = -8477.66 joule

c. q cold water = (103.03-78.52)*4.18*(35.7-8.3)

   = 2807.18 joule.

d. qfusion = qwarmwater - q cold water

            = 8477.66 - 2807.18 = 5670.48 joule = 5.67 kj

e. No of moles of ice melted = 24.51/18 = 1.36 mole

f. DH fusion = 5.67/1.36 = 4.17 kj/mol

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