Question

please show the answer in steps.

1)What is the change in entropy when a) 100 g of ice is melted at 0 ℃ b) 100 g of water is vaporized to 100 ℃

Use the values for each step to calculate the change in entropy when 100 g of ice at 0 ℃ is transformed to water vapor at 100 at ℃ . Suppose that the changes are brought about by a heater that supplies at a constant rate and sketch a graph showing a) the change in temperature of the system b) the enthalpy of the system, c) the entropy of the system as a function of time.

2)The average of the enthalpy of vaporization of propanone, C₃H₆O over the temperature range of 280 to 340 K is 30.2 kJ/mol. The vapor pressure is 30.600 kPa at 298.15 K. Estimate the normal boiling temperature of propanone.

Answer 1

1). (a). Mass of ice = 100 g

Temperature = 0 oC = 273.15 K

Heat required to melt the ice, q = m *

= 100 g * 333.55 J/g

= 33355 J

Entropy change = q / T

= 33355 / 273.15

= 122.13 J/K

(b). Mass of ice = 100 g

Moles of ice = 100 / 18 = 5.55

Final temperature = 100 oC = 373.15

Entropy change = Entropy change for melting ice at 0 oC () + Entropy change for increasing temperature from 0 oC to 100 oC () + Entropy change for vaporising ice at 100 oC ()

We have,

= 122.17 J/K   (Calculated in part a)

= n Cp,m ln (Tf / Ti) = 5.55 * 0.07529 * ln (373.15 / 273.15)

= 130.38 J/K

= n / T = 5.55 * 40.7 / 373.15 = 605.44 J/K

Now,

= 122.13 + 130.38 + 605.44

= 857.95 J/K