Question

PartA

Calculate the amount of heat required to completely sublime 97.0 g of solid dry ice (CO2) at its sublimation temperature. The heat of sublimation for carbon dioxide is 32.3 kJ/mol.

Part B

How much heat is evolved in converting 1.00 mol of steam at 155.0 ∘C to ice at -55.0 ∘C? The heat capacity of steam is 2.01 J/(g⋅∘C) and of ice is 2.09 J/(g⋅∘C).

Answer 1

Answer:::::

PART-A

Amount of CO₂ =97.0 g

molar mass of CO₂ =44 g/mol

no.of moles of CO2 = 97.0 g  mass of CO2

= 97.0/44 mol

= 2.205 mol

according to question ,

for 1 mole of CO2 need energy of 32.3 kJ

simillary 2.202 mole of CO2 need energy of = 32.3 * 2.205 kJ

=71.21 kj

so the total heat required to completely is 71.21 KJ

PART -B::

  capacity of steam is 2.01 J/(g⋅∘C)

capacity of   ice is 2.09 J/(g⋅∘C).

heat is evolved in converting of steam at 155.0 ∘C to ice at -55.0 ∘C and its 1 mole

(1.00 mol H2O) x (18.01532 g H2O/mol) = 18.015 g H2O

and  from cooling the steam to its boiling point

=(2.01 J/(g⋅°C)) x (18.015 g) x (155.0 - 100.0)°C = 1992 J..........(1)

from condensing the steam

=(2257 J/g) x (18.015 g) = 40660 J

from cooling the condensed steam to its freezing point

(4.184 J/g·°C) x (18.015 g) x (100 - 0) °C) = 7537 J

from freezing the water

=(2.09 J/(g⋅°C) x (18.015 g) x (0 - (-55.0))°C = 2070.9  j =2071 j

total= 1992 J + 40660 J + 7537 J + 6010 J + 2071 J = 57893 = 58270 J

=58.27 KJ

58.27 KJ    heat is evolved in converting 1.00 mol of steam at 155.0 ∘C to ice at -55.0 ∘C.