Question

# 1) Calculate the amount of energy in kilojoules needed to change 207 g of water ice...

1)

Calculate the amount of energy in kilojoules needed to change 207 g of water ice at −10 ∘C to steam at 125 ∘C. The following constants may be useful:

Cm (ice)=36.57 J/(mol⋅∘C)

Cm (water)=75.40 J/(mol⋅∘C)

Cm (steam)=36.04 J/(mol⋅∘C)

ΔHfus=+6.01 kJ/mol

ΔHvap=+40.67 kJ/mol

2) Given the values of ΔH∘rxn, ΔS∘rxn, and T below, determine ΔSuniv.

Part A

ΔH∘rxn= 90 kJ , ΔSrxn= 144 J/K , T= 304 K

Part B

ΔH∘rxn= 90 kJ , ΔSrxn= 144 J/K , T= 753 K

Part C

ΔH∘rxn= 90 kJ , ΔSrxn=− 144 J/K , T= 304 K

Part D

ΔH∘rxn=− 90 kJ , ΔSrxn= 144 J/K , T= 409 K

Part E

Predict whether or not the reaction in part A will be spontaneous.

Predict whether or not the reaction in part A will be spontaneous.

spontaneous

nonspontaneous

Part F

Predict whether or not the reaction in part B will be spontaneous.

Predict whether or not the reaction in part B will be spontaneous.

spontaneous

nonspontaneous

Part G

Predict whether or not the reaction in part C will be spontaneous.

Predict whether or not the reaction in part C will be spontaneous.

spontaneous

nonspontaneous

Part H

Predict whether or not the reaction in part D will be spontaneous.

Predict whether or not the reaction in part D will be spontaneous.

 spontaneous nonspontaneous

Q = heat change for conversion of ice at -15 oC to ice at 0 oC + heat change for
conversion of ice at 0oC to water at 0oC + heat change for conversion of water at 0oC to water at 100 oC +heat change for conversion of water at 100 oC to vapour at 100 oC+ heat change for conversion of vapour at 100 oC to vapour at 146 oC

Amount of heat absorbed ,

Q = mcdt + mL + mc'dt + mL' + mc"dt"
= m(cdt + L + c'dt' + L' + c"dt" )
Where
m = mass of water = 207g
c” = Specific heat of steam = 2.1 J/g degree C
c' = Specific heat of water = 4.186 J/g degree C
c = Specific heat of ice= 2.09 J/g degree C
L’ = Heat of Vaporization of water = 2260 J/g
L= Heat of fusion of ice = 334.9 J/g
dt’’ = 125-100 = 25oC
dt' = 100 -0 =100 oC
dt = 0-(-10)=10oC
Plug the values we get

Q = m(cdt + L + c'dt '+ L' + c"dt" )

=638.9x103J

=638.(9kJ

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