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Question
Calculate the enthalpy change of fusion of 1 mol of water at 5oC to ice at −5oC.
Given : ΔfusH=6.03 kJmol−1 at 0oC
Cp[H2O(l)]=75.3Jmol−1K−1
Cp[H2O(s)]=36.8Jmol−1K−1
Calculate the enthalpy change of fusion of 1 mol of water at 5oC to ice at −5oC.
Given : ΔfusH=6.03 kJmol−1 at 0oC
Cp[H2O(l)]=75.3Jmol−1K−1
Cp[H2O(s)]=36.8Jmol−1K−1
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Solution
The correct option is D −6.59 kJmol−1
The change of 5℃ water to −5℃ ice takes place in three steps.1) 5℃ water to 0℃ water.2) 0℃ water to 0℃ ice.3) 0℃ ice to −5℃ ice.
Enthalpy change for step 1 isH1=Cp[H2O(l)]×ΔTH1=75.3×(−5)H1=−376.5Jmol−1=−0.376kJmol−1
Enthalpy change for step 2 H2=−ΔfusHH2=−6.03kJmol−1
Enthalpy change for step 3 is H3=Cp[H2O(s)]×ΔTH3=36.8×(−5)H3=−184Jmol−1=−0.184kJmol−1
So, total enthalpy change=H1+H2+H3=−0.376−6.03−0.184=−6.59kJmol−1So, correct answer is option A.
The change of 5℃ water to −5℃ ice takes place in three steps.
1) 5℃ water to 0℃ water.
2) 0℃ water to 0℃ ice.
3) 0℃ ice to −5℃ ice.
Enthalpy change for step 1 is
H1=Cp[H2O(l)]×ΔTH1=75.3×(−5)H1=−376.5Jmol−1=−0.376kJmol−1
Enthalpy change for step 2
H2=−ΔfusHH2=−6.03kJmol−1
Enthalpy change for step 3 is
H3=Cp[H2O(s)]×ΔTH3=36.8×(−5)H3=−184Jmol−1=−0.184kJmol−1
So, total enthalpy change
=H1+H2+H3=−0.376−6.03−0.184=−6.59kJmol−1
So, correct answer is option A.
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