Calculate the change in pressure required to change the freezing point of water at 1°C and 0°C if heat of fusion of ice is 333.5 j/cm^3 and density of water is 0.998g/cm^3 and of ice is 0.9168g/cm^3?

I'm getting #-"7.53 atm"# to cause #DeltaT_"fus" = +"1 K"# from #0^@ "C"# to #1^@ "C"#, indicating the slope is NEGATIVE.

I assume you mean, calculate the change in pressure associated with a change in the freezing point of water from #0^@ "C"# to #1^@ "C"#.

I will forgo the derivation and present the Clapeyron Equation:

#(DeltaP)/(DeltaT) = (DeltabarS_(tr))/(DeltabarV_(tr)) = (DeltabarH_(tr))/(T_(tr)DeltabarV_"tr")#

where:

• #P# is pressure in #"atm"#
• #T_"tr"# is the normal phase transition temperature in #"K"#.
• #barS# is molar entropy in #"J/mol"cdot"K"#.
• #barV# is molar volume in #"L/mol"#.
• #barH# is molar enthalpy in #"J/mol"#.

First, let's evaluate the right-hand side. The molar volumes are:

#barV_"ice" = [(0.9168 cancel"g")/cancel("cm"^3) xx (1000 cancel("cm"^3))/("1 L") xx ("1 mol")/(18.015 cancel"g water")]^(-1) = "0.01965 L/mol"#

#barV_"water" = [(0.998 cancel"g")/cancel("cm"^3) xx (1000 cancel("cm"^3))/("1 L") xx ("1 mol")/(18.015 cancel"g water")]^(-1) = "0.01805 L/mol"#

Therefore,

#DeltabarV_("ice"->"water") = "0.01805 L/mol" - "0.01965 L/mol" = -"0.0016 L/mol"#

i.e. ice contracts when it melts. Hence, the right-hand side is:

#(DeltabarH_(tr))/(T_(tr)DeltabarV_"tr") = (333.5 cancel"J""/"cancel"g" xx (18.015 cancel"g")/(cancel"1 mol"))/("273.15 K" cdot (-0.0016 cancel"L""/"cancel"mol")) xx (cancel"1 L"cdot"atm")/(101.3cancel"J")#

#= -"7.53 atm/K"#

So, given the change in temperature is #"1 K"#,

#color(blue)(DeltaP) = DeltaT(-"7.53 atm/K") = color(blue)(-"7.53 atm")#