Tutored Practice Problem 13.5.1 Calculate enthalpy of sublimation from heats of

Question

Tutored Practice Problem 13.5.1 Calculate enthalpy of sublimation from heats of fusion and vaporization (a) The heat of fusion of propane (C,H) at its normal melting point of-188 'C is 3.5 /mol, while its heat of vaporization at its normal beiling point of 42'C is 190mo Use these data to cakculate the heat of sublimation for Cyl Is your answer precise or is it appeoximate? Close kJ/mol (b) A1 298 K, the standard heat of formation of CHg0 is-121.0J/mol whale the standand heat of Sormation of CyH(g) is-104.0 imol. Use this to calculate the heat of vaporization of propase LI mol Is your result for part (b) larger or smaller than the value given in the statement of the problem?D Show Approach

Explanation / Answer

For first question,C3H8

Cp,m(C3H8(l)) = 0.098kJ/mol K

Hfus(C3H8)=3.5 kJ/mol at -188ºC

Hvap(C3H8)= 19.0kJ/mol at -42ºC

The process we’re interested in is: C3H8(s), -188ºC C3H8(l), -188ºC C3H8(l), -42ºC C3H8(g), -42ºC

The total heat required is the sum of 3 enthalpy changes:

q = H= nHfus +n Cp,mT + nHvap

for n= 1, on putting value we get,

q= 3.5 + (188-42) *.098 +19.0 = 22.5 + 146*0.098 = 36.808kJ/mol

heat of sublimation for one molecule is 36.808kJ/mol

2) for second qestion,

The process is: C3H8(l) C3H8(g)

For 298.15 K, use heat of formation data at 298.15 K to obtain the enthalpy of the above reaction, which is equal to the enthalpy of vaporization:

Hvap= HfC3H8(g) - HfC3H8(l)

as given in table,

HfC3H8(l)= -121.0kJ/mol

HfC3H8(g)= -104.0kJ/mol

On putting value, we get

Hvap= -104.0 + 121.0 =17.0kJ/mol

so, heat of vaporisation is equal to 17.0kJ/mol

hope you get it what you want.

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