Use the References to access important values if needed for

Use the References to access important values if needed for this question The heat of fusion of heptane (CH16) at its normal melting point of -91°C is 14.0 kJ/mol, while its heat of vaporization at its normal boiling point of 98°C is 31.9 kJ/mol. (a) Use these data to calculate the heat of sublimation for CH16 Is your answer precise or is it approximate?\' B kJ/mol (b) At 298 K, the standard heat of formation of C,H1-224.2 kJ/mol while the standard heat of formation of C7H16(g) is-187.6 mol. Use this to calculate the heat of vaporization of heptane. Is your result for part (b) larger or smaller than the value given in the statement of the problem? kJ/mol Submit Answer

Solution

a) DHsub = DHfus + DHvap

         = 14.0+31.9

         = 45.9 kj/mol

it is precise value.

b) C7H16(l) <----> C7H16(g)

   DHrxn = (-187.6)-(-224.2)= 36.6 kj/mol