A) Hexanol (C6H14O) is a liquid alcohol similar to the ones used in the experime

Question

A) Hexanol (C6H14O) is a liquid alcohol similar to the ones used in the experiment. In another bomb calorimetry experiment 0.8278 g of hexanol is burned and the temperature of the calorimeter increased from 16.834?C to 19.203?C. The heat capacity of the calorimeter is 13.52 kJ ?C-1 . Calculate the enthalpy of formation of hexanol in kJ/mol C6H14O. Compare this to the accepted value of -377.5 kJ mol-1 .

B) What kind of changes in enthalpies of combustion did you observe as the chain length of the alcohols increased? What was the approximate change for each CH2 unit added?

Explanation / Answer

A)
No of moles of Hexanol = 0.8278/102.174 = 0.0081 mole.

DT = 19.203-16.834 = 2.369 C

Heat capacity of calorimeter = 13.52 kj/C

Heat released = 13.52*2.369 = 32.029 kj.

0.0081 mole of Hexanol = - 32.029 kj

1 mole of Hexanol = -32.029*1/0.0081 = -3954.2 kj/mol

it is approximately tentimes higher than accepted value.

if the mass of Hexanol taken is 8.278 grams

then its approximately equals to accepted value.

B)
By increasing the carbon chain enthalpies increases.

By increasing the each CH2 Enthalpy increased by approximately +623.57 kj

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