Question

The compound cyclohexanol, C₆H₁₂O, is a good fuel. It is a liquid at ordinary temperatures. When the liquid is burned, the reaction involved is

2 C₆H₁₂O(ℓ) + 17 O₂(g)---->12 CO₂(g) + 12 H₂O(g)  

The standard enthalpy of formation of liquid cyclohexanol at 25 °C is -348.2 kJ mol^-1; other relevant enthalpy of formation values in kJ mol^-1 are:
  C₆H₁₂O(g) = -286.2 ; CO₂(g) = -393.5 ; H₂O(g) = -241.8

(a) Calculate the enthalpy change in the burning of 3.000 mol liquid cyclohexanol to form gaseous products at 25°C. State explicitly whether the reaction is endothermic or exothermic.

ΔH° =  kJ

_________endothermic/exothermic

(b) Would more or less heat be evolved if gaseous cyclohexanol were burned under the same conditions?

_____more/less

What is the standard enthalpy change for vaporizing 3.000 mol C₆H₁₂O(l) at 25°C?

ΔH° =  kJ

Calculate the enthalpy change in the burning of 3.000 mol gaseous cyclohexanol to form gaseous products at 25°C.

ΔH° =  kJ

Answer 1

For the given reaction,

(a) dHrxn = (12 x -393.5 + 12 x -241.8) - (2 x -348.2)

               = -6927.2 kJ/mol

So for 3 mol

dHrxn = 3 x -6927.2 = -20781.6 kJ is released

Reaction is exothermic

(b) If gaseous cyclohexanol is burnt under similar condition, more heat would be released.

More

(c) dHrxn for vaporizing 3 mol cyclohexanol(l)

dHrxn = 3 x (-286.2 - (-348.2) =+186 kJ

(d) dHrxn = 3((12 x -393.5 + 12 x -241.8) - (2 x -286.2))

               = -3744 kJ