Question

When a 3.80-g sample of liquid octane (C8H18) is burned in a bomb calorimeter, the temperature of the calorimeter rises by 26.5 ∘C. The heat capacity of the calorimeter, measured in a separate experiment, is 6.21 kJ/∘C . You may want to reference (Page 265) Section 6.5 while completing this problem. Part A Determine ΔE for octane combustion in units of kJ/mol octane. Express your answer using three significant figures.

Answer 1

Combustion reaction at 298 K of octane

2 C₈H₁₈ (l)  + 25 O₂ (g)  16 CO₂ (g) + 18 H₂O (l)

Now, heat change in calorimeter( Qcal)  = C_m × temperature change

{Cm is heat capacity of calorimeter. }

Then, Q_cal = 6.21× 26.5 = 164.565 KJ.

Now, Q_combustion = - Q_cal

then Q_combustion = - 164.565 KJ

Moles of octane = mass ÷ molar mass = 3.80÷114 = 0.03247

Then, = - 164.565÷0.03247 = - 5068.21 KJ/mole

Now,

= + RT. (1)

At , T =298 K

R = 8.314 ×10^-3 KJ/mole K

= total moles of gaseous products - total moles of gaseous reactants

n =( 16 - 25)

n = - 9

Then, putting the above values in Eq.1

H = E - 9× (8.314×298)×10^-3 (KJ/mole)

Or, - 5068.21 = E - (9×8.314×298)×10^-3

Or, E = (- 5068.21) + (9×8.314×298)×10^-3

= ( -5068.21 + 22.29 ) KJ/mole

Or, E = - 5045.92 KJ /mole

Or, E = - 5.05×10³ KJ/mole ( upto 3 significant figures)