Answer: [tex]5.15\times 10^3 kJ[/tex]
Explanation:-
As we know heat released will be same as heat absorbed by the calorimeter
[tex]q_{cal}=c_{cal}\times \Delta T[/tex]
[tex]q_{cal}[/tex] = Heat gained by bomb calorimeter
[tex]c_{cal}[/tex] =Heat capacity of bomb calorimeter=12.66 kJ/°C
Change in temperature = ΔT= [tex](28.78-22.36)^0C =6.42^0C[/tex]
[tex]q_{cal}=c_{cal}\times \Delta T=12.66kJ/^0C\times 6.42^0C =81.3kJ[/tex]
Let the heat released during reaction be q.
q = Heat released = 81.3 kJ
1.800 g of octane releases = 81.3 kJ of heat
Moles of octane =[tex]\frac{1.800g}{114g/mol}=0.01579mol[/tex]
0.01579 moles of octane releases 81.3 kJ of heat
1 mole of octane releases =[tex]\frac{81.3}{0.01579}\times 1=5.15\times 10^3 kJ[/tex] of heat
Thus [tex]\Delta H[/tex] for the combustion of one mole of octane is [tex]5.15\times 10^3 kJ[/tex]
