Question

A 1.800 g sample of octane, C8H18, is burned a calorimeter whose total heat capacity is 12.66 kJ/°C. If the temperature of the calorimeter increased from 22.36 °C to 28.78 °C, then what is the ΔH for the combustion of one mole of octane? Do not add the unit in the answer.

Answer 1

Answer:
`5.15* 10^3 kJ`

Explanation:-

As we know heat released will be same as heat absorbed by the calorimeter

`q_(cal)=c_(cal)* \Delta T`

`q_(cal)` = Heat gained by bomb calorimeter

`c_(cal)` =Heat capacity of bomb calorimeter=12.66 kJ/°C

Change in temperature = ΔT=
`(28.78-22.36)^0C =6.42^0C`

`q_(cal)=c_(cal)* \Delta T=12.66kJ/^0C* 6.42^0C =81.3kJ`

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 =
`(1.800g)/(114g/mol)=0.01579mol`

0.01579 moles of octane releases 81.3 kJ of heat

1 mole of octane releases =
`(81.3)/(0.01579)* 1=5.15* 10^3 kJ` of heat

Thus
`\Delta H` for the combustion of one mole of octane is
`5.15* 10^3 kJ`