Question

The combustion of 1.771 g of propanol ( C 3 H 7 OH ) increases the temperature of a bomb calorimeter from 298.00 K to 302.34 K . The heat capacity of the bomb calorimeter is 13.70 kJ/K . Determine Δ H for the combustion of propanol to carbon dioxide gas and liquid water.

Answer 1

Answer: ΔH for the combustion of propanol to carbon dioxide gas and liquid water is 1980 kJ

Step-by-step explanation:

The quantity of heat required to raise the temperature of a substance by one degree Celsius is called the specific heat capacity.

`Q=C* \Delta T`

Q = Heat absorbed by calorimeter =?

C = heat capacity of calorimeter = 13.70 kJ/K

Initial temperature of the calorimeter =
`T_i` = 298.00 K

Final temperature of the calorimeter =
`T_f` = 302.34 K

Change in temperature ,
`\Delta T=T_f-T_i=(302.34-298.00)K=4.34K`

Putting in the values, we get:

`Q=13.70kJ/K* 4.34K=59.4kJ`

As heat absorbed by calorimeter is equal to heat released by combustion of propanol

`Q=q`

`\text{Moles of propanol}=\frac{\text{given mass}}{\text{Molar Mass}}=(1.771g)/(60g/mol)=0.030mol`

Heat released by 0.030 moles of propanol = 59.4 kJ

Heat released by 1 mole of propanol =
`(59.4)/(0.030)* 1=1980kJ`

ΔH for the combustion of propanol to carbon dioxide gas and liquid water is 1980 kJ/mol