Question

Assuming that the bath contains 250.0 g of water and that the calorimeter itself absorbs a negligible amount of heat, calculate combustion energies (ΔE) for benzene in both kilojoules per gram and kilojoules per mole.

Answer 1

To calculate the combustion energy for benzene, we can use the equation q = mCΔT. Given the mass of benzene, the temperature change of the calorimeter, and the heat capacity of the bomb calorimeter, we can calculate the heat produced. The combustion energy can be expressed in kilojoules per gram and kilojoules per mole.

Step-by-step explanation:

To calculate the combustion energy (ΔE) for benzene, we need to use the equation q = mCΔT. Where q is the heat produced, m is the mass of benzene, C is the heat capacity of the bomb calorimeter, and ΔT is the temperature change.

Given that 0.963 g of benzene is burned, the temperature change of the calorimeter is 8.39 °C, and the heat capacity of the bomb is 784 J/°C, we can calculate the heat produced as follows:

q = (0.963 g) * (784 J/°C) * (8.39 °C)

q ≈ 22.67 kJ.

The combustion energy can also be expressed in kilojoules per gram. To convert the heat produced to kilojoules per gram, we divide the heat produced by the mass of benzene:

ΔE = q / m = (22.67 kJ) / (0.963 g) ≈ 23.54 kJ/g.

Finally, to calculate the combustion energy in kilojoules per mole, we need to know the molar mass of benzene. The molar mass of benzene (C6H6) is approximately 78.11 g/mol. Therefore, we can calculate the combustion energy per mole as:

ΔE = (23.54 kJ/g) * (78.11 g/mol) ≈ 1838.06 kJ/mol.

Answer 2

`-3272` kJ/mol

Step-by-step explanation:

Given and known facts

Mass of Benzene
`= 0.187` grams

Mass of water
`= 250` grams

Standard heat capacity of water
`= 4.18` J/g∙°C

Change in temperature ΔT
`= 7.48`°C

Heat

`=250 * 4.18 * 7.48\\=7816.6 \\=7.82`

Heat released by benzine is - 7.82 kJ

Now, we know that

`0.187` grams of benzene release
`= -7.82` kJ heat

So,
`1` g benzine releases

`( -7.82 )/(0.187)\\= -41.8`

kJ/g

`0.187 * (1)/(78.108)=0.00239` mol C6H6

Heat released

`= (-7.82)/( 0.00239)`

`=-3272` kJ/mol