Question

A bomb calorimeter has a heat capacity of 2.47 kJ/K including the water. When a 0.111 g sample of ethylene (C 2H 4) was burned in this calorimeter, the temperature increased by 2.26 K. Calculate the energy of combustion for one mole of ethylene. a. -50.3 kJ/mol b. -1.41 x 103 kJ/mol c. -0.274 kJ/mol d. -624 kJ/mol e. -5.29 kJ/mol

Answer 1

Answer : The correct option is, (b)
`-1.41* 10^3kJ/mol`

Explanation :

First we have to calculate the heat produced.

`q=c* \Delta T`

where,

q = heat produced = ?

c = specific heat capacity =
`2.47kJ/K`

`\Delta T` = Change in temperature = 2.26 K

Now put all the given values in the above formula, we get:

`q=2.47kJ/K* 2.26K`

`q=5.5822kJ`

Now we have to calculate the energy of combustion for one mole of ethylene.

`\Delta H=(q)/(n)`

where,

`\Delta H` = energy of combustion for one mole of ethylene = ?

q = heat released = 5.5822 kJ

m = mass of
`C_2H_4` = 0.111 g

Molar mass of
`C_2H_4` = 28 g/mol

`\text{Moles of }C_2H_4=\frac{\text{Mass of }C_2H_4}{\text{Molar mass of }C_2H_4}=(0.111g)/(28g/mole)=0.00396mole`

Now put all the given values in the above formula, we get:

`\Delta H=(5.5822kJ)/(0.0396mole)`

`\Delta H=1409.6kJ/mol=1.41* 10^3kJ/mol`

Negative sign indicate the energy is released.

Therefore, the energy of combustion for one mole of ethylene is,
`-1.41* 10^3kJ/mol`