Question

A 0.468 g sample of pentane, C 5H 12, was burned in a bomb calorimeter. The temperature of the calorimeter and 1.00 kg of water in it rose from 20.45 °C to 23.65 °C. The heat capacity of the calorimeter by itself is 2.21 kJ/°C and the specific heat capacity of water is 4.184 J/g.°C What is the heat of combustion per mole of pentane?

Answer 1

`Q_(lost)` per mole of pentane = 3157.53 kJ/mol

Step-by-step explanation:

Given:

Mass of pentane, m = 0.468 gram

Molar mass of pentane, M = 72.15

Now, mol of pentane, n = mass/M = 0.468/72.15 = 0.00648 mol of C5H12

Now,

ΔT = 23.65 - 20.45 = 3.2°C

Heat capacity of the calorimeter, C = 2.21 kJ/°C

Specific heat capacity of the water, Cp = 4.184 J/g.°C

Now,

the heat gained = the heat lost

`Q_(gained) = -Q_(lost)`

also,

`Q_(gained) = Q_(water) + Q_(calorimeter)`

`Q_(water) = m* C*(T_f-T_i)`

or

`Q_(water) = 1000*4.184*(23.65-20.45) = 13388.8\ J`

and

`Q_(calorimeter) = C*\Delta T = 2.21*1000*3.2 = 7072\ J`

Now,

`Q_(total) = 13388.8 +7072 = 20460.8\ J`

we have,

`Q_(lost) = -Q_(gain) = - 20460.8\ J` (Here negative sign depicts the release of the heat)

`Q_(lost)` per mole of pentane =-20460.8/(0.00648 ) = 3157.53 kJ/mol