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

Answer:

$Q_{lost}$ per mole of pentane = 3157.53 kJ/mol

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,

$Q_{gained} = Q_{water} + Q_{calorimeter}$

$Q_{water} = m\times C\times(T_f-T_i)$

or

$Q_{water} = 1000\times4.184\times(23.65-20.45) = 13388.8\ J$

and

$Q_{calorimeter} = C\times\Delta T = 2.21\times1000\times3.2 = 7072\ J$

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