Question

In an experiment, a 0.4351 g sample of benzil (C14H10O2) is burned completely in a bomb calorimeter. The calorimeter is surrounded by 1.153×103 g of water. During the combustion the temperature increases from 25.10 to 27.60 °C. The heat capacity of water is 4.184 J g-1°C-1. The heat capacity of the calorimeter was determined in a previous experiment to be 994.1 J/°C. Assuming that no energy is lost to the surroundings, calculate the molar heat of combustion of benzil based on these data. C14H10O2(s) + (31/2) O2(g) 5 H2O(l) + 14 CO2(g) + Energy Molar Heat of Combustion = kJ/mol g

Answer 1

Answer:

The enthalpy change during the reaction is -7020.09 kJ/mole.

Explanation:

First we have to calculate the heat gained by the calorimeter.

$q=c\times (T_{final}-T_{initial})$

where,

q = heat gained = ?

c = specific heat = $994.1 J/^oC$

$T_{final}$ = final temperature = $27.60^oC$

$T_{initial}$ = initial temperature = $25.10^oC$

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

$q=994.1 J/^oC\times (27.60-25.10)^oC$

$q= 2,485.25 J$

The heat gained by water present in calorimeter. = q'

$q'=mc\times (T_{final}-T_{initial})$

where,

q' = heat gained = ?

m = mass of water = $1.153\times 10^3 g$

c' = specific heat of water = $4.184 J/^oC$

$T_{final}$ = final temperature = $27.60^oC$

$T_{initial}$ = initial temperature = $25.10^oC$

$q'=1.153\times 10^3 g\times 4.184 J/^oC\times (27.60-25.10)^oC$

q ' = 12,060.38 J

Now we have to calculate the enthalpy change during the reaction.

$\Delta H=-\frac{Q}{n}$

where,

$\Delta H$ = enthalpy change = ?

Q = heat gained = -(q+q') = -(2,485.25 J + 12,060.38 J)= -14,545.63 J

Q = -14.54563 kJ

n = number of moles fructose = $\frac{\text{Mass of benzil}}{\text{Molar mass of benzil}}=\frac{0.4351 g}{210 g/mol}=0.002072 mole$

$\Delta H=-\frac{-14.54563 kJ}{0.002072 mole}=-7020.09 kJ/mole$

Therefore, the enthalpy change during the reaction is -7020.09 kJ/mole.