Question

A 2.15g sample of benzene (C_6H_6) is burned in a bomb calorimeter, and the temperature rises from 22.46 degree C to 34.34 degree C. Calculate the heat capacity of the bomb calorimeter. Note the following thermochemical equation: C_6H_6(I) + 15/2 O_2 (g) rightarrow 6CO_2 (g) + 3H_2O (g) Delta H degree = -3267.5 kJ

Answer 1

Answer:

The heat capacity of the bomb calorimeter is 7.58 J/°C.

Explanation:

$C_6H_6(I) + \frac{15}{2} O_2 (g) \rightarrow 6CO_2 (g) + 3H_2O (g) ,\Delta H^o = -3267.5 kJ$

First, we will calculate energy released on combustion:

$\Delta H$ = enthalpy change =  -3267.5 kJ/mol

q = heat energy released

n = number of moles benzene= $\frac{\text{Mass of benzene}}{\text{Molar mass of benzene}}=\frac{2.15 g g}{78 g /mol}=0.02756 mol$

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

$q=\Delta H\times n =-3267.5 kJ/mol\times 0.02756 mol=-90.0657 kJ$

q = -90.0657 kJ = -90,065.7 J

Now we  calculate the heat gained by the calorimeter let it be Q.

Q = -q= -(-90,065.7 J) = 90,065.7 J (conservation of energy)

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

where,

Q = heat gained by calorimeter

c = specific heat capacity of calorimeter =?

$T_{final}$ = final temperature = $34.34^oC$

$T_{initial}$ = initial temperature = $22.46^oC$

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

$90,065.7 J=c\times (34.34-22.46)^oC$

$c=\frac{90,065.7 J}{(34.34-22.46)^oC}=7.58 J/^oC$

The heat capacity of the bomb calorimeter is 7.58 J/°C.