Question

A 1.000 gram sample of the rocket fuel hydrazine (N2H4) is burned in a bomb calorimeter. The temperature rises from 24.62°C to 28.16°C. The heat capacity of the calorimeter (including the water) is 5860 J/°C. Calculate the molar heat of combustion of hydrazine, in kJ/mole.

Answer 1

Answer:

The molar heat of combustion of hydrazine is -663.82 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 = $5860 J/^oC$

$T_{final}$ = final temperature = $28.16^oC$

$T_{initial}$ = initial temperature = $24.62^oC$

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

$q=5860 J/^oC\times (28.16-24.62)^oC$

$q=20,744.4 J=20.7444 kJ$

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

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

where,

$\Delta H$ = enthalpy change = ?

q = heat gained = 20.7444 kJ

n = number of moles fructose = $\frac{\text{Mass of hydrazine}}{\text{Molar mass of hydrazine}}=\frac{1.000 g}{32 g/mol}=0.03125 mole$

$\Delta H=-\frac{20.7444 kJ}{0.03125 mole}=-663.82 kJ/mole$

The molar heat of combustion of hydrazine is -663.82 kJ/mole.