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: -20.7

Step-by-step explanation:

qrxn=−(qwater+qbomb)qrxn=−(cwatermΔT+ccalorimeterΔT)qrxn=−[(4.184Jg∘C)(1200. g)(3.54∘C)+(840. J∘C)(3.54∘C)]

qrxn=−20,700 J=−20.7 kJ

Answer 2

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

Step-by-step explanation:

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

`q=c* (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* (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=-(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}}=(1.000 g)/(32 g/mol)=0.03125 mole`

`\Delta H=-(20.7444 kJ)/(0.03125 mole)=-663.82 kJ/mole`

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