Hydrazine is used as a rocket fuel. When 1.50 g of hydrazine, N2H4, was burned in a bomb calorimeter, temperature increased from 23.57 °C to 28.84 °C. The heat capacity of the calorimeter was 5.510 kJ/°C. Determine the enthalpy of combustion of octane.
q = Ccal x delta T = 5.510 kJ/c x (delta T) = ??
That gives you q kJ/1.5 g. Usually it is expressed as kJ/mol. Convert thre 1.5 g to moles.
To determine the enthalpy of combustion of hydrazine, we need to use the heat capacity of the calorimeter and the change in temperature caused by the combustion.
The heat absorbed or released during a chemical reaction can be determined using the equation:
q = C × ΔT
where q is the heat exchanged, C is the heat capacity of the calorimeter, and ΔT is the change in temperature.
In this case, the heat capacity of the calorimeter (C) is given as 5.510 kJ/°C, and the change in temperature (ΔT) is the difference between the final and initial temperatures, which is 28.84 °C - 23.57 °C = 5.27 °C.
Substituting the given values into the equation, we can find the heat exchanged during the combustion of 1.50 g of hydrazine:
q = (5.510 kJ/°C) × (5.27 °C)
= 28.9917 kJ
Now, to determine the enthalpy of combustion of hydrazine (N2H4), we can assume that it was the only substance burned in the calorimeter. So the heat absorbed (q) during the combustion of 1.50 g of hydrazine is equal to the enthalpy change (ΔH) for the combustion reaction:
q = ΔH
Therefore, the enthalpy of combustion of hydrazine is 28.9917 kJ.
However, the question asks for the enthalpy of combustion of octane, not hydrazine. It seems like there has been an error in the question. If you need assistance with the enthalpy of combustion calculation for octane, please let me know.