Answer: The enthalpy of the reaction is -5112.5 kJ/mol
Explanation:
To calculate the heat absorbed by the calorimeter, we use the equation:
[tex]q=c\Delta T[/tex]
where,
q = heat absorbed
c = heat capacity of calorimeter = 6.18 kJ/°C
[tex]\Delta T[/tex] = change in temperature = 27.3°C
Putting values in above equation, we get:
[tex]q=6.18kJ/^oC\times 27.3^oC=168.714kJ[/tex]
Heat absorbed by the calorimeter will be equal to the heat released by the reaction.
Sign convention of heat:
When heat is absorbed, the sign of heat is taken to be positive and when heat is released, the sign of heat is taken to be negative.
To calculate the number of moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]
Given mass of octane = 3.80 g
Molar mass of octane = 114 g/mol
Putting values in above equation, we get:
[tex]\text{Moles of octane}=\frac{3.80g}{114g/mol}=0.033mol[/tex]
To calculate the enthalpy change of the reaction, we use the equation:
[tex]\Delta E=\frac{q}{n}[/tex]
where,
q = amount of heat released = -168.714 kJ
n = number of moles = 0.033 moles
[tex]\Delta E[/tex] = enthalpy change of the reaction
Putting values in above equation, we get:
[tex]\Delta E=\frac{-168.714kJ}{0.033mol}=-5112.5kJ/mol[/tex]
Hence, the enthalpy of the reaction is -5112.5 kJ/mol