Question #f6c38

1 Answer
Dec 6, 2014

I would say 16709J, or 16.7kJ.

Let's start with the balanced equation

C12H22O11+12O212CO2+11H2O

The heat of reaction can be defined as the amount of heat the system gives off to its surroundings so that it can return to its initial temperature. The heat of reaction is the negative value of the heat absorbed by the calorimeter and its contents

qreaction=qcalorimeter, where

qcalorimeter=C(changeintemperature), C representing the heat capacity of the calorimeter (in this case, 4.9kJdegreeC);

We can determine that the change in temperature, defined as TfinalTinitial, is equal to 3.41 degrees.

Therefore, qcalorimeter=4.9kJdegreeC3.41=16.7kJ, which determines the heat of combustion of sucrose to be

qsucrose=qcalorimeter=16.7kJ -> exothermic reaction

Now, since 1.01 grams of sucrose was burned, the heat of combustion of sucrose is approximately equal to 16.7kJg. In other words, when 1 gram of sucrose is burned, 16.7 kJ of heat are released from the reaction.

For one mole we would have

qsucrose=16.7kJg246gmole=4108kJmole,

246gmolebeing sucrose's molar mass.