# Question f6c38

Dec 6, 2014

I would say $- 16709 J$, or $- 16.7 k J$.

${C}_{12} {H}_{22} {O}_{11} + 12 {O}_{2} \to 12 C {O}_{2} + 11 {H}_{2} O$

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

${q}_{r e a c t i o n} = - {q}_{c a l o r i m e t e r}$, where

${q}_{c a l o r i m e t e r} = C \cdot \left(c h a n g e i n t e m p e r a t u r e\right)$, $C$ representing the heat capacity of the calorimeter (in this case, 4.9 (kJ)/(degree C #);

We can determine that the change in temperature, defined as ${T}_{f i n a l} - {T}_{i n i t i a l}$, is equal to $3.41$ degrees.

Therefore, ${q}_{c a l o r i m e t e r} = 4.9 \frac{k J}{\mathrm{de} g r e e C} \cdot 3.41 = 16.7 k J$, which determines the heat of combustion of sucrose to be

${q}_{s u c r o s e} = - {q}_{c a l o r i m e t e r} = - 16.7 k J$ -> exothermic reaction

Now, since $1.01$ grams of sucrose was burned, the heat of combustion of sucrose is approximately equal to $- 16.7 \frac{k J}{g}$. 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

${q}_{s u c r o s e} = - 16.7 \frac{k J}{g} \cdot 246 \frac{g}{m o l e} = - 4108 \frac{k J}{m o l e}$,

$246 \frac{g}{m o l e}$being sucrose's molar mass.