Question

Kemmi Major does some experimental work on the combustion of sucrose: C12H22O11(s) 12 O2(g) → 12 CO2(g) 11 H2O(g) She burns a 0.05392 g pellet of sucrose in a bomb calorimeter with excess oxygen. She determines the qrxn to be –916.6 J for the reaction. Calculate the ∆H value for the combustion reaction. (Round the answer to 3 significant digits, units of kJ, pay attention to positive or negative.

Answer 1

Answer: $5.81\times 10^6J/mol$

Explanation:

Heat of combustion is the amount of heat released when 1 mole of the compound is completely burnt in the presence of oxygen.

$C_{12}H_{22}O_{11}(s)+12O_2\rightarrow 12CO_2(g)+11H_2O(g)$

To calculate the moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}=\frac{0.05392g}{342g/mol}=1.577\times 10^{-4}moles$

Thus $1.577\times 10^{-4}moles$ of sucrose releases =  916.6 J of heat

1 mole of sucrose releases =$\frac{916.6}{1.577\times 10^{-4}}\times 1=5.81\times 10^6J$ of heat

Thus ∆H value for the combustion reaction is $5.81\times 10^6J/mol$