Question

Combustion of glucose takes place according to the equation
$\text{C}_{\text{6}} \text{H}_{\text{12}} \text{O}_{\text{6}} \, \text{+} \, \text{6} \, \text{O}_{\text{2}} \, \rightarrow \, \text{6} \, \text{CO}_{\text{2}} \, \text{+} \, \text{6} \, \text{H}_{\text{2}} \text{O;} \, \text{ΔH} \, \text{=} \, - \text{72} \, \text{kcal/mol} \text{.}$
How much energy will be released by the combustion of $\text{1} \text{.6} \, \text{g}$ of glucose (Molecular mass of glucose = $\text{180} \, \text{g/mol}$ )?

• A. 0.064 kcal
• B. 0.64 kcal
• C. 6.4 kcal
• D. 64 kcal

Answer 1

Answer: (B)

Combustion of glucose takes place according to the equation
$\text{C}_{\text{6}} \text{H}_{\text{12}} \text{O}_{\text{6}} \, \text{+} \, \text{6O}_{\text{2}} \, \rightarrow \, \text{6CO}_{\text{2}} \, \text{+} \, \text{6H}_{\text{2}} \text{O;} \, \text{ΔH} \, \text{=} \, \text{-72} \, \text{kcal}$
$\Delta H$ per $1.6 g =\frac{72 \times 1.6}{180}=0.64 kcal$