Q5.3-84E
Question
The oxidation of the sugar glucose, \({{\bf{C}}_{\bf{6}}}{{\bf{H}}_{{\bf{12}}}}{{\bf{O}}_{\bf{6}}}\), is described by the following equation:
\({{\bf{C}}_{\bf{6}}}{{\bf{H}}_{{\bf{12}}}}{{\bf{O}}_{\bf{6}}}{\bf{(s) + 6}}{{\bf{O}}_{\bf{2}}}{\bf{(g)}} \to {\bf{6C}}{{\bf{O}}_{\bf{2}}}{\bf{(g) + 6}}{{\bf{H}}_{\bf{2}}}{\bf{O(l)}};{\bf{ \Delta H = - 2816 kJ}}\)
The metabolism of glucose gives the same products, although the glucose reacts with oxygen in a series of steps in the body.
(a) How much heat in kilojoules can be produced by the metabolism of 1.0 g of glucose?
(b) How many calories can be produced by the metabolism of 1.0 g of glucose?
Step-by-Step Solution
Verified(a) Heat produced =-563200 kJ.
(b) Calories produced =-134604.8 kilocalorie.
The reaction is \({{\bf{C}}_{\bf{6}}}{{\bf{H}}_{{\bf{12}}}}{{\bf{O}}_{\bf{6}}}{\bf{(s) + 6}}{{\bf{O}}_{\bf{2}}}{\bf{(g)}} \to {\bf{6C}}{{\bf{O}}_{\bf{2}}}{\bf{(g) + 6}}{{\bf{H}}_{\bf{2}}}{\bf{O(l)}};{\bf{ \Delta H = - 2816 kJ}}.\)
In the above reaction, the enthalpy change has a value of -2816 kJ.
The molar mass of \({{\rm{C}}_{\rm{6}}}{{\rm{H}}_{{\rm{12}}}}{{\rm{O}}_{\rm{6}}}\) = 180.15 g/mol.
1 mole of \({{\rm{C}}_{\rm{6}}}{{\rm{H}}_{{\rm{12}}}}{{\rm{O}}_{\rm{6}}}\) produces -2816 kJ heat.
The number of moles is evaluated as:
\(\begin{array}{c}{\bf{Number of moles = }}\frac{{{\bf{Given mass}}}}{{{\bf{Molar mass}}}}\\{\rm{ = }}\frac{{\rm{1}}}{{{\rm{180}}{\rm{.15}}}}\\{\rm{ = 0}}{\rm{.005}}\;{\rm{mol}}{\rm{.}}\end{array}\)
The amount of heat in kilojoules is evaluated as:
0.005 mol. \(\begin{array}{l}{\rm{ = }}\frac{{{\rm{ - 2816 kJ}}}}{{{\rm{0}}{\rm{.005 mol}}}}\\{\rm{ = \; - 563200 kJ/mol}}{\rm{.}}\end{array}\)
1 g of glucose produces –563200 kJ heat.
Unit conversion:
1 calorie = 4.184 kJ
So, 1 kJ = \(\frac{{\rm{1}}}{{{\rm{4}}{\rm{.184}}}}{\rm{ calorie}}{\rm{.}}\)
-563200 kJ of heat produces a specific amount of calories, which is evaluated as:
\(\begin{array}{c}{\rm{Calories = \;\;}}\frac{1}{{{\rm{4}}{\rm{.184}}}}{\rm{ \times \; - 563200 }}\\{\rm{ = 0}}{\rm{.2390 }} \times {\rm{ - 563200 }}\\{\rm{ = \;134604}}{\rm{.8\;kilocalorie}}{\rm{.}}\end{array}\)