The oxidation of the sugar glucose, C6H12O6, is described by the following equation. C6H12O6(s) + 6 O2(g) → 6 CO2(g) + 6 H2O(l) ΔH = −2802.5 kJ/mol 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 18.1 g of glucose?

(b) How many Calories can be produced by the metabolism of 18.1 g of glucose?

We can multiply the enthalpy by the number of moles of glucose to get the heat produced by the metabolism. Grams of glucose will be converted to moles using the molar mass of glucose (180.156 g/mol):

(18.1 g)(mol/180.156g)(2802.5 kJ/mol) = 282 kJ

(b) Using the result we obtained above, kJ will be converted to Calories using the conversion factor of 4.184J = 1 cal. Calorie with a capital C is the same as a kilocalorie.

(282 kJ)(1 cal/4.184J) = 67.4 kcal = 67.4 Calories

Аноним Аноним · Answer 2 · 2019-12-20T11:00:46+01:00

Answer:

For a: The amount of heat produced for given amount of glucose is -283.05 kJ

For b: The amount of heat produced for given amount of glucose is -67648.9 Cal

Explanation:

To calculate the number of moles, we use the equation:

[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]

Given mass of glucose = 18.1 g

Molar mass of glucose = 180.16 g/mol

Putting values in above equation, we get:

[tex]\text{Moles of glucose}=\frac{18.1g}{180.16g/mol}=0.101mol[/tex]

The given chemical reaction follows:

[tex]C_6H_{12}O_6(s)+6O_2(g)\rightarrow 6CO_2(g)+6H_2O(l);\Delta H=-2802.5kJ/mol[/tex]

For a:

By Stoichiometry of the reaction:

When 1 mole of glucose is reacted, the amount of heat released is 2802.5 kJ

So, when 0.101 moles of glucose is reacted, the amount of heat released is [tex]\frac{2802.5}{1}\times 0.101=283.05kJ[/tex]

Hence, the amount of heat produced for given amount of glucose is -283.05 kJ

For b:

To convert the heat produced in kilo joules to calories, we use the conversion factor:

1 kJ = 239 Cal

So, [tex]-283.05kJ\times (\frac{239Cal}{1kJ})=-67648.9Cal[/tex]

Hence, the amount of heat produced for given amount of glucose is -67648.9 Cal