Question

Extension: If one mole of glucose contains 2870 KJ of energy and one mole of ATP releases 30.7 KJ of energy during a reaction, what is the percentage of energy in glucose that is available for the body to use?

Answer 1

40.6 % is the percentage of energy in glucose that is available for the body to use.

Step-by-step explanation:

We know that during cellular respiration one mole of glucose gives 38 molecules of ATP.

Given that 1 mole of ATP gives 30.7 KJ of energy.

So 38 molecules of ATP would give

38 x 30.7 KJ of energy.

= 1166 KJ would be yielded by 38 moles of ATP.

1 mole of glucose contains 2870 KJ of energy.

100 % of energy is released in glucose metabolism and x percent is used. It is given by the equation

2870: 100 = 1166:x

x=
`(1166 x 100)/(2870)`

= 40.6 %

40.6% of energy is available for the body to use and rest of the energy is dissipated in the form of heat when 1 mole of glucose undergoes combustion in cellular respiration producing 38 moles of ATP.

Answer 2

The percentage of energy in glucose that is available for the body to use is 40.6%.

Explanation:

Total used energy is the energy released by the hydrolysis of one mole of ATP and it is 30.7 kJ.

The total possible energy (used pus released as heat) is 2870 kJ.

Hydrolysis of 38 moles ATP gives = 38
`*` 30.7

= 1166.6 kJ of useful energy

The percentage of energy is transformed into useful energy, we can calculate by simple proportion,

2870 : 100 = 1166.6 : x

x = (1166.6
`*` 100) / 2870

x = 40.6% of energy is transformed into useful energy.