Chapter 23, Problem 95A

Interpretation Introduction

Interpretation:

The efficiency of cellular respiration in terms of the percentage of available energy that is stored in the chemical bonds of ATP needs to be determined.

Concept introduction:

Cellular respiration is a redox process. The balanced chemical equation for cellular respiration is,

$\begin{array}{l}{\text{C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}\text{+}{\text{6O}}_{\text{2}}\to {\text{6CO}}_{\text{2}}+{\text{6H}}_{\text{2}}\text{O}\text{+}\text{energy}\\ \text{Glucose}\text{Oxygen}\text{Carbon-dioxide}\text{Water}\end{array}$

Answer to Problem 95A

The efficiency of cellular respiration in terms of the percentage of available energy that is stored in the chemical bonds of ATP is 41%.

Explanation of Solution

Given,

38 ATP are formed during cellular respiration, heat of combustion of 1 mol of glucose is $\text{2}{\text{.82×10}}^{\text{3}}\text{kJ/mol}$ and each mol of ATP stores 30.5 kJ of energy.

Now,

$\begin{array}{c}\text{Energy}\text{of}\text{38}\text{moles}\text{of}\text{ATP=}\text{38}\text{mol}\text{ATP×}\frac{\text{30}\text{.5}\text{kJ}}{\text{1}\text{mol}\text{ATP}}\\ \text{=}\text{1159}\text{kJ}\end{array}$

Now to calculate the percentage we calculate the energy value with 100%.

$\begin{array}{c}\text{Percentage}\text{of}\text{avialable}\text{energy}\text{=}\frac{\text{1159}\text{kJ}}{\text{2}{\text{.82×10}}^{\text{3}}\text{kJ}}\text{×100\%}\\ \text{=}\text{41\%}\end{array}$

Hence, the percentage of energy is 41%.

Conclusion

The energy efficiency of cellular respiration in terms of the percentage is 41%.