Question

The Gibbs energy available from the complete combustion of 1 mol of glucose to carbon dioxide and water is $$\begin{array}{r} \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(\mathrm{aq})+6 \mathrm{O}_{2}(\mathrm{g}) \longrightarrow 6 \mathrm{CO}_{2}(\mathrm{g})+6 \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \\ \Delta G^{\circ}=-2870 \mathrm{kJ} \mathrm{mol}^{-1} \end{array}$$ (a) Under biological standard conditions, compute the maximum number of moles of ATP that could form from ADP and phosphate if all the energy of combustion of 1 mol of glucose could be utilized. (b) The actual number of moles of ATP formed by a cell under aerobic conditions (that is, in the presence of oxygen) is about \(38 .\) Calculate the efficiency of energy conversion of the cell. (c) Consider these typical physiological conditions. $$\begin{array}{l} P_{\mathrm{CO}_{2}}=0.050 \mathrm{bar} ; P_{\mathrm{O}_{2}}=0.132 \mathrm{bar} \\\ {[\mathrm{glucose}]=1.0 \mathrm{mg} / \mathrm{mL} ; \mathrm{pH}=7.0} \\ {[\mathrm{ATP}]=[\mathrm{ADP}]=\left[P_{\mathrm{i}}\right]=0.00010 \mathrm{M}} \end{array}$$ Calculate \(\Delta G\) for the conversion of 1 mol ADP to ATP and \(\Delta G\) for the oxidation of 1 mol glucose under these conditions. (d) Calculate the efficiency of energy conversion for the cell under the conditions given in part (c). Compare this efficiency with that of a diesel engine that attains \(78 \%\) of the theoretical efficiency operating with \(T_{\mathrm{h}}=1923 \mathrm{K}\) and \(T_{1}=873 \mathrm{K} .\) Suggest a reason for your result. [ Hint: See Feature Problem 95.]

Short answer

The maximum number of ATP mol is -94.1 and the percent efficiency of energy conversion of the cell is 40.4%. \(\Delta G\) values for the conversion of ADP to ATP and the oxidation of glucose should be calculated in step 3 and 4. Then a new efficiency value is calculated using the corrected \(\Delta G\) values. The differences in efficiencies between a cell and a diesel engine can be many, and some possible reasons are differences in energy extraction processes.

Step-by-step solution

Calculate the maximum number of moles of ATP

The standard free-energy change, \(\Delta G^{0'}\) for the synthesis of ATP from ADP and phosphate under cellular conditions is +30.5 kJ/mol. Considering all of the Gibbs energy from glucose combustion is used to form ATP, the number of moles of ATP formed is 1 mol of glucose * \(\frac{-2870 \,kJ}{30.5 \,kJ/mol}\) = -94.1 mol ATP.

Calculate the efficiency of energy conversion

The actual number of moles of ATP formed by a cell under aerobic conditions is about 38. Thus, the percent efficiency of energy conversion of the cell can be calculated with:\( \frac{38\, ATP}{94.1\, ATP} * 100\% = 40.4\% \)

Calculate \(\Delta G\) for the conversion of ADP to ATP

To find \(\Delta G\) for the conversion of 1 mol ADP to ATP under physiological conditions, use the equation \(\Delta G=\Delta G^{\circ}+RT\,ln(Q)\), where \(Q=[\mathrm{ATP}]/[\mathrm{ADP}][P_{\mathrm{i}}]\) and replace the terms with given values. After substituting the values, the final \(\Delta G\) can be calculated.

Calculate \(\Delta G\) for the oxidation of glucose

To calculate the oxidation of glucose under these conditions, use the equation \(\Delta G=\Delta G^{\circ}+RT\,ln(Q)\), where \(Q=[\mathrm{CO}_2]^6/[\mathrm{glucose}][\mathrm{O}_2]^6\). After substituting the provided values for these quantities, \(\Delta G\) can be calculated.

Calculate the efficiency under the given conditions

By repeating step 2 with the corrected \(\Delta G\) values, the new efficiency can be calculated.

Compare and find the reason

After obtaining the new efficiency, compare it with the efficiency of a diesel engine and think about reasons why the efficiency for a cell might differ from the efficiency of a diesel engine. Some possible reasons could involve the differences in energy extraction processes between combustion engines and chemical reactions in cells.

Key concepts

Combustion Reaction

The combustion reaction of glucose is a fundamental process in both biological and chemical systems. In the given reaction, glucose (\( \mathrm{C}_6\mathrm{H}_{12}\mathrm{O}_6 \)) reacts with oxygen (\( \mathrm{O}_2 \)) to produce carbon dioxide (\( \mathrm{CO}_2 \)) and water (\( \mathrm{H}_2\mathrm{O} \)). This is a typical exothermic reaction, releasing energy in the form of heat and Gibbs energy. The Gibbs energy change (\( \Delta G^{\circ} = -2870 \mathrm{kJ} \mathrm{mol}^{-1} \)) signifies that the process is spontaneous and releases a significant amount of energy.

Combustion reactions are crucial for energy production in biological systems. In cells, this chemical energy is transformed into usable biochemical energy, such as ATP, to fuel various cellular processes. Understanding the Gibbs energy in these reactions helps us comprehend the efficiency and feasibility of biological energy transformation.

ATP Synthesis

ATP synthesis is an indispensable process in all living cells, allowing for energy to be stored and used as needed. The reaction transforms ADP (adenosine diphosphate) and inorganic phosphate into ATP (adenosine triphosphate), the energy currency of the cell. Under standard conditions, the synthesis of 1 mole of ATP from ADP and phosphate has a Gibbs energy change (\( \Delta G^{0'} \)) of +30.5 \( \mathrm{kJ/mol} \).

In the context of glucose combustion, the energy released can potentially generate multiple ATP molecules. However, it's theoretically possible for such an amount of energy released by the complete combustion of 1 mole of glucose to form about 94 moles of ATP under ideal conditions. Naturally, cells cannot convert all available energy, leading to a lower actual number of ATP molecules produced.

Biological Thermodynamics

Biological thermodynamics involves the study of energy transformations and exchanges within biological systems. The fundamental concern is how living organisms utilize energy to maintain structure and function. These energy transformations are guided by the principles of thermodynamics, particularly the changes in Gibbs energy within chemical reactions.

In biological systems, thermodynamic calculations help evaluate the feasibility and efficiency of biochemical pathways. Consider the conversion of ADP to ATP under physiological conditions, where the actual Gibbs energy change might differ from the standard Gibbs energy due to varying concentrations and environmental conditions. This requires the use of the equation: \( \Delta G = \Delta G^{\circ} + RT \ln(Q) \) to determine the actual energy changes happening in vivo.

Energy Conversion Efficiency

Energy conversion efficiency in cells pertains to how effectively available energy from biochemical reactions is harnessed for cellular work. For glucose metabolism, the initial step of combustion provides energy, theoretically enough to produce around 94 moles of ATP from 1 mole of glucose. However, practical constraints and biological necessities mean only about 38 moles of ATP are produced in reality.

This translates to an energy conversion efficiency of approximately 40.4%, as not all the released energy can be utilized. Interestingly, when comparing this to non-biological systems, like a diesel engine which can reach efficiencies as high as 78%, we see that biological systems are not solely optimized for maximum efficiency. Instead, they balance multiple factors, including regulation, speed, and flexibility, to sustain life processes effectively. The efficiency calculation involves comparing the theoretical maximum ATP yield to the actual yield obtained in aerobic conditions, highlighting the compromises and adaptations necessitated by cellular functioning.