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.]

Step by step solution

01

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.

02

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:\( \frac{38\, ATP}{94.1\, ATP} * 100\% = 40.4\% \)

03

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.

04

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.

05

Calculate the efficiency under the given conditions

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

06

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.

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