Gycolysis is the process by which glucose is metabolized to lactic acid according to the equation $$ \begin{gathered} \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(a q) \longrightarrow 2 \mathrm{C}_{3} \mathrm{H}_{6} \mathrm{O}_{3}(a q) \\ \Delta G^{6}=-198 \mathrm{~kJ} \text { at } \mathrm{pH} 7.0 \text { and } 25^{\circ} \mathrm{C} \end{gathered} $$ Glycolysis is the source of energy in human red blood cells. In these cells, the concentration of glucose is \(5.0 \times 10^{-3} M_{3}\) while that of lactic acid is \(2.9 \times 10^{-3} M\). Calculate \(\Delta G\) for glycolysis in human blood cells under these conditions. Use the equation \(\Delta G=\Delta G^{\circ}+\mathrm{RT} \ln Q\), where \(Q\) is the concentration quotient, analogous to \(K\).

Step by step solution

01

1. Identify the given information

We have the following information: - The glycolysis equation: \(\mathrm{C}_{6}\mathrm{H}_{12}\mathrm{O}_{6}(aq) \longrightarrow 2\mathrm{C}_{3}\mathrm{H}_{6}\mathrm{O}_{3}(aq)\) - The standard free energy change: \(\Delta G^{o} = -198\,\mathrm{kJ}\) - The concentration of glucose: \([\mathrm{C}_{6}\mathrm{H}_{12}\mathrm{O}_{6}] = 5.0 \times 10^{-3} M\) - The concentration of lactic acid: \([\mathrm{C}_{3}\mathrm{H}_{6}\mathrm{O}_{3}] = 2.9 \times 10^{-3} M\) - Formula: \(\Delta G = \Delta G^{o} + \mathrm{RT}\ln Q\)

02

2. Calculate the reaction quotient (Q)

The reaction quotient \(Q\) is calculated using the concentrations of the reactant (glucose) and the product (lactic acid): $$ Q = \frac{[\mathrm{C}_{3}\mathrm{H}_{6}\mathrm{O}_{3}]^2}{[\mathrm{C}_{6}\mathrm{H}_{12}\mathrm{O}_{6}]} $$ Substitute the given concentrations: $$ Q = \frac{(2.9 \times 10^{-3})^2}{(5.0 \times 10^{-3})} $$

03

3. Evaluate Q

Calculating the value of Q: $$ Q = \frac{8.41 \times 10^{-6}}{5.0 \times 10^{-3}} = 1.682 \times 10^{-3} $$

04

4. Calculate ΔG using the formula

To determine ΔG, we use the formula provided: $$ \Delta G = \Delta G^{o} + \mathrm{RT}\ln Q $$ At 25°C, the value of the gas constant R (in kJ/mol) and temperature T (in Kelvin) are: - \(R = 8.314 \times 10^{-3}\,\mathrm{kJ/mol \cdot K}\) - \(T = 273 + 25 = 298\,\mathrm{K}\) Substituting the values of \(\Delta G^{o}\), R, T, and Q in the formula: $$ \Delta G = -198\,\mathrm{kJ} + (8.314 \times 10^{-3}\,\mathrm{kJ/mol \cdot K})(298\,\mathrm{K})(\ln(1.682 \times 10^{-3})) $$

05

5. Calculate the final value of ΔG

Evaluating the expression for ΔG: $$ \Delta G = -198\,\mathrm{kJ} + (-13.02\,\mathrm{kJ}) = -211.02\,\mathrm{kJ} $$ The free energy change \((\Delta G)\) for glycolysis in human blood cells under these conditions is approximately \(-211.02\,\mathrm{kJ}\).

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