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Chapter 18: Problem 36

In the metabolism of glucose, the first step is the conversion of glucose to glucose 6 -phosphate: glucose \(+\mathrm{H}_{3} \mathrm{PO}_{4} \longrightarrow\) glucose 6 -phosphate \(+\mathrm{H}_{2} \mathrm{O}\) $$ \Delta G^{\circ}=13.4 \mathrm{~kJ} / \mathrm{mol} $$ Because \(\Delta G^{\circ}\) is positive, this reaction does not favor the formation of products. Show how this reaction can be made to proceed by coupling it with the hydrolysis of ATP. Write an equation for the coupled reaction and estimate the equilibrium constant for the coupled process.

Short Answer

Expert verified
The coupled reaction of glucose conversion to glucose-6-phosphate and ATP hydrolysis proceeds spontaneously with a \(\Delta G^{\circ}\) of -17.1 kJ/mol. The equilibrium constant for the reaction, \(K_{eq}\), can be found using the relation \(\Delta G^{\circ} = -RTlnK_{eq}\) and substituting relevant values.

Step by step solution

01

Understanding ATP hydrolysis

The hydrolysis of ATP (adenosine triphosphate) into ADP (adenosine diphosphate) and inorganic phosphate is a spontaneous reaction that releases energy. The reaction is given by the equation: ATP \(\rightarrow\) ADP + \(Pi\) with \(\Delta G^{\circ} = -30.5 \mathrm{kJ/mol}\).
02

Coupling reactions

A coupled reaction involves mixing a spontaneous and non-spontaneous reaction to make the overall process spontaneous. The conversion of glucose to glucose-6-phosphate is coupled with ATP hydrolysis. The equation for the coupled reaction is: glucose + ATP \(\rightarrow\) glucose 6-phosphate + ADP + \(H_{2}O\)
03

Calculating the \(\Delta G^{\circ}\) for the coupled process

The \(\Delta G^{\circ}\) for the coupled process equals the sum of the \(\Delta G^{\circ}\) values of the individual processes. For this coupled process, \(\Delta G^{\circ}_{coupled} = \Delta G^{\circ}_{glucose \rightarrow glucose-6-phosphate} + \Delta G^{\circ}_{ATP \rightarrow ADP + Pi} = 13.4 \mathrm{kJ/mol} + (-30.5 \mathrm{kJ/mol}) = -17.1 \mathrm{kJ/mol}\). Because \(\Delta G^{\circ}_{coupled}\) is negative, the coupled reaction is spontaneous and will proceed.
04

Find the equilibrium constant

From the equation \(\Delta G^{\circ} = -RTlnK_{eq}\), we solve for \(K_{eq}\) to find the equilibrium constant. Input \(\Delta G^{\circ}_{coupled} = -17.1 \mathrm{kJ/mol}\), \(R = 8.314 \mathrm{J/K \cdot mol}\), and \(T = 298 K\) (standard temperature). Calculating, we get \(K_{eq} = e^{(\Delta G^{\circ}_{coupled}/RT)}\). Note that the units of \(\Delta G^{\circ}\) should be converted to J/mol before performing the calculation.

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Most popular questions from this chapter

State whether the sign of the entropy change expected for each of the following processes will be positive or negative, and explain your predictions. (a) \(\mathrm{PCl}_{3}(l)+\mathrm{Cl}_{2}(g) \longrightarrow \mathrm{PCl}_{5}(s)\) (b) \(2 \mathrm{HgO}(s) \longrightarrow 2 \mathrm{Hg}(l)+\mathrm{O}_{2}(g)\) (c) \(\mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{H}(g)\) (d) \(\mathrm{U}(s)+3 \mathrm{~F}_{2}(g) \longrightarrow \mathrm{UF}_{6}(s)\)

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