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Chapter 3: Problem 6

For the process \(A \equiv B, K_{c g}(A B)\) is 0.02 at \(37^{\circ} \mathrm{C}\). For the process \(\mathrm{B} \rightleftharpoons \mathrm{C}, K_{\mathrm{eq}}(\mathrm{BC})=1000\) at \(37^{\circ} \mathrm{C}\) a. Determine \(K_{\mathrm{rg}}(\mathrm{AC}),\) the equilibrium constant for the overall process \(A \rightleftharpoons C,\) from \(K_{c q}(A B)\) and \(K_{c g}(B C)\) b. Determine standardstate free energy changes for all three processes, and \(\mathrm{us}=\Delta G^{\circ}(\mathrm{AC})\) to determine \(K_{\mathrm{rg}}(\mathrm{AC}) .\) Make sure that this value agrees with that determined in part a of this problem.

Short Answer

Expert verified
The equilibrium constant for the overall process \(A \rightleftharpoons C\) is 20. The standard state free energy changes for the processes \(A \rightleftharpoons B\), \(B \rightleftharpoons C\), and \(A \rightleftharpoons C\) can be calculated using the expression -RTlnK. Additional analysis shows that the calculated equilibrium constant from the overall free energy change agrees with the directly calculated equilibrium constant.

Step by step solution

01

Determine the Overall Equilibrium Constant

According to the law of mass action, the equilibrium constant for a net reaction made up of two or more steps is equal to the product of the equilibrium constants for the individual steps. Therefore, we can represent the overall equilibrium constant for the process \(A \rightleftharpoons C\) as \(K_{c g}(A C)=K_{c g}(A B) \times K_{c g}(B C)\). Substituting the provided values into this expression, we find \(K_{c g}(A C)=0.02 \times 1000 = 20\)
02

Calculating Free Energy Changes

The equation relating standard free energy change (\(\Delta G^{\circ}\)) and equilibrium constant (K) is \(\Delta G^{\circ}=-R T \ln K\), where R is the universal gas constant (8.314 J /(mol \cdot K)) and T is the temperature in Kelvin (310K for 37°C). Plugging in the provided values of K for each process, we can calculate \(\Delta G^{\circ}(A B)=-R T \ln K_{c g}(A B)\) and \(\Delta G^{\circ}(B C)=-R T \ln K_{c g}(B C)\). The overall free energy change \(\Delta G^{\circ}(A C)\) is the sum of these two values.
03

Use Delta G AC to Confirm Overall Equilibrium Constant

The equilibrium constant derived from the calculated \(\Delta G^{\circ}(A C)\) should match the \(K_{c g}(A C)\) obtained in the first step. We can use the equation \(\Delta G^{\circ}=-R T \ln K_{c g}(A C)\) to calculate \(K_{c g}(A C)\) from the derived \(\Delta G^{\circ}(A C)\). We find that the calculated \(K_{c g}(A C)\) is indeed equal to 20, confirming accuracy.

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Most popular questions from this chapter

Calculate the free energy of hydrolysis of ATP in a rat liver cell in which the ATP, ADP, and \(P\), concentrations are \(3.4,1.3,\) and \(4.8 \mathrm{m} M\) respectively.

An enzymatic hydrolysis of fructose- \(1-P\) \\[\text { Fructose- } 1-\mathrm{P}+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \text { fructose }+\mathrm{P}_{1}\\] was allowed to proceed to equilibrium at \(25^{\circ} \mathrm{C}\). The original concentration of fructose-1-P was \(0.2 \mathrm{M}\), but when the system had reached equilibrium the concentration of fructose-1-P was only \(6.52 \times 10^{-5} \mathrm{M}\). Calculate the equilibrium constant for this reaction and the free energy of hydrolysis of fructose- \(1-P\).

ATP hydrolysis at pH 7.0 is accompanicd by release of a hydrogen ion to the medium \\[\mathrm{ATP}^{6-}+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{ADP}^{3-}+\mathrm{HPO}_{4}^{2-}+\mathrm{H}^{+}\\] If the \(\Delta G^{\circ}\) for this reaction is \(-30.5 \mathrm{kJ} / \mathrm{mol}\), what is \(\Delta G^{*}\) (that is, the free energy change for the same reaction with all components, including \(\mathrm{H}^{+},\) at a standard state of \(1 \mathrm{M}\) )?

The acyl-Co.A synthetase reaction activates fatty acids for oxidation in cells: \\[ \mathrm{R}-\mathrm{COO}^{-}+\mathrm{CaASH}+\mathrm{ATP} \longrightarrow \mathrm{R}-\mathrm{COSCOA}+\mathrm{AMP}+\text { pyrophosphate } \\] The reaction is driven forward in part by hydrolysis of ATP to AMP and pyrophosphate. However, pyrophosphate undergoes further cleavage to yield two phosphate anions. Discuss the energetics of this reaction both in the presence and absence of pyrophosphate cleavage.

The standard-state free energy of hydrolysis for acetyl phosphate is \\[ \begin{aligned} \Delta G^{\circ}=-42.3 \mathrm{kJ} / \mathrm{mol} \\ \text { Actyl-P }+\mathrm{H}_{2} \mathrm{O} \longrightarrow \text { acctate }+\mathrm{P}_{\mathrm{i}} \end{aligned} \\] Calculate the free energy change for acetyl phosphate hydrolysis in a solution of \(2 \mathrm{m} M\) acctate, \(2 \mathrm{m} M\) phosphate, and \(3 \mathrm{n} M\) acetyl phosphate.
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