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The direction of a reaction with a very large or very small value of \(K_{\text {eq }}\) is difficult, though not impossible, to alter by changing the mass action ratio. Explain.

Short Answer

Expert verified
Very large or small \(K_{eq}\) means extreme favoring, making changes difficult to achieve.

Step by step solution

01

Understanding Equilibrium Constant

The equilibrium constant, denoted as \(K_{eq}\), is a measure of the position of equilibrium in a reaction. It indicates the ratio of concentrations of products to reactants at equilibrium. A very large \(K_{eq}\) implies that products are much more favored, while a very small \(K_{eq}\) implies reactants are favored.
02

Understanding Mass Action Ratio

The mass action ratio (Q) is the ratio of concentrations of products to reactants at any given point in time. If \(Q < K_{eq}\), the reaction will proceed forward to produce more products. If \(Q > K_{eq}\), the reaction will proceed in the reverse direction to produce more reactants.
03

Effect of Changing Mass Action Ratio

By altering the mass action ratio (Q), you can shift the reaction toward products or reactants. However, if \(K_{eq}\) is extremely large or small, the system already deeply favors one side. The amount of change in Q needed to significantly affect the reaction direction is quite substantial, making it difficult to achieve with practical concentrations and conditions.
04

Conclusion

While changing the mass action ratio can theoretically alter reaction direction, in cases of very large or small \(K_{eq}\), significant changes in concentration are required. This makes altering reaction direction difficult but not impossible.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Mass Action Ratio
The mass action ratio, commonly referred to as Q, is a snapshot of the state of a reaction at any point in time. It represents the ratio of the concentrations of products to reactants. This can be compared to the equilibrium constant, \( K_{eq} \), to understand which direction a reaction will shift to reach equilibrium.
When Q equals \( K_{eq} \), the reaction is at equilibrium. However, if Q is less than \( K_{eq} \), the system is not at equilibrium and will try to increase the concentration of products; thus, the reaction moves forward. Conversely, if Q is greater than \( K_{eq} \), the system will shift towards the reactants, reversing the reaction.
  • Q < \( K_{eq} \): Reaction favors forward direction, increasing products.
  • Q = \( K_{eq} \): Reaction is at equilibrium, no shift occurs.
  • Q > \( K_{eq} \): Reaction favors reverse direction, increasing reactants.
Understanding Q is crucial because it provides insight into how close a system is to equilibrium and which way the reaction tends to go.
Reaction Direction
The direction in which a reaction proceeds can be influenced by comparing the mass action ratio (Q) with the equilibrium constant (\( K_{eq} \)). Since \( K_{eq} \) is a fixed value for a given reaction at a specific temperature, any variation in the reaction's Q compared to \( K_{eq} \) offers insights into its direction and behavior.
A reaction with a large \( K_{eq} \) signifies that, at equilibrium, the concentration of products is significantly higher than that of reactants, favoring product formation. If Q is significantly lower than this \( K_{eq} \), the reaction will proceed strongly in the forward direction. On the other hand, a small \( K_{eq} \) would indicate reactants are favored, and a high Q would push the reaction in reverse.

Although changing Q can theoretically alter reaction paths, especially when \( K_{eq} \) is extremely large or small, the practical application of this approach involves overcoming substantial barriers. These barriers could arise from the sheer magnitudes of concentration changes required to significantly influence reaction direction.
Concentration Changes
Concentration changes are one of the key factors that can influence the mass action ratio, and subsequently, the direction of a reaction. Making meaningful modifications to concentrations is easier said than done, especially when dealing with reactions having very large or small \( K_{eq} \).
In practice, altering concentrations means adding or removing products or reactants from the system. For reactions where \( K_{eq} \) is extremely large, you would need to drastically increase the amount of reactants to shift the reaction backward, given that the equilibrium heavily favors products. On the flip side, with very small \( K_{eq} \), the reactants are inherently favored, calling for a substantial increase in products to drive the reaction forward.
  • Large \( K_{eq} \): Need significant increases in reactants to affect the direction.
  • Small \( K_{eq} \): Need substantial increases in products for forward movement.
Therefore, while changes in concentration can potentially affect reaction pathways, the magnitude of change required for reactions with extreme \( K_{eq} \) values is often impractical under normal conditions.

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Most popular questions from this chapter

The multi-component enzyme aspartate transcarbamoylase catalyzes the formation of \(N\)-carbamoylaspartate from carbamoyl phosphate and aspartate. Arthur Pardee has demonstrated that this reaction is the first step unique to the biosynthesis of pyrimidines, including cytosine, thymine and uracil, major components of nucleic acids. Aspartate transcarbamoylase has at least two stable folded conformations, known as \(R\) (high substrate affinity) and T (low substrate affinity). Interestingly, the relative stability of the \(\mathrm{T}\) and \(\mathrm{R}\) states is affected by the binding of ATP (a purine) to R and CTP (a pyrimidine) to T, a topic covered in Chapter 7. Measurement of the standard state free energy difference between \(\mathrm{R}\) and \(\mathrm{T}\) in the absence of ATP and CTP yielded the value \(3.3 \mathrm{kcal} \mathrm{mol}^{-1}\). Calorimetric determination of \(\Delta \mathrm{H}^{\circ}\) for the transition was \(-6 \mathrm{kcal} \mathrm{mol}^{-1}\). Calculate the standard state entropy change for the \(\mathrm{T} \rightarrow \mathrm{R}\) transition.

Calculate \(K_{\mathrm{eq}}\) for the hydrolysis of the following compounds at neutral \(\mathrm{pH}\) and \(25^{\circ} \mathrm{C}\) : phosphoenolpyruvate \(\left(\Delta G^{\circ \prime}=-61.9\right.\) \(\mathrm{kJ} \mathrm{mol}^{-1}\) ), pyrophosphate \(\left(\Delta G^{\circ}=-33.5 \mathrm{~kJ} \mathrm{~mol}^{-1}\right)\), and glucose1-phosphate \(\left(\Delta G^{\circ}=-20.9 \mathrm{~kJ} \mathrm{~mol}^{-1}\right) .\) Assume that the equilibrium constant includes water, accounting for the possibility that the water concentration is relatively low, as in the cell. These compounds are involved in the glycolytic pathway.

Calculate \(\Delta G^{\circ}\left(25{ }^{\circ} \mathrm{C}\right)\) for \(K_{\mathrm{eq}}=0.001,0.01,0.1,1,10,100\), and 1000 .

\(\Delta G\) cannot generally be equated with \(\Delta G^{\circ}\). To a very good first approximation \(\Delta H\) can be equated with \(\Delta H^{\circ}\). Explain.

When a photon in the visible range is absorbed in the retina by rhodopsin, the photoreceptor in rod cells, 11-cis-retinal is converted to the all-trans isomer. Light energy is transformed into molecular motion. The efficiency of photons to initiate the reaction is about \(20 \%\) at \(500 \mathrm{~nm}\left(57 \mathrm{kcal} \mathrm{mol}^{-1}\right)\). About \(50 \%\) of the absorbed energy is available for the next signaling step. This process takes about \(10 \mathrm{~ms}\). In the absence of light, spontaneous isomerization of 11 -cis- retinal is very slow, on the order of \(0.001 \mathrm{yr}^{-1}\) ! Experimental studies have shown that the equilibrium energetics of retinal isomerization are \(\Delta S^{\circ}=4.4 \mathrm{cal}\) \(\mathrm{mol}^{-1} \mathrm{~K}^{-1}\) and \(\Delta \mathrm{H}^{\circ}=150 \mathrm{cal} \mathrm{mol}^{-1}\). Calculate the equilibrium constant for the reaction.

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