91Ó°ÊÓ

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.

Step by step solution

01

Understand the Formula for Equilibrium Constant

The equilibrium constant \( K \) is related to the changes in Gibbs free energy \( \Delta G^{\circ} \) by the equation:\[\Delta G^{\circ} = -RT \ln K\]where \( R \) is the universal gas constant \( 1.987 \, \text{cal mol}^{-1} \text{K}^{-1} \) and \( T \) is the temperature in Kelvin. To find \( K \), we first need \( \Delta G^{\circ} \).

02

Calculate ΔG° Using Enthalpy and Entropy

First, calculate \( \Delta G^{\circ} \) using the formula:\[\Delta G^{\circ} = \Delta H^{\circ} - T \Delta S^{\circ}\]Substitute \( \Delta H^{\circ} = 150 \, \text{cal mol}^{-1} \), \( \Delta S^{\circ} = 4.4 \, \text{cal mol}^{-1} \text{K}^{-1} \), and because most biochemical reactions are at body temperature, we use \( T = 298 \, \text{K} \):\[\Delta G^{\circ} = 150 - 298 \times 4.4 = 150 - 1311.2 = -1161.2 \, \text{cal mol}^{-1}\]

03

Use ΔG° to Find the Equilibrium Constant K

Using the relationship \( \Delta G^{\circ} = -RT \ln K \), substitute the values \( \Delta G^{\circ} = -1161.2 \, \text{cal mol}^{-1} \), \( R = 1.987 \, \text{cal mol}^{-1} \text{K}^{-1} \), and \( T = 298 \, \text{K} \):\[-1161.2 = -1.987 \times 298 \times \ln K\]First solve for \( \ln K \):\[\ln K = \frac{-1161.2}{-1.987 \times 298} \approx 1.968\]Now, find \( K \) by solving the exponential:\[K = e^{1.968} \approx 7.16\]

04

Conclusion

The calculated equilibrium constant \( K \approx 7.16 \) shows how favorable the formation of the all-trans isomer is compared to the 11-cis form under standard conditions.

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Key Concepts

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

Thermodynamics

Thermodynamics is a branch of physics that deals with the relations between heat and other forms of energy. In the context of chemical reactions, it helps us understand how energy is transformed and utilized.
When light energy is absorbed by rhodopsin in the retina, it is converted into molecular motion — this is a thermodynamic process. We often examine these transformations using parameters like energy, enthalpy, and entropy.

• Energy exchange: How energy is gained or lost.
• Work done: How energy is used to perform work.
• System and surroundings: The system is what we are focused on, while surroundings are everything else.

For our exercise, we look at how light, as a form of energy, drives the isomerization reaction of retinal in thermodynamic terms.

Rhodopsin Photoreception

Rhodopsin is a light-sensitive receptor protein found in the retina. It plays a crucial role in visual photoreception, particularly in low-light conditions.
Upon absorption of a photon, rhodopsin undergoes a structural change, specifically in its retinal component, converting from 11-cis-retinal to all-trans-retinal.

• This change is crucial for activating the phototransduction cascade, which ultimately results in visual perception.
• Rhodopsin's efficiency: In this exercise, we are given that only 20% of absorbed photons effectively start this reaction.

Understanding rhodopsin photoreception offers insight into the bio-thermodynamic processes that convert light into a chemical signal, emphasizing the importance of these biological systems.

Gibbs Free Energy

Gibbs Free Energy (\(\Delta G\)) is a concept in thermodynamics that helps us determine the spontaneity of a reaction. It reflects the balance of enthalpy (heat content) and entropy (disorder) at constant temperature and pressure.
This quantity tells us whether a reaction will proceed on its own without any external energy input.

• If \(\Delta G < 0\), the reaction is spontaneous and will proceed forward.
• If \(\Delta G = 0\), the reaction is at equilibrium, with no net change.
• If \(\Delta G > 0\), the reaction is non-spontaneous as written.

In our problem, we used the relationship \(\Delta G^{\circ} = \Delta H^{\circ} - T \Delta S^{\circ}\). Calculating \(\Delta G^{\circ}\) enabled us to find the equilibrium constant, giving us insight into the reaction's favorability.

Enthalpy and Entropy

Enthalpy (\(\Delta H\)) and Entropy (\(\Delta S\)) are two fundamental concepts in thermodynamics helping describe the energy landscape of chemical reactions.
Enthalpy refers to the total heat content in a system and reflects the energy needed to break and form bonds during a reaction.

• A positive \(\Delta H\) means the reaction absorbs energy (endothermic).
• A negative \(\Delta H\) indicates the release of energy (exothermic).

Entropy, on the other hand, measures the degree of randomness or disorder in a system.

• A positive \(\Delta S\) reflects an increase in disorder.
• A negative \(\Delta S\) suggests a decrease in disorder.

In our exercise, \(\Delta H^{\circ} = 150\)cal/mol indicates some heat absorption, while \(\Delta S^{\circ} = 4.4\)cal/mol/K suggests an increase in disorder. These values help inform the calculation of \(\Delta G^{\circ}\) and determine the equilibrium position of the reaction.