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As an approximation, we can assume that proteins exist either in the native (or physiologically functioning) state and the denatured state $$\text { native } \rightleftharpoons \text { denatured }$$ The standard molar enthalpy and entropy of the denaturation of a certain protein are \(512 \mathrm{~kJ} / \mathrm{mol}\) and \(1.60 \mathrm{~kJ} / \mathrm{K} \cdot \mathrm{mol}\), respectively. Comment on the signs and magnitudes of these quantities, and calculate the temperature at which the process favors the denatured state.

Short Answer

Expert verified
The positive standard molar enthalpy and entropy suggest that the denaturation of the protein is an endothermic and disorder-increasing process. The protein denaturation process begins to favor the denatured state at temperatures greater than 320 K.

Step by step solution

01

Understand the Significance of Given Quantities

The given standard molar enthalpy (\( \Delta H \)) of denaturation is positive, which indicates that the process is endothermic. This means energy is absorbed when the protein shifts from the native to denatured state. The positive standard molar entropy (\( \Delta S \)) suggests an increase in disorder or randomness when the protein denatures.
02

Understanding Gibbs Free Energy and Spontaneity

The condition for spontaneity of a reaction is determined by the Gibbs free energy change (\( \Delta G \)). If \( \Delta G \) is negative, the process is spontaneous. The relationship between \( \Delta G \), \( \Delta H \), and \( \Delta S \) is given by the equation: \( \Delta G = \Delta H - T \Delta S \). We focus on the situation where the denaturation process is spontaneous, or \( \Delta G < 0 \). Therefore, we set up the inequality: \( 0 > \Delta H - T \Delta S \).
03

Calculate the Critical Temperature

The inequality in Step 2 can be rearranged to: \( T > \Delta H / \Delta S \). By substituting the values of \( \Delta H = 512 \, kJ/mol \) (or 512000 \( J/mol \)) and \( \Delta S = 1.60 \, kJ/K\cdot mol \) (or 1600 \( J/K\cdot mol \)), the temperature at which the denaturation process becomes spontaneous is \( T > 512000 / 1600 \) which simplifies to \( T > 320 \, K \). This is the temperature at which the process begins to favor the denatured state.

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Key Concepts

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

Enthalpy in Protein Denaturation
The process of protein denaturation involves the unfolding and loss of the complex structure of a protein. Understanding enthalpy changes during this process is crucial. Enthalpy, represented as \(\Delta H\), can be thought of as the total heat content of a system. When a protein denatures, its enthalpy change is positive, meaning the process is endothermic and requires energy input. For the denaturation of a certain protein, if we consider an enthalpy change of \(512 \text{kJ/mol}\), it tells us that this amount of energy is absorbed to break down the intricate structures of the protein.

Since the enthalpy change is positive, it implies various interactions like hydrogen bonds, ionic bonds, and van der Waals forces that stabilize the protein's native state are being disrupted. The higher the value of \(\Delta H\), the more energy is required to overcome these stabilizing forces which contribute to the protein's structured state. In practical terms, the magnitude of \(\Delta H\) can give insights into the stability and robustness of the protein's folded state and also help predict the conditions under which the protein will denature.
Entropy in Protein Denaturation
Entropy, denoted by \(\Delta S\), is a measure of randomness or disorder in a system. Protein denaturation often leads to an increase in entropy, as seen through the positive value of \(1.60\text{ kJ/K}\cdot\text{mol}\). When a protein unfolds from its highly ordered native state to a less ordered denatured state, its entropy increases because there are more ways to arrange the denatured protein's atoms in space.

A higher entropy value indicates that the protein's molecules have more freedom to move and there are greater possibilities for their arrangements. This is especially important when considering protein function because the structured, low-entropy native state is often necessary for the protein's biological activity. During denaturation, this functional structure is lost, leading to increased entropy. Understanding the entropy changes provides valuable information on not only the protein's behavior upon denaturation but also on the balance between energy usage and disorder within the biological systems.
Gibbs Free Energy and Spontaneity
Gibbs Free Energy (\(\Delta G\)) is a thermodynamic quantity that measures the balance between enthalpy and entropy to determine the spontaneity of a process. A process is spontaneous if it occurs without being driven by an external force. In the context of protein denaturation, we're particularly interested in the temperature at which it becomes spontaneous.

Using the equation \(\Delta G = \Delta H - T\Delta S\), we can assess the spontaneity. At temperatures where \(\Delta G\) is negative, the denaturation is spontaneous, meaning it will occur without external energy input. To find this critical temperature (\(T\)), we rearrange the equation to \(T > \frac{\Delta H}{\Delta S}\) using the provided enthalpy and entropy values. This calculation yields a temperature above which the protein's denaturation is favored. It's fascinating to wield such thermodynamic equations to predict biological behavior, revealing an intersection between physics and biology that is both intricate and immensely practical for understanding life at a molecular level.

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Most popular questions from this chapter

For reactions carried out under standard-state conditions, Equation (18.10) takes the form \(\Delta G^{\circ}=\Delta H^{\circ}-\) \(T \Delta S^{\circ} .\) (a) Assuming \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) are independent of temperature, derive the equation $$\ln \frac{K_{2}}{K_{1}}=\frac{\Delta H^{\circ}}{R}\left(\frac{T_{2}-T_{1}}{T_{1} T_{2}}\right)$$ where \(K_{1}\) and \(K_{2}\) are the equilibrium constants at \(T_{1}\) and \(T_{2}\), respectively. (b) Given that at \(25^{\circ} \mathrm{C} K_{\mathrm{c}}\) is \(4.63 \times 10^{-3}\) for the reaction $$ \mathrm{N}_{2} \mathrm{O}_{4}(g) \rightleftharpoons 2 \mathrm{NO}_{2}(g) \quad \Delta H^{\circ}=58.0 \mathrm{~kJ} / \mathrm{mol} $$ calculate the equilibrium constant at \(65^{\circ} \mathrm{C}\)

How does the entropy of a system change for each of the following processes? (a) A solid melts. (b) A liquid freezes. (c) A liquid boils. (d) A vapor is converted to a solid. (e) A vapor condenses to a liquid. (f) A solid sublimes. (g) Urea dissolves in water.

From the values of \(\Delta H\) and \(\Delta S,\) predict which of the following reactions would be spontaneous at \(25^{\circ} \mathrm{C}\) : Reaction \(\mathrm{A}: \Delta H=10.5 \mathrm{~kJ} / \mathrm{mol}, \Delta S=30 \mathrm{~J} / \mathrm{K} \cdot \mathrm{mol}\) reaction \(\mathrm{B}: \Delta H=1.8 \mathrm{~kJ} / \mathrm{mol}, \Delta S=-113 \mathrm{~J} / \mathrm{K} \cdot \mathrm{mol}\). If either of the reactions is nonspontaneous at \(25^{\circ} \mathrm{C}\), at what temperature might it become spontaneous?

Hydrogenation reactions (for example, the process of converting \(\mathrm{C}=\mathrm{C}\) bonds to \(\mathrm{C}-\mathrm{C}\) bonds in food industry) are facilitated by the use of a transition metal catalyst, such as Ni or \(\mathrm{Pt}\). The initial step is the adsorption, or binding, of hydrogen gas onto the metal surface. Predict the signs of \(\Delta H, \Delta S,\) and \(\Delta G\) when hydrogen gas is adsorbed onto the surface of Ni metal.

Give a detailed example of each of the following, with an explanation: (a) a thermodynamically spontaneous process; (b) a process that would violate the first law of thermodynamics; (c) a process that would violate the second law of thermodynamics; (d) an irreversible process; (e) an equilibrium process.

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