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As an approximation, we can assume that proteins exist either in the native (physiologically functioning) state or the denatured state. 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 denaturation becomes spontaneous.

Short Answer

Expert verified
Denaturation becomes spontaneous at 320 K.

Step by step solution

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01

Understanding the Problem

We need to find the temperature at which the denaturation process becomes spontaneous, given the enthalpy and entropy values for a specific protein. The process is spontaneous when the change in Gibbs free energy, \(\Delta G\), is zero.
02

Introducing Gibbs Free Energy Equation

The equation for Gibbs free energy change is \(\Delta G = \Delta H - T\Delta S, \\) where \(\Delta H\) is the change in enthalpy, \(\Delta S\) is the change in entropy, and \(T\) is the temperature in Kelvin.
03

Signs and Magnitudes

The enthalpy change, \(\Delta H = 512\ \mathrm{kJ/mol}\), is positive, indicating that denaturation is an endothermic process. The entropy change, \(\Delta S = 1.60\ \mathrm{kJ/K \cdot mol}\), is also positive, suggesting an increase in disorder as the protein denatures.
04

Setting \\(\Delta G = 0\\) for Spontaneity

To find the temperature at which the process becomes spontaneous (\(\Delta G = 0\)), set the equation to zero: \(0 = 512\ \mathrm{kJ/mol} - T\times1.60\ \mathrm{kJ/K \cdot mol}\).
05

Solving for Temperature

Rearrange the equation to solve for \(T\):\[T = \frac{512\ \mathrm{kJ/mol}}{1.60\ \mathrm{kJ/K \cdot mol}}\]
06

Calculating the Spontaneous Temperature

Perform the division to find \(T\):\[T = \frac{512}{1.60} = 320\ \mathrm{K}\]

Key Concepts

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

Thermodynamics
Thermodynamics is at the heart of understanding protein denaturation. This branch of physics describes how energy is transferred and transformed in systems. In the case of proteins, we're interested in how proteins move from one state to another, specifically from their native, functional form to a denatured form.
This process involves changes in energy, particularly heat energy, which is precisely what thermodynamics deals with.
When considering the denaturation process, we focus on two specific thermodynamic properties: enthalpy and entropy. Both of these properties help us understand the energy changes and disorder in the system during denaturation.
  • Enthalpy refers to the heat content of the system, which changes during denaturation.
  • Entropy measures the disorder or randomness in the system, which usually increases upon denaturation.
As we proceed through these concepts, you'll see how thermodynamics principles apply to the spontaneity of the denaturation, revealed through calculations involving Gibbs free energy.
Gibbs Free Energy
Gibbs free energy is a crucial concept when studying the spontaneity of chemical reactions, including protein denaturation. It combines enthalpy and entropy to determine if a process will occur spontaneously. The formula used is \[\Delta G = \Delta H - T\Delta S\]where \(\Delta G\) is the change in Gibbs free energy, \(\Delta H\) is the change in enthalpy, \(T\) is the temperature in Kelvin, and \(\Delta S\) is the change in entropy.
A spontaneous process implies that it can occur without any energy input from outside the system. This happens when \(\Delta G\) is less than zero. However, the denaturation process is considered spontaneous when \(\Delta G = 0\), after which the process can proceed without further energy.
The calculation of the exact temperature where denaturation becomes spontaneous involves setting \(\Delta G\) equal to zero and solving for \(T\), giving us a clear threshold based on entropy and enthalpy values.
Enthalpy
Enthalpy, denoted as \(\Delta H\), represents the heat content of a system. During protein denaturation, \(\Delta H\) is positive, signifying that the process absorbs energy from its surroundings - an endothermic reaction.
This absorption means the protein requires an input of thermal energy to unfold and move into a denatured state. In the exercise, the positive value of \(512\ \mathrm{kJ/mol}\) reveals that it takes a significant amount of energy to disrupt the structured state of the protein.
Understanding enthalpy in this context is important because it sets the stage for why energy needs to be added for denaturation. This value helps identify the energy threshold that needs to be overcome for the process to reach a point of spontaneity or equilibrium, where enthalpy changes balance out with temperature and entropy.
Entropy
Entropy, symbolized as \(\Delta S\), describes the level of disorder or randomness in a system. In the process of protein denaturation, \(\Delta S\) is positive, suggesting an increase in disorder. As the protein unfolds, it transitions from a highly ordered state to a less ordered one.
This change is indicated by an entropy value of \(1.60\ \mathrm{kJ/K \cdot mol}\). This increase in entropy is a key driver for the spontaneity of the process, as systems tend to move towards states of higher disorder naturally.
By considering entropy, alongside enthalpy, we can predict whether the unfolding process will happen spontaneously at a particular temperature. The interplay of enthalpy and entropy influences the Gibbs free energy, and through this lens, we can pinpoint the exact conditions under which denaturation becomes a spontaneous event.

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Most popular questions from this chapter

Ammonium nitrate \(\left(\mathrm{NH}_{4} \mathrm{NO}_{3}\right)\) dissolves spontaneously and endothermically in water. What can you deduce about the sign of \(\Delta S\) for the solution process?

.Under what conditions does a substance have a standard entropy of zero? Can an element or a compound ever have a negative standard entropv?

Consider two carboxylic acids (acids that contain the \(-\) COOH group \(): \mathrm{CH}_{3} \mathrm{COOH}\) (acetic acid, \(\left.K_{\mathrm{a}}=1.8 \times 10^{-5}\right)\) and \(\mathrm{CH}_{2} \mathrm{ClCOOH}\) (chloroacetic acid, \(K_{\mathrm{a}}=1.4 \times 10^{-3}\) ). (a) Calculate \(\Delta G^{\circ}\) for the ionization of these acids at \(25^{\circ} \mathrm{C}\). (b) From the equation \(\Delta G^{\circ}=\Delta H^{\circ}-T \Delta S^{\circ},\) we see that the contributions to the \(\Delta G^{\circ}\) term are an enthalpy term \(\left(\Delta H^{\circ}\right)\) and a temperature times entropy term \(\left(T \Delta S^{\circ}\right)\). These contributions are listed here for the two acids: \begin{tabular}{lcc} & \(\Delta \boldsymbol{H}^{\circ}(\mathbf{k} \mathbf{J} / \mathbf{m o l})\) & \(T \Delta S^{\circ}(\mathbf{k} \mathbf{J} / \mathbf{m o l})\) \\ \hline \(\mathrm{CH}_{3} \mathrm{COOH}\) & -0.57 & -27.6 \\\ \(\mathrm{CH}_{2} \mathrm{ClCOOH}\) & -4.7 & -21.1 \end{tabular} Which is the dominant term in determining the value of \(\Delta G^{\circ}\) (and hence \(K_{\mathrm{a}}\) of the acid)? (c) What processes contribute to \(\Delta H^{\circ} ?\) (Consider the ionization of the acids as a Bronsted acid-base reaction.) (d) Explain why the \(T \Delta S^{\circ}\) term is more negative for \(\mathrm{CH}_{3} \mathrm{COOH} .\)

Why is it more convenient to predict the direction of a reaction in terms of \(\Delta G_{\mathrm{sys}}\) instead of \(\Delta S_{\text {univ }}\) ? Under what conditions can \(\Delta G_{\mathrm{sys}}\) be used to predict the spontaneity of a reaction?

Which of the following thermodynamic functions are associated only with the first law of thermodynamics: \(S, U, G,\) and \(H ?\)

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