Problem 14 - Biochemistry

Question

You are studying the various components of the venom of a poisonous lizard. One of the venom components is a protein that appears to be temperature sensitive. When heated, it denatures and is no longer toxic. The process can be described by the following simple equation: \\[ \mathbf{T}(\text { toxic }) \rightleftharpoons \mathrm{N} \text { (nontoxic) } \\] There is only enough protein from this venom to carry out two equilibrium measurements. At \(298 \mathrm{K}\), you find that \(98 \%\) of the protein is in its toxic form. However, when you raise the temperature to \(320 \mathrm{K},\) you find that only \(10 \%\) of the protein is in its toxic form. a. Calculate the equilibrium constants for the T to N conversion at these two temperatures. b. Use the data to determine the \(\Delta H^{\circ}, \Delta S^{\circ},\) and \(\Delta G^{\circ}\) for this process.

Step-by-Step Solution

Answer

The equilibrium constants at 298 K and 320 K are 0.0204 and 9, respectively. The calculated values for the changes are: \(\Delta H = 57.63 kJ/mol\), \(\Delta S = 152 J/mol.K\), \(\Delta G_{298} = 14.86 kJ/mol\), and \(\Delta G_{320} = 8.73 kJ/mol\).

Step 1: Calculate the equilibrium constants

We can calculate the equilibrium constant,\(K\), as \(K = [N]/[T]\) where [N] and [T] are the concentrations of nontoxic and toxic protein, respectively. At 298 K, 98% of the protein is in its toxic form. Therefore, \(K_{298} = (100-98)/98 = 0.0204\). Similarly, at 320 K, 10% of the protein is in its toxic form. So, \(K_{320} = (100-10)/10 = 9\).

Step 2: Use the Van't Hoff equation

We can use the Van't Hoff equation to relate the equilibrium constants at two temperatures with the change in enthalpy (\(\Delta H\)). The equation is given by \(\ln(K_{2}/K_{1}) = -\Delta H/R(1/T_{2} - 1/T_{1})\), where \(K_{1}\) and \(K_{2}\) are equilibrium constants at temperatures \(T_{1}\) and \(T_{2}\), respectively, and \(R\) is the universal gas constant. Substituting given values, we obtain- \(\ln(9/0.0204) = -\Delta H/8.314(1/320 - 1/298)\). Simplifying and solving for \(\Delta H\) gives \(\Delta H = 57.63 kJ/mol\).

Step 3: Calculate the change in entropy

Now, with the calculated \(\Delta H\), we can determine \(\Delta S\) by dividing \(\Delta H\) by the absolute temperature (in Kelvin), as shown in the relationship \(K = e^{(-\Delta H/\Delta SRT)}\) where R is the universal gas constant. At 298 K, \(K_{298} = 0.0204\) so \(\Delta S = \Delta H / (298 K * \ln 0.0204) = 152 J/mol.K\)

Step 4: Calculate the change in Gibbs free energy

\(\Delta G\) can be calculated using the Gibbs-Helmholtz equation, \(\Delta G = \Delta H - T\Delta S\). Substituting our known values at 298 K: \(\Delta G_{298} = 57.63 kJ/mol - 298 K * 152 J/mol.K = 14.86 kJ/mol\). Similarly, at 320 K, \(\Delta G_{320} = 57.63 kJ/mol - 320 K * 152 J/mol.K = 8.73 kJ/mol\).

Key Concepts

Understanding Protein Denaturation and Equilibrium Constants

When studying reactions such as the denaturation of a protein, a key concept to grasp is the idea of equilibrium constants. These constants, often denoted as K, provide a quantitative measure of the position of equilibrium in a reversible reaction.

In the context of protein denaturation, we consider the shift from a toxic form (T) to a nontoxic form (N) upon heating. The equilibrium constant for this shift can be calculated using the ratio of the concentrations of the nontoxic to the toxic forms, represented by \(K = [N]/[T]\). When most of the protein is in its toxic form, the equilibrium constant is low, indicating that the reaction favors the toxic state. However, as temperature increases, denaturation occurs, and K increases, suggesting a shift in favor of the nontoxic form.

To better understand the calculations, consider a situation where 98% of the protein is toxic at 298 K. The equilibrium constant \(K_{298} = (100-98)/98 = 0.0204\). At 320 K, with only 10% of the protein being toxic, the equilibrium constant significantly increases to \(K_{320} = (100-10)/10 = 9\). This demonstrates the sensitive dependence of the protein structure on temperature, as well as the use of equilibrium constants to describe this relationship.

Deciphering the Van't Hoff Equation and its Role in Protein Behavior

The Van't Hoff equation is immensely valuable for understanding the thermal sensitivity of proteins, as it relates the equilibrium constants at two different temperatures to the change in enthalpy (\(\text{Δ}H\)) of the reaction. This equation is expressed as \(\text{ln}(K_{2}/K_{1}) = -\text{Δ}H/R(1/T_{2} - 1/T_{1})\), where \(K_{1}\) and \(K_{2}\) are the equilibrium constants at respective temperatures \(T_{1}\) and \(T_{2}\), and R is the universal gas constant.

Using the values from the protein denaturation example, we could derive the change in enthalpy, which reflects the amount of heat absorbed or released during the denaturation process. A calculated ΔH indicates how much heat energy is involved in disrupting the protein's structure. After performing the calculations, we find \(ΔH = 57.63 kJ/mol\), signifying that this amount of energy is needed for the toxic form to convert to the nontoxic form.

The Van't Hoff equation thus reveals the enthalpic (heat-related) aspect of the protein's thermal sensitivity and enables predictions about how the equilibrium shifts with temperature changes.

Gibbs Free Energy: The Driving Force Behind Protein Stability

Gibbs free energy (\(ΔG\)) is a fundamental concept in thermodynamics, particularly when assessing the spontaneity of a reaction. It combines the system's enthalpy (\(\text{Δ}H\)) and entropy (\(\text{Δ}S\)) to determine whether a reaction will proceed on its own.

The Gibbs free energy change is defined by the equation \(ΔG = ΔH - TΔS\), where T is the absolute temperature in Kelvin. A negative value of \(ΔG\) signals that the reaction is spontaneous, while a positive value implies non-spontaneity. In the case of protein denaturation, by using the earlier calculated values, we find \(ΔG_{298} = 14.86 kJ/mol\) and \(ΔG_{320} = 8.73 kJ/mol\), indicating that at both temperatures, the denaturation process is not spontaneous due to positive values of \(ΔG\).

Understanding the Gibbs free energy changes during protein denaturation is crucial as it helps explain the balance between enthalpic and entropic forces that govern the protein's state. With increasing temperature, the reduction in \(ΔG\) suggests that denaturation becomes more favorable, aligned with the increasing equilibrium constants observed.