The protein lysozyme unfolds at a transition temperature of \(75.5^{\circ} \mathrm{C}\) and the standard enthalpy of transition is \(509 \mathrm{k}) \mathrm{mol}^{-1} .\) Calculate the entropy of unfolding of lysozyme at \(25.0^{\circ} \mathrm{C}\), given that the difference in the constantpressure heat capacities upon unfolding is \(6.28 \mathrm{~kJ} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\) and can be assumed to be independent of temperature. Hint. Imagine that the transition at \(25.0^{\circ} \mathrm{C}\) occurs in three steps: (i) heating of the folded protein from \(25.0^{\circ} \mathrm{C}\) to the transition temperature, (ii) unfolding at the transition temperature, and function, the entropy change at \(25.0^{\circ} \mathrm{C}\) is equal to the sum of the entropy changes of the steps. (iii) cooling of the unfolded protein to \(25.0^{\circ} \mathrm{C}\). Because the entropy is a state

Short Answer

Expert verified
The entropy of unfolding of lysozyme at 25.0°C is calculated as the sum of the entropy changes of heating the protein to the transition temperature, unfolding it, and then cooling the unfolded protein back to 25.0°C.

Step by step solution

01

Convert temperatures to Kelvin

Before we calculate the entropy changes for each step, we need to convert the Celsius temperatures to Kelvin. For this conversion, we use the formula: Kelvin = Celsius + 273.15.
02

Calculate the entropy change for heating the folded protein

The change in entropy for heating the folded protein from 25.0°C to 75.5°C is calculated using the heat capacity at constant pressure. The formula to use is \( \Delta S_1 = C_p \cdot \ln(\frac{T_2}{T_1}) \), where \( C_p \) is the heat capacity, \( T_1 \) is the initial temperature, and \( T_2 \) is the final temperature. Both \( T_1 \) and \( T_2 \) should be in Kelvin.
03

Calculate the entropy change for unfolding at the transition temperature

The entropy change for the unfolding of the protein at the transition temperature can be calculated by the formula \( \Delta S_2 = \frac{\Delta H}{T_{\text{transition}}} \), where \( \Delta H \) is the standard enthalpy of transition and \( T_{\text{transition}} \) is the transition temperature in Kelvin.
04

Calculate the entropy change for cooling the unfolded protein

Now, we must calculate the entropy change for cooling the unfolded protein from the transition temperature back to 25.0°C using the same formula as in Step 2: \( \Delta S_3 = C_p \cdot \ln(\frac{T_1}{T_2}) \), with the unfolded protein's constant pressure heat capacity. Since the heat capacity is assumed to be independent of temperature, we use the same value, but we swap \( T_1 \) and \( T_2 \) because the protein is cooling.
05

Calculate the total entropy change

The total entropy change for the entire process is the sum of the entropy changes for each step: \( \Delta S_{\text{total}} = \Delta S_1 + \Delta S_2 - \Delta S_3 \). The sum is determined by the principle that entropy is a state function and the net change is path independent.

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Key Concepts

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

Enthalpy of Transition
The enthalpy of transition refers to the energy absorbed or released during a phase transition of a substance at a constant temperature and pressure. In the context of protein thermodynamics, this concept becomes critical when studying protein folding and unfolding events. For instance, lysozyme unfolding occurs with a standard enthalpy of transition of 509 kJ/mol. This figure represents the amount of energy required to disrupt the structural conformation of the protein from its folded state to an unfolded one at the transition temperature.

To put this into perspective for students, imagine the protein structure as a complex origami creation. The enthalpy of transition would be analogous to the energy you need to input to unfold every crease and return the paper to its flat, original state. In the thermal process of proteins, this correlates closely with the protein's stability and its interaction with the surrounding environment.
Constant-Pressure Heat Capacity
Heat capacity at constant pressure, often denoted as Cp, is an important physical quantity that tells us how much heat is required to raise the temperature of a substance by one degree Kelvin at constant pressure. When it comes to proteins like lysozyme, the difference in Cp upon unfolding is significant because it signals changes in the protein's interaction with the solvent when transitioning from a folded to an unfolded state.

The provided problem mentions that the heat capacity change upon unfolding is 6.28 kJ/K mol. It's useful to imagine this as a measure of the 'thermal responsiveness' of the protein. After lysozyme unfolds, its increased heat capacity implies that the unfolded state is more responsive to temperature changes – it absorbs more heat to increase in temperature than when it was in its folded state. This change is crucial for accurately calculating the unfolding process' overall entropy change.
Entropy Change Calculation
Entropy change is a way to quantify the disorder or randomness in a system. The calculation of entropy change for the unfolding of lysozyme involves considering the protein's transition in stages and accounting for the heat exchanged during heating, unfolding, and cooling processes.

For students, a useful analogy is to think of entropy as a measure of how many ways you can rearrange the pieces of a puzzle without affecting the big picture. When lysozyme unfolds, it goes from a highly-ordered state to a less-ordered one, increasing the system's entropy. By incrementally computing the entropy changes for each stage of the temperature change and unfolding process, as seen in the exercise steps, one can determine the total entropy change of the system. This step-wise approach helps simplify and understand a complex concept, ensuring that the entropy change at any temperature can be calculated.
Protein Thermodynamics
Protein thermodynamics encompasses the study of energetic aspects of protein folding and stability. It deals with how proteins respond to changes in temperature and other environmental factors, important for understanding processes like lysozyme unfolding. Thermodynamics provides the scientific base for understanding the energetics behind protein conformation changes – involving enthalpy, entropy, and heat capacity among other variables.

In discussing protein thermodynamics, remember that each protein's folding path is unique, yet governed by the universal principles of thermodynamics. Enthalpy changes reflect the interactions within the protein and with its environment, while entropy changes reveal the degrees of freedom or disorder. Heat capacity changes give insight into the protein's thermal stability. By piecing these elements together, as seen in the exercise solution, a comprehensive understanding of the protein unfolding process emerges, which is essential for fields such as biochemistry and molecular biology.

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Most popular questions from this chapter

The heat capacity of chloroform (trichloromethane, \(\mathrm{CHCl}_{3}\) ) in the range \(240 \mathrm{~K}\) to \(330 \mathrm{~K}\) is given by \(\mathrm{C}_{\mathrm{p}, \mathrm{m}} /\left(\mathrm{J} \mathrm{K}^{-1} \mathrm{~mol}^{-1}\right)=91.47+7.5 \times 10^{-2}(T / \mathrm{K})\). In a particular experiment, \(1.00 \mathrm{~mol} \mathrm{CHCl}_{3}\) is heated from \(273 \mathrm{~K}\) to \(300 \mathrm{~K}\). Calculate the change in molar entropy of the sample.

A gaseous sample consisting of \(1.00 \mathrm{~mol}\) molecules is described by the equation of state \(p \mathrm{~V}_{\mathrm{m}}=\mathrm{RT}(1+B p)\). Initially at \(373 \mathrm{~K}\), it undergoes JouleThomson expansion from \(100 \mathrm{~atm}\) to \(1.00 \mathrm{~atm}\). Given that \(C_{p, \mathrm{~m}}=R, \mu=0.21\) \(\mathrm{K} \mathrm{atm}^{-1}, B=-0.525(\mathrm{~K} / T) \mathrm{atm}^{-1}\), and that these are constant over the temperature range involved, calculate \(\Delta T\) and \(\Delta S\) for the gas.

Show that, for a perfect gas, \((\partial U / \partial S)_{v}=T\) and \((\partial U / \partial V)_{\mathrm{S}}=-p\).

Consider a perfect gas contained in a cylinder and separated by a frictionless adiabatic piston into two sections \(\mathrm{A}\) and \(\mathrm{B}\). All changes in \(\mathrm{B}\) is isothermal; that is, a thermostat surrounds \(\mathrm{B}\) to keep its temperature constant. There is \(2.00 \mathrm{~mol}\) of the gas in each section. Initially, \(T_{\mathrm{A}}==T_{\mathrm{B}}=300 \mathrm{~K}, V_{\mathrm{A}}=\) \(V_{\mathrm{B}}\) \(=2.00 \mathrm{dm}^{3}\). Energy is supplied as heat to Section A and the piston moves to the right reversibly until the final volume of Section B is \(1.00 \mathrm{dm}^{3}\). Calculate (a) \(\Delta S_{\mathrm{A}}\) and \(\Delta S_{\mathrm{B}}\), (b) \(\Delta A_{\mathrm{A}}\) and \(\Delta \mathrm{A}_{\mathrm{B}}\), (c) \(\Delta G_{\mathrm{A}}\) and \(\Delta G_{\mathrm{B}}\), (d) AS of the total system and its surroundings. If numerical values cannot be obtained, indicate whether the values should be positive, negative, or zero or are indeterminate from the information given. (Assume \(C_{\mathrm{v}, \mathrm{m}}=20 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\).)

The cycle involved in the operation of an internal combustion engine is called the Otto cycle. Air can be considered to be the working substance and can be assumed to be a perfect gas. The cycle consists of the following steps: (1) reversible adiabatic compression from \(\mathrm{A}\) to \(\mathrm{B},(2)\) reversible constantvolume pressure increase from \(\mathrm{B}\) to \(\mathrm{C}\) due to the combustion of a small amount of fuel, (3) reversible adiabatic expansion from \(\mathrm{C}\) to \(\mathrm{D}\), and \((4)\) reversible and constant-volume pressure decrease back to state A. Determine the change in entropy (of the system and of the surroundings) tor each step of the cycle and determine an expression for the efficiency of the cycle, assuming that the heat is supplied in Step \(2 .\) Evaluate the efficiency for a compression ratio of \(10: 1\). Assume that, in state \(\mathrm{A}, \mathrm{V}=4.00 \mathrm{dm}^{3}, p=1.00 \mathrm{~atm}\), and \(\mathrm{T}=300 \mathrm{~K}\), that \(\mathrm{V}_{\mathrm{A}}==10 \mathrm{~V}_{\mathrm{B}}, \mathrm{p}_{\mathrm{C}} / \mathrm{p}_{\mathrm{B}}=5\), and that \(C_{p m}=\frac{2}{2} R\).

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