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When most biologic enzymes are heated, they lose their catalytic activity. The change Original enzyme \(\longrightarrow\) new form that occurs on heating is endothermic and spontaneous. Is the structure of the original enzyme or its new form more ordered (has the smaller positional probability)? Explain.

Short Answer

Expert verified
The structure of the original enzyme is more ordered (has smaller positional probability) compared to its new form after heating. This is because the process of heating leads to an increase in entropy, which implies a more disordered state with higher positional probability.

Step by step solution

01

What happens to enzymes upon heating

When most biologic enzymes are heated, they lose their catalytic activity. This occurs because heat causes the enzyme to denature, which means its molecular structure changes. The original enzyme transitions to a new form with altered molecular structure and reduced enzyme activity. This process is endothermic (heat is absorbed) and spontaneous (occurs without external input).
02

Discuss spontaneity

A spontaneous process is one that occurs naturally without any external input of energy. In thermodynamics, spontaneity is governed by the concept of Gibbs free energy (G). A process is spontaneous when \(\Delta G < 0\). Gibbs free energy is a function of enthalpy (H), entropy (S), and temperature (T), and can be calculated using the formula \(\Delta G = \Delta H - T\Delta S\). In this problem, since the process is endothermic and spontaneous, it means that \(\Delta H > 0\) and \(\Delta S > 0\).
03

Entropy and positional probability

Entropy (S) is a measure of the disorder or randomness in a system. A system with higher entropy has a higher positional probability, which means its molecules are more disordered and less constrained in their movement. Conversely, a system with lower entropy has lower positional probability and is more ordered. In this problem, when the enzyme undergoes a change due to heating, the entropy of the system increases (\(\Delta S > 0\)).
04

Determine which enzyme form is more ordered

Since the increase in entropy (\(\Delta S > 0\)) suggests that the final state has higher positional probability (more disordered) than the initial state, we can conclude that the structure of the original enzyme is more ordered (has smaller positional probability) compared to its new form after heating.

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Key Concepts

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

Entropy
Entropy is a measure of the disorder or randomness within a system. When an enzyme is heated, its structure becomes more random and less organized, which is reflected by an increase in entropy. This means that the molecules that make up the enzyme have more freedom in their movement. As a result, the positional probability, which refers to the likelihood of molecules being in a particular arrangement, is higher. When the entropy increases, it indicates that the system has moved toward a state with greater disorder.
  • This change represents a higher degree of randomness in molecular positioning.
  • Higher entropy translates to more microstates or arrangements that the system's molecules can assume.
In the context of the enzyme, its more structured original form has lower entropy and fewer possible molecular arrangements. When heated, it moves to a state of higher entropy, meaning more disordered and with greater positional probability.
Gibbs Free Energy
Gibbs free energy is a crucial concept in determining the spontaneity of chemical and physical processes. It is denoted by the symbol G and combines enthalpy (H) and entropy (S), along with temperature (T), in the formula: \[ \Delta G = \Delta H - T\Delta S \]For a process to be spontaneous, the change in Gibbs free energy (\(\Delta G\)) must be negative. In the case of enzyme denaturation:
  • The process is endothermic, so \(\Delta H > 0\), meaning heat is absorbed.
  • The process is spontaneous, indicating \(\Delta G < 0\).
  • The positive \(\Delta S\) observes an increase in entropy.
Since the overall \(\Delta G\) is negative, it reveals that the increase in entropy significantly contributes to the spontaneity of the enzyme's structural transformation. This demonstrates how entropy, despite the endothermic nature, drives the reaction forward.
Spontaneity
Spontaneity in chemical and physical processes describes whether they occur naturally without external intervention. For enzyme denaturation by heat to be spontaneous, the Gibbs free energy change (\(\Delta G\)) must be negative, as stated by the equation:\[ \Delta G = \Delta H - T\Delta S \]When Gibbs free energy is negative:
  • The process occurs without needing added energy.
  • The system may release or absorb energy, but the net effect is a natural progression.
In enzyme denaturation, even though the process is endothermic (absorbs heat), the increase in entropy (\(\Delta S > 0\)) is enough to drive \(\Delta G\) to be negative, thus making the reaction spontaneous. This shows that the increase in disorder (entropy) outweighs the energy absorption, favoring the spontaneity.
Positional Probability
Positional probability is intricately linked to entropy, referring to the likelihood of molecules in a system being in particular spatial arrangements. In simpler terms, it is the number of possible ways molecules can be arranged while still composing the system:
  • Higher positional probability corresponds to higher entropy and more disordered or random molecular arrangements.
  • Lower positional probability correlates with lower entropy, indicating a more ordered system.
When an enzyme, originally in a specific structured and ordered form, is heated, it undergoes denaturation, transitioning to a form with higher entropy. This results in an increase in positional probability as molecules adopt more varied and less constrained positions. Consequently, the original enzyme structure is more ordered, possessing a smaller positional probability.

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Most popular questions from this chapter

Two crystalline forms of white phosphorus are known. Both forms contain \(\mathrm{P}_{4}\) molecules, but the molecules are packed together in different ways. The \(\alpha\) form is always obtained when the liquid freezes. However, below \(-76.9^{\circ} \mathrm{C}\), the \(\alpha\) form spontaneously converts to the \(\beta\) form: $$\mathrm{P}_{4}(s, \alpha) \longrightarrow \mathrm{P}_{4}(s, \beta)$$ a. Predict the signs of \(\Delta H\) and \(\Delta S\) for this process. b. Predict which form of phosphorus has the more ordered crystalline structure (has the smaller positional probability).

Cells use the hydrolysis of adenosine triphosphate, abbreviated as ATP, as a source of energy. Symbolically, this reaction can be written as $$\mathrm{ATP}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{ADP}(a q)+\mathrm{H}_{2} \mathrm{PO}_{4}^{-}(a q)$$ where ADP represents adenosine diphosphate. For this reaction, \(\Delta G^{\circ}=-30.5 \mathrm{~kJ} / \mathrm{mol}\) a. Calculate \(K\) at \(25^{\circ} \mathrm{C}\). b. If all the free energy from the metabolism of glucose $$\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(s)+6 \mathrm{O}_{2}(g) \longrightarrow 6 \mathrm{CO}_{2}(g)+6 \mathrm{H}_{2} \mathrm{O}(l)$$ goes into forming ATP from ADP, how many ATP molecules can be produced for every molecule of glucose? c. Much of the ATP formed from metabolic processes is used to provide energy for transport of cellular components. What amount (mol) of ATP must be hydrolyzed to provide the energy for the transport of \(1.0 \mathrm{~mol} \mathrm{~K}^{+}\) from the blood to the inside of a muscle cell at \(37^{\circ} \mathrm{C}\) as described in Exercise \(78 ?\)

Predict the sign of \(\Delta S_{\text {surr }}\) for the following processes. a. \(\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{H}_{2} \mathrm{O}(g)\) b. \(I_{2}(g) \longrightarrow I_{2}(s)\)

A green plant synthesizes glucose by photosynthesis, as shown in the reaction $$6 \mathrm{CO}_{2}(g)+6 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(s)+6 \mathrm{O}_{2}(g)$$ Animals use glucose as a source of energy: $$\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(s)+6 \mathrm{O}_{2}(g) \longrightarrow 6 \mathrm{CO}_{2}(g)+6 \mathrm{H}_{2} \mathrm{O}(l)$$ If we were to assume that both these processes occur to the same extent in a cyclic process, what thermodynamic property must have a nonzero value?

The Ostwald process for the commercial production of nitric acid involves three steps: a. Calculate \(\Delta H^{\circ}, \Delta S^{\circ}, \Delta G^{\circ}\), and \(K\) (at \(298 \mathrm{~K}\) ) for each of the three steps in the Ostwald process (see Appendix 4 ). b. Calculate the equilibrium constant for the first step at \(825^{\circ} \mathrm{C}\), assuming \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) do not depend on temperature. c. Is there a thermodynamic reason for the high temperature in the first step, assuming standard conditions?

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