Question

When most biological enzymes are heated, they lose their catalytic activity. The change

$\mathit{O}\mathit{r}\mathit{i}\mathit{g}\mathit{i}\mathit{n}\mathit{a}\mathit{l}\mathit{e}\mathit{n}\mathit{z}\mathit{y}\mathit{m}\mathit{e}\mathbf{\to }\mathit{n}\mathit{e}\mathit{w}\mathit{f}\mathit{o}\mathit{r}\mathit{m}$

that occurs upon heating is endothermic and spontaneous. Is the structure of the original enzyme or its new form more ordered (has the smaller positional probability)? Explain your answer

Short answer

In comparison to the new form, the original enzyme's structure is more ordered.

Step-by-step solution

Step 1: Introduction to the Concept

The ratio of thermal energy to the temperature at which work cannot be done is known as entropy. It's also known as the measure of a system's molecular disarray. It has a lot of properties and state functions.

The number of microstates for a system is connected to entropy, and the microstate is defined as the number of possible arrangements for the system.

Solution Explanation

The given reaction: $Originalenzyme\to newform$

Endothermic reactions are those in which heat is absorbed, and they are not spontaneous because an external source of heat is necessary for the reaction to occur. However, if there is no rise in heat, the spontaneous process implies that there is an increase in entropy, as shown by the following reaction:

$\mathit{\Delta }\mathit{G}\mathbf{=}\mathit{\Delta }\mathit{H}\mathbf{-}\mathit{T}\mathit{\Delta }\mathit{S}$

The value of $∆\mathrm{G}$ must be smaller than zero for the change to be spontaneous. Because the value of $∆\mathrm{H}$ is positive in an endothermic process, the value of $∆\mathrm{S}$ must be negative to obtain a negative valueof $∆\mathrm{G}$.

Entropy now quantifies the disorder in the process and increases entropy, implying that the new form of enzyme will be less ordered than the original one. As a result, the original form has a more ordered structure, lowering entropy.