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Chapter 18: Problem 3

The rate of a certain biochemical reaction at physiological temperature \((T)\) occurs \(10^{6}\) times faster with enzyme than without. The change in the activation energy upon adding enzyme is: (a) \(-6(2.303) \mathrm{RT}\) (b) \(-6 \mathrm{RT}\) (c) \(+6(2.303) \mathrm{RT}\) (d) \(+6 \mathrm{RT}\)

Short Answer

Expert verified
Option (a) \(-6(2.303) \mathrm{RT}\).

Step by step solution

01

Understanding the Problem

We are given the rate increase of a biochemical reaction when an enzyme is added. The rate increases by a factor of \(10^{6}\). We need to find how the activation energy changes when the enzyme is added.
02

Using the Arrhenius Equation

The Arrhenius equation is \( k = A e^{-E_{a}/RT} \), where \( k \) is the rate constant, \( A \) is the pre-exponential factor, \( E_{a} \) is the activation energy, \( R \) is the gas constant, and \( T \) is the temperature in Kelvin.
03

Relating Rate Change to Activation Energy

Since the rate is \(10^{6}\) times faster with the enzyme, the new rate constant \( k' \) is \(10^{6}k\). Therefore, \( k' = A e^{-E_{a}'/RT} = 10^{6}k = 10^{6}(A e^{-E_{a}/RT}) \).
04

Setting Up the Equations

We can write two equations: \( k = A e^{-E_{a}/RT} \) and \( k' = A e^{-E_{a}'/RT} \), where \( E_{a}' \) is the new activation energy with the enzyme. According to the given, \( k'/k = 10^{6} \).
05

Solving for Activation Energy Change

Using \( k'/k = 10^{6} \), we have \( \frac{A e^{-E_{a}'/RT}}{A e^{-E_{a}/RT}} = 10^{6} \). This simplifies to \( e^{-(E_{a}' - E_{a})/RT} = 10^{6} \). Taking the natural logarithm of both sides, we get \(-(E_{a}' - E_{a})/RT) = \ln(10^{6}) \).
06

Calculating the Change in Activation Energy

We know that \( \ln(10^{6}) = 6 \times \ln(10) = 6 \times 2.303 \). Therefore, \(-(E_{a}' - E_{a})/RT = 6 \times 2.303 \). Rearranging to find \( E_{a}' - E_{a} \), we have \( E_{a}' - E_{a} = -6 \times 2.303 \times RT \).
07

Selecting the Correct Answer

From the calculation, the change in activation energy \( E_{a}' - E_{a} \) is \(-6(2.303) RT\). Therefore, the correct option is \( \text{(a)} \) \(-6(2.303) \mathrm{RT}\).

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Key Concepts

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

Enzyme Kinetics
Enzyme kinetics is the study of how enzymes speed up biochemical reactions. Enzymes are biological catalysts that can increase the reaction rate by lowering the activation energy. This means that reactions can occur more quickly and with less energy input.
Key aspects of enzyme kinetics include:
  • **Rate of reaction**: Enzymes can increase reaction rates by factors of millions, as shown in our problem where the rate is accelerated by a factor of \(10^6\) when an enzyme is added. This means the enzyme makes the reaction a million times faster.
  • **Reaction mechanism**: Enzymes interact with substrates to form an enzyme-substrate complex, leading to the formation of products.
  • **Michaelis-Menten kinetics**: This model describes how the rate of reaction depends on the concentration of substrate and the concentration of enzyme.
Understanding these core concepts in enzyme kinetics helps explain how biochemical processes in living organisms are efficiently regulated and controlled.
Arrhenius Equation
The Arrhenius equation is a formula that gives insight into the effect of temperature on the rates of chemical reactions.
It is expressed as:
  • \[ k = A e^{-E_{a}/RT} \]
  • In this equation, \( k \) is the rate constant that indicates the speed of reaction, \( A \) is the pre-exponential factor that encompasses the frequency of collisions with the correct orientation, \( E_{a} \) is the activation energy, \( R \) is the universal gas constant, and \( T \) is the temperature in Kelvin.

Role of Activation Energy

Activation energy is the minimum energy required to initiate a chemical reaction. The Arrhenius equation helps us predict how lowering the activation energy (as enzymes do) will drastically increase the reaction rate.

Application in Biochemical Reactions

In the given problem, by using the Arrhenius equation, the relationship between the initial and new activation energy is analyzed to understand how enzymes make reactions exponentially faster.
This lays the groundwork for understanding enzyme efficiency in various physiological processes.
Biochemical Reactions
Biochemical reactions are the chemical processes that occur in living organisms, driven by enzymes that act as catalysts.
These types of reactions are vital for maintaining life and enabling growth, reproduction, and cellular processes.

Characteristics of Biochemical Reactions

  • **Specificity**: Reactions are specific to certain enzymes, meaning only particular substrates will bind with particular enzymes to produce a reaction.
  • **Efficiency**: Enzymes make biochemical reactions feasible at body temperature by drastically reducing the activation energy required.
  • **Regulation**: Biochemical reactions are heavily regulated to meet the needs of the organism, ensuring that reactions occur only when necessary and at controlled rates.
Understanding these reactions helps visualize how organisms handle complex chemical processes efficiently and sustain life through a network of interrelated biochemical pathways.

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Most popular questions from this chapter

In the nuclear transmutation $$ { }_{4}^{9} \mathrm{Be}+X \rightarrow{ }_{4}^{8} \mathrm{Be}+Y $$ \((\mathrm{X}, \mathrm{Y})\) is (are) (a) \((\gamma, \mathrm{n})\) (b) (p, D) (c) (n, D) (d) \((\gamma, \mathrm{p})\)

According to the Arrhenius equation, (a) a high activation energy usually implies a fast reaction. (b) rate constant increases with increase in temperature. This is due to a greater number of collisions whose energy exceeds the activation energy. (c) higher the magnitude of activation energy, stronger is the temperature dependence of the rate constant. (d) the pre-exponential factor is a measure of the rate at which collisions occur, irrespective of their energy.

An experiment requires minimum beta activity product at the rate of 346 beta particles per minute. The half life period of \({ }_{42}^{99} \mathrm{Mo}\), which is a beta emitter is \(66.6\) hours. Find the minimum amount of \({ }_{42}^{99}\) Mo required to carry out the experiment in \(6.909\) hours.

In the Arrhenius equation for a certain reaction, the value of \(A\) and \(E_{a}\) (activation energy) are \(4 \times 10^{13} \mathrm{sec}^{-1}\) and \(98.6 \mathrm{~kJ} \mathrm{~mol}^{-1}\) respectively. If the reaction is of first order, at what temperature will its half-life period be ten minutes?

At constant temperature and volume, \(X\) decomposes as [2005 - 4 Marks] \(2 \mathrm{X}(\mathrm{g}) \rightarrow 3 \mathrm{Y}(\mathrm{g})+2 \mathrm{Z}(\mathrm{g}) ; P_{X}\) is the partial pressure of \(X\) \begin{tabular}{|c|c|c|} \hline Observation No. & Time (in minute) & \(P_{X}\) (in mm of \(\mathrm{Hg}\) ) \\\ \hline 1 & 0 & 800 \\ \hline 2 & 100 & 400 \\ \hline 3 & 200 & 200 \\ \hline \end{tabular} (i) What is the order of reaction with respect to \(X ?\) (ii) Find the rate constant. (iii) Find the time for \(75 \%\) completion of the reaction. (iv) Find the total pressure when pressure of \(X\) is \(700 \mathrm{~mm}\) of \(\mathrm{Hg}\).
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