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Chapter 35: Problem 46

Catalase is an enzyme that promotes the conversion of hydrogen peroxide \(\left(\mathrm{H}_{2} \mathrm{O}_{2}\right)\) into water and oxygen. The diffusion constant and radius for catalase are \(6.0 \times 10^{-7} \mathrm{cm}^{2} \mathrm{s}^{-1}\) and 51.2 A. For hydrogen peroxide the corresponding values are \(1.5 \times 10^{-5} \mathrm{cm}^{2} \mathrm{s}^{-1}\) and \(r \sim 2.0\) A. The experimentally determined rate constant for the conversion of hydrogen peroxide by catalase is \(5.0 \times 10^{6} \mathrm{M}^{-1} \mathrm{s}^{-1} .\) Is this a diffusioncontrolled reaction?

Short Answer

Expert verified
Briefly explain your answer. Answer: Yes, the conversion of hydrogen peroxide by catalase is a diffusion-controlled reaction. This is because the rate constant for the reaction (k = 5.0 x 10^6 M^-1 s^-1) is approximately equal or less than the diffusion rate constant (k_D = 1.40 x 10^10 M^-1 s^-1), which indicates that the reaction is limited by the diffusion process.

Step by step solution

01

To calculate k_D, we will use the Smoluchowski equation, which is: $$k_D = 4\pi(D_{cat} + D_{H_2O_2})(R_{cat} + R_{H_2O_2})$$ Where: \(D_{cat}\) is the diffusion constant of catalase (\(6.0 \times 10^{-7} \mathrm{cm}^{2} \mathrm{s}^{-1}\)) \(D_{H_2O_2}\) is the diffusion constant of hydrogen peroxide (\(1.5 \times 10^{-5} \mathrm{cm}^{2} \mathrm{s}^{-1}\)) \(R_{cat}\) is the radius of catalase (51.2 A = \(5.12 \times 10^{-8}\) cm) \(R_{H_2O_2}\) is the radius of hydrogen peroxide (2.0 A = \(2.0 \times 10^{-8}\) cm) Remember to convert 脜ngstr枚m (A) to centimeters (cm) by multiplying by 1*10^(-8). #Step 2: Plug values into Smoluchowski equation#

We now plug the values into the equation: $$k_D = 4\pi[(6.0 \times 10^{-7}) + (1.5 \times 10^{-5})][(5.12 \times 10^{-8}) + (2.0 \times 10^{-8})]$$ #Step 3: Calculate k_D#
02

After performing the calculations, we find: $$k_D = 4\pi(1.56 \times 10^{-5})(7.12 \times 10^{-8})$$ $$k_D = 1.40 \times 10^{10} \mathrm{M}^{-1} \mathrm{s}^{-1}$$ #Step 4: Compare k_D to the rate constant (k) of the reaction#

The rate constant for the conversion of hydrogen peroxide by catalase is given as \(k = 5.0 \times 10^{6} \mathrm{M}^{-1} \mathrm{s}^{-1}\). Let's compare it to k_D: \(k = 5.0 \times 10^{6} \mathrm{M}^{-1} \mathrm{s}^{-1}\) \(k_D = 1.40 \times 10^{10} \mathrm{M}^{-1} \mathrm{s}^{-1}\) Since the value of k is approximately equal or less than k_D, the conversion of hydrogen peroxide by catalase is a diffusion-controlled reaction.

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Key Concepts

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

Smoluchowski Equation
The Smoluchowski equation plays a crucial role in understanding diffusion-controlled reactions. It is used to determine the rate constant of a reaction by considering the diffusion coefficients and sizes (radii) of the reacting species. The general formula is given by:\[ k_D = 4\pi(D_1 + D_2)(R_1 + R_2) \]Here:
  • \( D_1 \) and \( D_2 \) are the diffusion constants of the two reacting particles
  • \( R_1 \) and \( R_2 \) are their respective radii
This equation helps in assessing whether a reaction is limited by how fast molecules can diffuse through a medium and encounter each other. When using this equation, real-world applications often involve converting units appropriately, such as 脜ngstr枚ms (a unit of length) to centimeters for consistency in calculations.
Catalase Enzyme
Catalase is a highly efficient enzyme found in nearly all living organisms that are exposed to oxygen. Its primary role is to accelerate the decomposition of hydrogen peroxide, a potentially harmful by-product of many metabolic reactions, into harmless water and oxygen. This reaction is vital because it helps protect cells from oxidative damage caused by hydrogen peroxide. Catalase works by providing an alternate pathway for reactions, significantly reducing the activation energy required.
  • It contains a heme group similar to hemoglobin, which helps in the catalytic process.
  • It has a remarkable turnover number, capable of processing millions of hydrogen peroxide molecules per second.
Catalase is a large enzyme, and its efficiency makes it crucial for the detoxification processes in cells, highlighting its importance in maintaining cellular health.
Hydrogen Peroxide Conversion
The conversion of hydrogen peroxide (\( H_2O_2 \)) involves breaking it down into water (\( H_2O \)) and oxygen (\( O_2 \)) through the action of catalase. The chemical reaction is represented as:\[ 2H_2O_2 \rightarrow 2H_2O + O_2 \]Hydrogen peroxide itself is used by cells as a signaling molecule in low concentrations, but at higher concentrations, it can be damaging. Therefore, its breakdown is essential to protecting cellular structures.
  • Catalase facilitates this breakdown efficiently, ensuring rapid conversion of hydrogen peroxide before it can accumulate.
  • Diffusion-controlled conditions suggest that the rate of conversion is largely limited by how quickly hydrogen peroxide molecules can diffuse and collide with catalase.
This reaction can be tested in laboratory settings to analyze its rate constants, as shown with the Smoluchowski equation, to determine whether it is diffusion-controlled. The given values for catalase and hydrogen peroxide highlight the significant role diffusion rates play in such biochemical processes.

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Most popular questions from this chapter

For a type II second-order reaction, the reaction is \(60 \%\) complete in 60 seconds when \([\mathrm{A}]_{0}=0.1 \mathrm{M}\) and \([\mathrm{B}]_{0}=0.5 \mathrm{M}\) a. What is the rate constant for this reaction? b. Will the time for the reaction to reach \(60 \%\) completion change if the initial reactant concentrations are decreased by a factor of \(2 ?\)

Bananas are somewhat radioactive due to the presence of substantial amounts of potassium. Potassium- 40 decays by two different paths: $$\begin{array}{l}_{19}^{40} \mathrm{K} \rightarrow_{20}^{40} \mathrm{Ca}+\beta^{-}(89.3 \%) \\ _{19}^{40} \mathrm{K} \rightarrow_{18}^{40} \mathrm{Ar}+\beta^{+}(10.7 \%)\end{array}$$ The half-life for potassium decay is \(1.3 \times 10^{9}\) years. Determine the rate constants for the individual channels.

The conversion of \(\mathrm{NO}_{2}(g)\) to \(\mathrm{NO}(g)\) and \(\mathrm{O}_{2}(g)\) can occur through the following reaction: $$\mathrm{NO}_{2}(g) \rightarrow 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g)$$ The activation energy for this reaction is \(111 \mathrm{kJ} \mathrm{mol}^{-1}\) and the pre-exponential factor is \(2.0 \times 10^{-9} \mathrm{M}^{-1} \mathrm{s}^{-1}\). Assume that these quantities are temperature independent. a. What is the rate constant for this reaction at \(298 \mathrm{K} ?\) b. What is the rate constant for this reaction at the tropopause where \(\mathrm{T}=225 \mathrm{K}\).

P35.10 (Challenging) The first-order thermal decomposition of chlorocyclohexane is as follows: \(\mathrm{C}_{6} \mathrm{H}_{11} \mathrm{Cl}(g) \rightarrow\) \(\mathrm{C}_{6} \mathrm{H}_{10}(g)+\mathrm{HCl}(g) .\) For a constant volume system the following total pressures were measured as a function of time: $$\begin{array}{rccr}\text { Time }(s) & P(\text { Torr }) & \text { Time }(s) & P(\text { Torr }) \\\\\hline 3 & 237.2 & 24 & 332.1 \\\6 & 255.3 & 27 & 341.1 \\\9 & 271.3 & 30 & 349.3 \\\12 & 285.8 & 33 & 356.9 \\\15 & 299.0 & 36 & 363.7 \\\18 & 311.2 & 39 & 369.9 \\\21 & 322.2 & 42 & 375.5\end{array}$$ a. Derive the following relationship for a first-order reaction: \\[P\left(t_{2}\right)-P\left(t_{1}\right)=\left(P\left(t_{\infty}\right)-P\left(t_{0}\right)\right) e^{-k t_{1}}\left(1e^{-k\left(t_{2}-t_{1}\right)}\right)\\] In this relation, \(P\left(t_{1}\right)\) and \(P\left(t_{2}\right)\) are the pressures at two specific times; \(\mathrm{P}\left(t_{0}\right)\) is the initial pressure when the reaction is initiated, \(P\left(t_{\infty}\right)\) is the pressure at the completion of the reaction, and \(k\) is the rate constant for the reaction. To derive this relationship do the following: i. Given the first-order dependence of the reaction, write the expression for the pressure of chlorocyclohexane at a specific time \(t_{1}\) ii. Write the expression for the pressure at another time \(t_{2},\) which is equal to \(t_{1}+\Delta\) where delta is a fixed quantity of time. iii. Write expressions for \(P\left(t_{\infty}\right)-P\left(t_{1}\right)\) and \(P\left(t_{\infty}\right) P\left(t_{2}\right)\) iv. Subtract the two expressions from part (iii). b. Using the natural log of the relationship from part (a) and the data provided in the table given earlier in this problem, determine the rate constant for the decomposition of chlorocyclohexane. (Hint: Transform the data in the table by defining \(t_{2}-t_{1}\) to be a constant value, for example, 9 s.

Consider the reaction A temperature-jump experiment is performed where the relaxation time constant is measured to be \(310 \mu\) s, resulting in an equilibrium where \(K_{e q}=0.70\) with \([\mathrm{P}]_{e q}=0.20 \mathrm{M}\) What are \(k\) and \(k^{\prime} ?\) (Watch the units!)
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