Cobra venom helps the snake secure food by binding to acetylcholine receptors on the diaphragm of a bite victim, leading to the loss of function of the diaphragm muscle tissue and eventually death. In order to develop more potent antivenins, scientists have studied what happens to the toxin once it has bound the acetylcholine receptors. They have found that the toxin is released from the receptor in a process that can be described by the rate law $$Rate \(=k[\text { acetylcholine receptor-toxin complex }]$$ If the activation energy of this reaction at \)37.0^{\circ} \mathrm{C}\( is \)26.2 \mathrm{kJ} /\( mol and \)A=0.850 \mathrm{s}^{-1},\( what is the rate of reaction if you have a 0.200-M solution of receptor-toxin complex at \)37.0^{\circ} \mathrm{C} ?$

Short Answer

Expert verified
The rate of reaction can be calculated using the Arrhenius equation and the given rate law expression. First, find the rate constant (k) using the activation energy (Ea = 26.2 kJ/mol), pre-exponential factor (A = 0.850 s^{-1}), temperature (T = 37.0°C or 310.15 K), and the gas constant (R = 8.314 J/mol K): \(k = 0.850 s^{-1} \cdot e^{\frac{-26200 J/mol}{(8.314 J/mol \cdot K)(310.15 K)}}\) Next, calculate the rate of reaction using the rate law expression and the concentration of the receptor-toxin complex (0.200 M): Rate = k * [0.200 M] After calculating the values, you can find the rate of the reaction at 37.0°C and 0.200-M solution of receptor-toxin complex.

Step by step solution

01

Calculate the rate constant (k) using the Arrhenius equation

The Arrhenius equation calculates the rate constant (k) using the activation energy (Ea), pre-exponential factor (A), and temperature (T) as follows: \(k = Ae^{\frac{-Ea}{RT}}\), where k = rate constant, A = pre-exponential factor, Ea = activation energy, R = gas constant (8.314 J/mol K), T = temperature in Kelvin (K). First, convert the activation energy from kJ/mol to J/mol and the temperature from °C to K: Ea = 26.2 kJ/mol × 1000 J/1 kJ = 26200 J/mol T = 37.0°C + 273.15 = 310.15 K Now substitute the values into the Arrhenius equation: \(k = 0.850 s^{-1} \cdot e^{\frac{-26200 J/mol}{(8.314 J/mol \cdot K)(310.15 K)}}\) Calculate the value of k.
02

Calculate the rate of reaction using the rate law expression

We are given the rate law expression as: Rate = k[acetylcholine receptor-toxin complex] We found the value of k in Step 1, and the concentration of the receptor-toxin complex is given as 0.200 M. Substitute these values into the expression: Rate = k * [0.200 M] Calculate the rate of the reaction. The rate of reaction at 37.0°C and 0.200-M solution of receptor-toxin complex is found using the given information and the Arrhenius equation.

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Key Concepts

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

Arrhenius Equation
Understanding how and why reactions occur with certain speeds is crucial in many scientific fields. The Arrhenius equation is a mathematical expression that links the rate constant of a chemical reaction with the temperature and activation energy. It is given by the formula:

(1) Siegfried Heydrich 1944 if( \(Ae^{\frac{-Ea}{RT}}\)Activation energy \(E_a\) is the energy required for reactants to reach the transition state, which is essentially the energy 'barrier' the reaction must overcome to occur. The pre-exponential factor \(A\) represents the frequency of reactant collisions that have the correct orientation for reaction. As temperature increases, so does the rate constant \(k\), generally leading to faster reactions, because more molecules have sufficient energy to overcome the activation energy barrier.

In our exercise, we use the Arrhenius equation to calculate the rate constant for the release of snake venom toxin from the acetylcholine receptor. This concept is critical for understanding the temperature dependency of reaction rates and how they can be manipulated, such as in the development of antivenins. To calculate the rate constant, we incorporated the activation energy, temperature, and the pre-exponential factor into the Arrhenius equation, providing us with a crucial variable to derive the overall rate of reaction.
Enzyme Kinetics
Enzyme kinetics is a field that studies the rates of enzyme-catalyzed reactions. In biological systems, enzymes act as catalysts to speed up chemical reactions without being consumed in the process. This acceleration is crucial, as many biochemical reactions would proceed too slowly without catalysis to sustain life.

Key parameters in enzyme kinetics include the affinity of an enzyme for its substrate, represented by the Michaelis-Menten constant \(K_M\), and the maximum rate of reaction, \(V_{max}\). The interaction between enzymes and their substrates can often be approximated by the Michaelis-Menten equation, which describes how the reaction rate depends on the concentration of both enzyme and substrate.

While our original exercise does not directly involve enzymes, the concepts can be analogous. For instance, the rate law described in the exercise reflects how the reaction rate is proportional to the concentration of the acetylcholine receptor-toxin complex. Similarly, in enzyme kinetics, the reaction rate is proportional to the enzyme-substrate complex concentration. Understanding enzyme kinetics is vital for biotechnology, pharmacology, and understanding metabolic processes within cells.
Chemical Kinetics
Chemical kinetics is the overall field that studies the rate at which chemical reactions proceed and the factors affecting these rates. It encompasses the concepts we have touched upon in the Arrhenius equation and enzyme kinetics, while also dealing with more diverse chemical reactions.

Factors affecting reaction rates include concentration of reactants, temperature, and the presence of a catalyst, whether it be a chemical catalyst or an enzyme. In the reaction rate law expression,
\( Rate = k[\text{reactant}]^n \), the rate constant \(k\) represents the proportionality between the rate and the reactant concentration, which is raised to the power of \(n\), the order of the reaction. The order provides insight into the reaction mechanism and is determined experimentally.

The exercise demonstrates a real-world application of chemical kinetics. By understanding the rate law and how to manipulate it, scientists can predict reaction behavior under different conditions, synthesize products more efficiently, and, in the context of our exercise, develop treatments such as antivenins by evaluating how a toxin interacts with its target.

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Most popular questions from this chapter

Hydrogen peroxide and the iodide ion react in acidic solution as follows: $$\mathrm{H}_{2} \mathrm{O}_{2}(a q)+3 \mathrm{I}^{-}(a q)+2 \mathrm{H}^{+}(a q) \longrightarrow \mathrm{I}_{3}^{-}(a q)+2 \mathrm{H}_{2} \mathrm{O}(l)$$ The kinetics of this reaction were studied by following the decay of the concentration of \(\mathrm{H}_{2} \mathrm{O}_{2}\) and constructing plots of \(\ln \left[\mathrm{H}_{2} \mathrm{O}_{2}\right]\) versus time. All the plots were linear and all solutions had \(\left[\mathrm{H}_{2} \mathrm{O}_{2}\right]_{0}=8.0 \times 10^{-4} \mathrm{mol} / \mathrm{L} .\) The slopes of these straight lines depended on the initial concentrations of \(\mathrm{I}^{-}\) and \(\mathrm{H}^{+} .\) The results follow: The rate law for this reaction has the form $$\text { Rate }=\frac{-\Delta\left[\mathrm{H}_{2} \mathrm{O}_{2}\right]}{\Delta t}=\left(k_{1}+k_{2}\left[\mathrm{H}^{+}\right]\right)\left[\mathrm{I}^{-}\right]^{m}\left[\mathrm{H}_{2} \mathrm{O}_{2}\right]^{n}$$ a. Specify the order of this reaction with respect to \(\left[\mathrm{H}_{2} \mathrm{O}_{2}\right]\) and \(\left[\mathrm{I}^{-}\right]\) b. Calculate the values of the rate constants, \(k_{1}\) and \(k_{2}\) c. What reason could there be for the two-term dependence of the rate on \(\left[\mathrm{H}^{+}\right] ?\)

A certain reaction has the following general form: $$\text { aA } \longrightarrow \mathrm{bB}$$ At a particular temperature and \([\mathrm{A}]_{0}=2.00 \times 10^{-2} \mathrm{M}, \)concentration versus time data were collected for this reaction, and a plot of \(\ln [\mathrm{A}]\) versus time resulted in a straight line with a slope value of \(-2.97 \times 10^{-2} \mathrm{min}^{-1}\)a. Determine the rate law, the integrated rate law, and the value of the rate constant for this reaction. b. Calculate the half-life for this reaction. c. How much time is required for the concentration of A to decrease to \(2.50 \times 10^{-3} \mathrm{M} ?\)

Describe at least two experiments you could perform to determine a rate law.

The activation energy for a reaction is changed from \(184 \space\mathrm{kJ} /\) mol to \(59.0 \space\mathrm{kJ} / \mathrm{mol}\) at \(600 .\) K by the introduction of a catalyst. If the uncatalyzed reaction takes about 2400 years to occur, about how long will the catalyzed reaction take? Assume the frequency factor \(A\) is constant, and assume the initial concentrations are the same.

The reaction $$\left(\mathrm{CH}_{3}\right)_{3} \mathrm{CBr}+\mathrm{OH}^{-} \longrightarrow\left(\mathrm{CH}_{3}\right)_{3} \mathrm{COH}+\mathrm{Br}^{-}$$ in a certain solvent is first order with respect to \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{CBr}\) and zero order with respect to OH \(^{-} .\) In several experiments, the rate constant \(k\) was determined at different temperatures. A plot of \(\ln (k)\) versus \(1 / T\) was constructed resulting in a straight line with a slope value of \(-1.10 \times 10^{4} \mathrm{K}\) and \(y\) -intercept of 33.5. Assume \(k\) has units of \(\mathrm{s}^{-1}\).a. Determine the activation energy for this reaction. b. Determine the value of the frequency factor \(A\) c. Calculate the value of \(k\) at \(25^{\circ} \mathrm{C}\)

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