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 antivenoms, 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 \(=k[\) acetylcholine receptor-toxin complex \(]\) If the activation energy of this reaction at \(37.0^{\circ} \mathrm{C}\) is \(26.2 \mathrm{~kJ} / \mathrm{mol}\) 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 at a concentration of 0.200 M receptor-toxin complex and at 37.0\(^\circ C\) is approximately \(3.1 \times 10^{-4} M s^{-1}\).

Step by step solution

01

Convert temperature to Kelvin

Convert the temperature from Celsius to Kelvin using the formula: \(T(K) = T(^\circ C) + 273.15\) The given temperature is: \(T = 37.0^\circ C\) So, \(T(K) = 37.0 + 273.15 = 310.15 K\)
02

Calculate the rate constant, k, using the Arrhenius equation

The Arrhenius equation is given as: \(k = Ae^{-\frac{Ea}{RT}}\) where k - rate constant A - pre-exponential factor Ea - activation energy (\(26.2 kJ/mol = 26200 J/mol\)) R - gas constant (\(8.314 J/mol K\)) T - temperature in Kelvin Now, plug in the values and compute k: \(k = (0.850 s^{-1})e^{-\frac{(26200 J/mol)}{(8.314 J/mol K)(310.15 K)}} \) Calculating the exponential term first: \(e^{-\frac{26200}{(8.314)(310.15)}} \approx 0.00182\) Now multiplying by A: \(k = 0.850 \times 0.00182 \approx 0.00155 s^{-1}\)
03

Calculate the rate of reaction at the given concentration of the receptor-toxin complex

We know that the rate law for this reaction is given as: Rate \(= k[\text{acetylcholine receptor-toxin complex}]\) Now, plug in the rate constant (k) and the given concentration (0.200 M) of the receptor-toxin complex: Rate \(= (0.00155 s^{-1})(0.200 M)\) Rate \(= 3.1 \times 10^{-4} M s^{-1}\) The rate of reaction at a concentration of 0.200 M receptor-toxin complex and at 37.0\(^\circ C\) is approximately \(3.1 \times 10^{-4} M s^{-1}\).

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Most popular questions from this chapter

Sulfuryl chloride undergoes first-order decomposition at \(320 .{ }^{\circ} \mathrm{C}\) with a half-life of \(8.75 \mathrm{~h}\). $$ \mathrm{SO}_{2} \mathrm{Cl}_{2}(g) \longrightarrow \mathrm{SO}_{2}(g)+\mathrm{Cl}_{2}(g) $$ What is the value of the rate constant, \(k\), in \(\mathrm{s}^{-1} ?\) If the initial pressure of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) is 791 torr and the decomposition occurs in a \(1.25\) -L container, how many molecules of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) remain after \(12.5 \mathrm{~h}\) ?

The reaction $$ 2 \mathrm{NO}(g)+\mathrm{Cl}_{2}(g) \longrightarrow 2 \mathrm{NOCl}(g) $$ was studied at \(-10^{\circ} \mathrm{C}\). The following results were obtained where $$ \text { Rate }=-\frac{\Delta\left[\mathrm{Cl}_{2}\right]}{\Delta t} $$

For enzyme-catalyzed reactions that follow the mechanism $$ \begin{array}{l} \mathrm{E}+\mathrm{S} \rightleftharpoons \mathrm{E} \cdot \mathrm{S} \\ \mathrm{E} \cdot \mathrm{S} \rightleftharpoons \mathrm{E}+\mathrm{P} \end{array} $$ a graph of the rate as a function of \([\mathrm{S}]\), the concentration of the substrate, has the following appearance: Note that at higher substrate concentrations the rate no longer changes with [S]. Suggest a reason for this.

The following data were obtained for the reaction \(2 \mathrm{ClO}_{2}(a q)+2 \mathrm{OH}^{-}(a q) \longrightarrow \mathrm{ClO}_{3}^{-}(a q)+\mathrm{ClO}_{2}^{-}(a q)+\mathrm{H}_{2} \mathrm{O}(l)\) where \(\quad\) Rate \(=-\frac{\Delta\left[\mathrm{ClO}_{2}\right]}{\Delta t}\)

Consider the general reaction $$ \mathrm{aA}+\mathrm{bB} \longrightarrow \mathrm{cC} $$ and the following average rate data over some time period \(\Delta t\) : $$ \begin{aligned} -\frac{\Delta \mathrm{A}}{\Delta t} &=0.0080 \mathrm{~mol} / \mathrm{L} \cdot \mathrm{s} \\ -\frac{\Delta \mathrm{B}}{\Delta t} &=0.0120 \mathrm{~mol} / \mathrm{L} \cdot \mathrm{s} \\ \frac{\Delta \mathrm{C}}{\Delta t} &=0.0160 \mathrm{~mol} / \mathrm{L} \cdot \mathrm{s} \end{aligned} $$ Determine a set of possible coefficients to balance this general reaction.

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