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Chapter 12: Problem 95

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 $$ \text {Rate} =k[\text { acetylcholine receptor-toxin complex }] $$ 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 \(\mathrm{a} 0.200-\mathrm{M}\) solution of receptor-toxin complex at \(37.0^{\circ} \mathrm{C} ?\)

Short Answer

Expert verified
The rate of the reaction for a 0.200 M solution of receptor-toxin complex at 37.0°C is approximately 2.74 × 10^-5 M/s.

Step by step solution

01

Convert the activation energy and temperature to the proper units

The activation energy (Ea) needs to be in Joules per mol (J/mol) and the temperature (T) should be in Kelvins (K). Ea = 26.2 kJ/mol × (1000 J / 1 kJ) = 26200 J/mol T = 37.0 °C + 273.15 = 310.15 K
02

Use the Arrhenius equation to find the rate constant (k)

The Arrhenius equation is: \(k = Ae^{\frac{-Ea}{RT}}\) where k is the rate constant, A is the pre-exponential factor, Ea is the activation energy, R is the gas constant (8.314 J/mol·K), and T is the temperature. Plugging in the given values: k = 0.850 s^-1 × exp(-26200 J/mol ÷ (8.314 J/mol·K × 310.15 K)) Calculate the value for k: k ≈ 1.37 × 10^-4 s^-1
03

Use the rate law equation to find the rate of reaction

Now that we have the rate constant (k), we can use the rate law equation to find the rate of the reaction: Rate = k[acetylcholine receptor-toxin complex] Plug in the values: Rate = (1.37 × 10^-4 s^-1) × 0.200 M Calculate the rate: Rate ≈ 2.74 × 10^-5 M/s So, the rate of the reaction is approximately 2.74 × 10^-5 M/s at 37.0°C for a 0.200 M solution of receptor-toxin complex.

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Most popular questions from this chapter

The activation energy for a reaction is changed from 184 \(\mathrm{kJ} /\) \(\mathrm{mol}\) to 59.0 \(\mathrm{kJ} / \mathrm{mol}\) at \(600 . \mathrm{K}\) 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 rate constant \((k)\) depends on which of the following (there may be more than one answer)? a. the concentration of the reactants b. the nature of the reactants c. the temperature d. the order of the reaction Explain.

A certain substance, initially at 0.10\(M\) in solution, decomposes by second- order kinetics. If the rate constant for this process is 0.40 $\mathrm{L} / \mathrm{mol} \cdot \min$ , how much time is required for the concentration to reach 0.020 \(\mathrm{M}\) ?

Consider the hypothetical reaction $\mathrm{A}_{2}(g)+\mathrm{B}_{2}(g) \longrightarrow\( \)2 \mathrm{AB}(g),$ where the rate law is: $$ -\frac{\Delta\left[\mathrm{A}_{2}\right]}{\Delta t}=k\left[\mathrm{A}_{2}\right]\left[\mathrm{B}_{2}\right] $$ The value of the rate constant at \(302^{\circ} \mathrm{C}\) is $2.45 \times 10^{-4} \mathrm{L} / \mathrm{mol}\( \)\mathrm{s},\( and at \)508^{\circ} \mathrm{C}\( the rate constant is 0.891 \)\mathrm{L} / \mathrm{mol} \cdot \mathrm{s}$ . What is the activation energy for this reaction? What is the value of the rate constant for this reaction at \(375^{\circ} \mathrm{C} ?\)

In the gas phase, the production of phosgene from chlorine and carbon monoxide is assumed to proceed by the following mechanism: $$ \mathrm{Cl}_{2} \stackrel{k_{1}}{\rightleftharpoons_{k_{1}}} 2 \mathrm{Cl} $$ $$ \mathrm{Cl}+\mathrm{CO} \stackrel{k_{2}}{\leftrightharpoons_{k-2}} \mathrm{COCl} $$ $$ \mathrm{COCl}+\mathrm{Cl}_{2} \stackrel{k_{3}}{\longrightarrow} \mathrm{COCl}_{2}+\mathrm{Cl} $$ $$ 2 \mathrm{Cl} \stackrel{k}{\longrightarrow} \mathrm{Cl}_{2} $$ Overall reaction: $\mathrm{CO}+\mathrm{Cl}_{2} \longrightarrow \mathrm{COCl}_{2}$ a. Write the rate law for this reaction. b. Which species are intermediates?
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