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The second-order rate constant for the dimerization of a protein (P) \(\mathrm{P}+\mathrm{P} \longrightarrow \mathrm{P}_{2}\) is \(6.2 \times 10^{-3} / M \cdot \mathrm{s}\) at \(25^{\circ} \mathrm{C}\). If the concentration of the protein is \(2.7 \times 10^{-4} M,\) calculate the initial rate \((M / \mathrm{s})\) of formation of \(\mathrm{P}_{2}\). How long (in seconds) will it take to decrease the concentration of \(\mathrm{P}\) to \(2.7 \times 10^{-5} \mathrm{M}\) ?

Short Answer

Expert verified
The initial rate is \(4.52 \times 10^{-10} \, M/s\). It takes approximately 5379 seconds.

Step by step solution

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01

Identify Known Variables

The problem provides the second-order rate constant \( k = 6.2 \times 10^{-3} / M \cdot s \) and the initial concentration of the protein \( \mathrm{[P]}_0 = 2.7 \times 10^{-4} \ M \).
02

Rate Law for Second-Order Reaction

For a second-order reaction, the rate law is given by \( \, \text{Rate} = k \cdot [\mathrm{P}]^2 \,\).
03

Calculate Initial Rate

Substitute the known values into the rate law:\[\text{Rate} = 6.2 \times 10^{-3} \, M^{-1}\cdot s^{-1} \times (2.7 \times 10^{-4} \, M)^2\]Calculate:\[\text{Rate} = 6.2 \times 10^{-3} \times 7.29 \times 10^{-8}\]\[\text{Rate} = 4.52 \times 10^{-10} \, M \cdot s^{-1}\]
04

Second-Order Integrated Rate Equation

The integrated rate law for a second-order reaction is \( \frac{1}{[\mathrm{P}]} = \frac{1}{[\mathrm{P}]_0} + kt \).
05

Rearrange for Time

Rearrange the formula to solve for time \( t \):\[t = \frac{1}{k} \left( \frac{1}{[\mathrm{P}]} - \frac{1}{[\mathrm{P}]_0} \right)\]
06

Substitute Known Values into Time Equation

Plug in the values:\[t = \frac{1}{6.2 \times 10^{-3}} \left( \frac{1}{2.7 \times 10^{-5}} - \frac{1}{2.7 \times 10^{-4}} \right)\]Calculate:\[t = 161.29 \times (37037 - 3703.7) \approx 161.29 \times 33333.3\]\[t \approx 5379 \, s\]

Key Concepts

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

Second-order reactions
Chemical kinetics often involves determining how reaction rates are affected by concentrations of reactants. For a second-order reaction, two reactant molecules combine in a single reaction step. This is typically represented as:

- For the given problem, two protein molecules (P) combine to form a dimer (\( \mathrm{P}_2 \))
- These reactions can have the form \( A + B \to \, C \), but in our case, it simplifies to \( P + P \to P_2 \)
This indicates that the rate of reaction is proportional to the square of the concentration of a single reactant. Therefore, understanding the nature of second-order kinetics is crucial for predicting how fast or slow a reaction will proceed under various conditions.

Second-order reactions typically involve two reactants or two molecules of the same reactant. The concentration of reactants directly affects how quickly the products form. Because the reaction depends on two molecules of P reacting together, the reaction is more sensitive to changes in its concentration compared to first-order reactions. This sensitivity is expressed mathematically in the related rate laws and equations.
Rate law
The rate law is an equation that links the rate of reaction to the concentration of reactants. For a second-order reaction, the rate law is expressed as:\[ \text{Rate} = k [P]^2 \]Here, \( k \) is the rate constant, and \([P]\) represents the concentration of reactants. The rate constant is unique for every reaction and is affected by factors such as temperature and the nature of the reactants.

- For the protein dimerization given, \( k = 6.2 \times 10^{-3} \) M\(^{-1}\)s\(^{-1}\)
- The initial concentration \([P]_0 = 2.7 \times 10^{-4}\) M
The rate law allows us to calculate how quickly the dimer (\( P_2 \)) forms by considering both the rate constant and the initial concentrations. By substituting values into the rate law expression, we can determine the initial rate of the reaction. Understanding this relationship is key to predicting reaction behavior and manipulating conditions to achieve desired outcomes.
Integrated rate equations
Integrated rate equations are derived from the basic rate laws and they allow for the calculation of reactant concentrations over time. For a second-order reaction like our protein dimerization, the integrated rate equation is:\[ \frac{1}{[P]} = \frac{1}{[P]_0} + kt \]In this equation, \([P]_0\) is the initial concentration, \([P]\) is the concentration at time \( t \), and \( k \) is the rate constant.

- This formula is especially useful when you need to calculate how long it takes for a certain reaction concentration to reduce from one initial concentration to a lower one.
By rearranging the formula for \( t \), we can calculate the time taken for the concentration of protein P to decrease to a specific level. The calculation involves plugging in all known values and solving for \( t \). Integrated rate equations thus offer a powerful tool for understanding the dynamics of reactant degradation or product formation over time. By mastering these equations, students can apply them to a wide array of chemical systems, enhancing their understanding of reaction mechanisms and conditions.

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Most popular questions from this chapter

The concentrations of enzymes in cells are usually quite small. What is the biological significance of this fact?

The rate law for the reaction \(2 \mathrm{NO}_{2}(g) \longrightarrow \mathrm{N}_{2} \mathrm{O}_{4}(g)\) is rate \(=k\left[\mathrm{NO}_{2}\right]^{2}\). Which of the following changes will change the value of \(k ?\) (a) The pressure of \(\mathrm{NO}_{2}\) is doubled. (b) The reaction is run in an organic solvent. (c) The volume of the container is doubled. (d) The temperature is decreased. (e) A catalyst is added to the container.

A gas mixture containing \(\mathrm{CH}_{3}\) fragments, \(\mathrm{C}_{2} \mathrm{H}_{6}\) molecules, and an inert gas (He) was prepared at \(600 \mathrm{~K}\) with a total pressure of 5.42 atm. The elementary reaction $$ \mathrm{CH}_{3}+\mathrm{C}_{2} \mathrm{H}_{6} \longrightarrow \mathrm{CH}_{4}+\mathrm{C}_{2} \mathrm{H}_{5} $$ has a second-order rate constant of \(3.0 \times 10^{4} / M \cdot \mathrm{s} .\) Given that the mole fractions of \(\mathrm{CH}_{3}\) and \(\mathrm{C}_{2} \mathrm{H}_{6}\) are 0.00093 and 0.00077 , respectively, calculate the initial rate of the reaction at this temperature.

The thermal decomposition of phosphine \(\left(\mathrm{PH}_{3}\right)\) into phosphorus and molecular hydrogen is a first-order reaction: $$ 4 \mathrm{PH}_{3}(g) \longrightarrow \mathrm{P}_{4}(g)+6 \mathrm{H}_{2}(g) $$ The half-life of the reaction is \(35.0 \mathrm{~s}\) at \(680^{\circ} \mathrm{C}\). Calculate (a) the first-order rate constant for the reaction and (b) the time required for 95 percent of the phosphine to decompose.

The first-order rate constant for the decomposition of dimethyl ether: $$ \left(\mathrm{CH}_{3}\right)_{2} \mathrm{O}(g) \longrightarrow \mathrm{CH}_{4}(g)+\mathrm{H}_{2}(g)+\mathrm{CO}(g) $$ is \(3.2 \times 10^{-4} \mathrm{~s}^{-1}\) at \(450^{\circ} \mathrm{C}\). The reaction is carried out in a constant-volume flask. Initially only dimethyl ether is present and the pressure is \(0.350 \mathrm{~atm} .\) What is the pressure of the system after 8.0 min? Assume ideal behavior.

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