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A certain first-order reaction is 35.5 percent complete in 4.90 min at \(25^{\circ} \mathrm{C}\). What is its rate constant?

Short Answer

Expert verified
The rate constant is approximately 0.0894 min^{-1}.

Step by step solution

01

Understand the first-order reaction equation

A first-order reaction follows the equation \( [A]_t = [A]_0 e^{-kt} \), where \([A]_t\) is the concentration at time \(t\), \([A]_0\) is the initial concentration, \(k\) is the rate constant, and \(t\) is time in minutes.
02

Determine the extent of reaction completion

35.5% of the reaction is complete, meaning that 64.5% of the reactant remains, i.e., \( [A]_t = 0.645 [A]_0 \).
03

Insert the values into the equation

Substitute \(t = 4.90 \text{ min}\), and \([A]_t = 0.645 [A]_0\) into the first-order reaction equation: \[ 0.645 = e^{-k imes 4.90}. \]
04

Solve for the rate constant \(k\)

Take the natural logarithm of both sides of the equation:\[ \ln(0.645) = -4.90k. \]Calculate \( \ln(0.645) \) and divide by \(-4.90\) to find \(k\):\[ k = -\frac{\ln(0.645)}{4.90}. \]
05

Calculate \(k\)

Compute \(k\):\( \ln(0.645) \approx -0.438 \), hence \[ k \approx \frac{0.438}{4.90} = 0.0894 \text{ min}^{-1}. \]
06

Finalize the result

The rate constant \(k\) for the reaction at \(25^{\circ} \text{C}\) is approximately \(0.0894 \text{ min}^{-1}.\)

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Key Concepts

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

Rate Constant Calculation
In a first-order reaction, the rate constant, denoted by the symbol \(k\), is a critical factor. It tells us how fast a reaction proceeds over time. Calculating this constant helps us predict concentrations at various times during the reaction.

To determine \(k\), we use the equation for a first-order reaction:
  • \([A]_t = [A]_0 e^{-kt}\)
  • \([A]_t\) is the concentration at time \(t\).
  • \([A]_0\) is the initial concentration.
  • \(e\) is the base of the natural logarithm.
  • \(t\) is the time in minutes.
The natural logarithm can simplify exponential expressions in calculations. For our problem, we use it to isolate \(k\) in the equation by rearranging: \ \[ k = -\frac{\ln(\text{fraction remaining})}{t} \] By substituting known values into the formula, we can determine the numerical value for \(k\). In this case with 35.5% completed in 4.90 minutes, \([A]_t = 0.645 [A]_0\) was plugged into the formula, enabling us to solve for \(k\) and find it to be approximately \(0.0894 \, \text{min}^{-1}\).
Reaction Kinetics
Reaction kinetics is a branch of chemistry that examines the rates of chemical processes and the factors affecting them. For first-order reactions, kinetics are described by a straightforward relation with concentration and time, involving a rate constant \(k\).

This relation shows that the rate of the reaction is directly proportional to the concentration of one reactant. As a reaction progresses, the concentration of the reactant decreases. Consequently, the rate of the reaction decreases as well, but in a predictable and calculable manner.

Key factors influencing reaction kinetics include:
  • Reactant concentrations: Higher concentrations typically increase the rate of reaction.
  • Temperature: As temperature rises, molecules move faster and collide more often, speeding up reactions.
  • Catalysts: These substances increase reaction rates without themselves undergoing permanent changes.
For students studying first-order reactions, understanding these factors is crucial. They allow you to recognize how changing such conditions can alter the rate at which a reaction proceeds.
Exponential Decay Formula
In the context of reaction kinetics, the exponential decay formula is a vital tool used to describe the decrease in concentration over time for first-order reactions. The general equation is given by:
  • \([A]_t = [A]_0 e^{-kt}\)
This formula represents the concentration \([A]_t\) at any time \(t\) in relation to its initial concentration \([A]_0\), with \(e^{-kt}\) indicating the exponential nature of decay.

Here's why the formula is essential:
  • It elegantly captures the nature of first-order processes.
  • It allows for straightforward calculations of remaining concentrations after certain timing intervals.
  • It helps to determine the rate constant, showing how fast the reaction is occurring.
Learning to apply this formula involves straightforward algebraic manipulations, including taking natural logarithms to extract the rate constant. With practice, manipulating this formula will become second nature, providing a deeper understanding of how reactant concentrations shrink over time in a predictable exponential manner.

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Most popular questions from this chapter

The decomposition of dinitrogen pentoxide has been studied in carbon tetrachloride solvent \(\left(\mathrm{CCl}_{4}\right)\) at a certain temperature: $$ 2 \mathrm{~N}_{2} \mathrm{O}_{5} \longrightarrow 4 \mathrm{NO}_{2}+\mathrm{O}_{2} $$ $$ \begin{array}{cc} {\left[\mathrm{N}_{2} \mathrm{O}_{5}\right](M)} & \text { Initial Rate }(M / \mathrm{s}) \\ \hline 0.92 & 0.95 \times 10^{-5} \\ 1.23 & 1.20 \times 10^{-5} \\ 1.79 & 1.93 \times 10^{-5} \\ 2.00 & 2.10 \times 10^{-5} \\ 2.21 & 2.26 \times 10^{-5} \end{array} $$ Determine graphically the rate law for the reaction and calculate the rate constant.

Write an equation relating the concentration of a reactant \(\mathrm{A}\) at \(t=0\) to that at \(t=t\) for a first-order reaction. Define all the terms and give their units.

For the reaction $$ \mathrm{NO}(g)+\mathrm{O}_{3}(g) \longrightarrow \mathrm{NO}_{2}(g)+\mathrm{O}_{2}(g) $$ the frequency factor \(A\) is \(8.7 \times 10^{12} \mathrm{~s}^{-1}\) and the activation energy is \(63 \mathrm{~kJ} / \mathrm{mol}\). What is the rate constant for the reaction at \(75^{\circ} \mathrm{C} ?\)

The reaction of \(\mathrm{G}_{2}\) with \(\mathrm{E}_{2}\) to form \(2 \mathrm{EG}\) is exothermic, and the reaction of \(\mathrm{G}_{2}\) with \(\mathrm{X}_{2}\) to form \(2 \mathrm{XG}\) is endothermic. The activation energy of the exothermic reaction is greater than that of the endothermic reaction. Sketch the potential energy profile diagrams for these two reactions on the same graph.

For each of these pairs of reaction conditions, indicate which has the faster rate of formation of hydrogen gas: (a) sodium or potassium with water, (b) magnesium or iron with \(1.0 \mathrm{M} \mathrm{HCl}\), (c) magnesium rod or magnesium powder with \(1.0 \mathrm{M} \mathrm{HCl}\) (d) magnesium with \(0.10 \mathrm{M} \mathrm{HCl}\) or magnesium with \(1.0 M \mathrm{HCl}\)

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