/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 209 \(\mathrm{t}_{1}\) can be taken ... [FREE SOLUTION] | 91影视

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Chapter 10: Problem 209

\(\mathrm{t}_{1}\) can be taken as the time taken for the concentration of a reactant to drop to \(\frac{3}{4}\) of its initial value. If the rate constant for a first order reaction is then \(\mathrm{t}_{1 / 4}\) can be written as (a) \(0.1 / k\) (b) \(0.29 / k\) (c) \(0.69 / k\) (d) \(0.75 / k\)

Short Answer

Expert verified
The correct answer is (b) \(0.29/k\).

Step by step solution

01

Understand First Order Reactions

In a first-order reaction, the rate of the reaction depends linearly on the concentration of only one reactant. The rate law is given by \( \text{Rate} = k[A] \), where \([A]\) is the concentration of the reactant and \(k\) is the rate constant.
02

Use First Order Reaction Formula

The formula for a first-order reaction concentration over time is \([A] = [A]_0 e^{-kt}\). We need the time when the concentration \([A]\) becomes \( \frac{3}{4} [A]_0 \).
03

Solve for \( t_1 \)

Set the equation as \( \frac{3}{4}[A]_0 = [A]_0 e^{-kt_1} \). Divide both sides by \([A]_0\), resulting in \( \frac{3}{4} = e^{-kt_1} \).
04

Use Natural Logarithm

Take the natural logarithm of both sides: \(\ln(\frac{3}{4}) = -kt_1\). This simplifies to \(t_1 = -\frac{\ln(\frac{3}{4})}{k}\).
05

Numerical Calculation for \( t_1 \)

Calculate \( \ln(\frac{3}{4}) \), which is \( \ln(1) - \ln(\frac{4}{3}) \). \( \ln(\frac{4}{3}) \approx 0.2877 \), so \(-\ln(\frac{4}{3}) \approx -0.2877\). Therefore, \( t_1 \approx \frac{0.2877}{k} \).
06

Conclusion and Option Match

Comparing \(t_1 = \frac{0.2877}{k}\) with the options, it most closely matches option (b) \(0.29/k\).

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Key Concepts

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

Rate Constant
The rate constant, denoted by \(k\), is a crucial element in chemical reactions, specifically first-order reactions. In a first-order reaction, the rate of reaction is directly proportional to the concentration of one reactant. This means that the reaction rate is dictated by the change in concentration of the reactant. The formula is \(\text{Rate} = k[A]\), where \([A]\) represents the concentration of the reactant.

  • For first-order reactions, the unit of the rate constant \(k\) is \(\text{s}^{-1}\), which stands for per second.
  • The rate constant provides essential information about the speed of the reaction.
  • The larger the \(k\), the faster the reaction occurs.
  • Since \(k\) is specific to each reaction and condition, factors such as temperature can influence its value.
The rate constant aids in determining how long a reaction will take to reach a certain concentration level, which is why it plays a crucial role when calculating concentration over time.
Concentration Over Time
In a first-order reaction, understanding how concentration changes over time helps in predicting the behavior of the reaction. The concentration of the reactant decreases exponentially with respect to time. This property is captured by the equation for concentration over time: \[ [A] = [A]_0 e^{-kt} \] where:- \([A]\) is the concentration of the reactant at time \(t\).- \([A]_0\) is the initial concentration of the reactant.- \(e\) is the base of the natural logarithm.- \(k\) is the rate constant.
  • This formula allows you to compute the concentration at any given time during the reaction process.
  • It provides an understanding of how quickly the reactant concentration is reduced to a desired level.
  • In our exercise, you need to find when the concentration becomes \( \frac{3}{4} \) of its initial value.
By setting up the equation \( \frac{3}{4}[A]_0 = [A]_0 e^{-kt_1} \) and solving for \(t_1\), you can calculate the time required for this change in concentration.
Natural Logarithm Calculation
Logarithms, specifically natural logarithms, are an invaluable tool in chemistry for solving exponential relations seen in concentration time equations. A natural logarithm, denoted \(\ln\), uses the constant \(e\) (approximately 2.718) as its base. In the context of a first-order reaction, natural logarithms are used to linearize the exponential decay equation:\[ \frac{3}{4} = e^{-kt_1} \]Taking the natural logarithm of both sides simplifies this equation:\[ \ln\left( \frac{3}{4} \right) = -kt_1 \]
  • Natural logarithms convert multiplicative processes into additive ones, making them easier to solve algebraically.
  • In our exercise, calculating \( \ln\left( \frac{3}{4} \right) \) leads to solving for \(t_1\), where \(t_1\) represents the time for the concentration to drop to \( \frac{3}{4} \) of its initial value.
  • By solving \( t_1 = -\frac{\ln\left( \frac{3}{4} \right)}{k} \), we find that \(t_1\) approximately equals \( \frac{0.2877}{k} \), thus matching one of the given exercise options.
Natural logarithms hence play a crucial role in simplifying the complex relationships between variables in reaction equations.

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Most popular questions from this chapter

A catalyst increases rate of reaction by (a) decreasing enthalpy (b) decreasing activation energy (c) decreasing internal energy (d) increasing activation energy

The following data pertains to the reaction between \(\mathrm{A}\) and \(\mathrm{B}\) $$ \begin{array}{llll} \hline \text { S. No. } & {[\mathrm{A}] \mathrm{mol} \mathrm{L}^{-1}} & {[\mathrm{~B}] \mathrm{mol} \mathrm{L}^{-1}} & \begin{array}{l} \text { Rate mol } \\ \mathrm{L}-\mathbf{S}^{-1} \end{array} \\ \hline \text { 1. } & 1 \times 10^{-2} & 2 \times 10^{-2} & 2 \times 10^{-4} \\\ 2 . & 2 \times 10^{-2} & 2 \times 10^{-2} & 4 \times 10^{-4} \\ 3 . & 2 \times 10^{-2} & 4 \times 10^{-2} & 8 \times 10^{-8} \\ \hline \end{array} $$ Which of the following inferences are drawn from the above data? (1) rate constant of the reaction is \(10^{-4}\) (2) rate law of the reaction is \([\mathrm{A}][\mathrm{B}]\) (3) rate of reaction increases four times by doubling the concentration of each reactant. Select the correct answer the codes given below: (a) 1 and 3 (b) 2 and 3 (c) land 2 (d) 1,2 and 3

Which of the following statements is correct? (1) order of a reaction can be known from experimental results and not from the stoichiometry of a reaction. (2) molecularity a reaction refers to (i) each of the elementary steps in (an overall mechanism of) a complex reaction or (ii) a single step reaction. (3) overall molecularity of a reaction may be determined in a manner similar to overall order of reaction. (4) overall order of a reaction \(\mathrm{A}^{\mathrm{m}}+\mathrm{B}^{\mathrm{n}} \longrightarrow \mathrm{AB}_{\mathrm{x}}\) is \(\mathrm{m}+\mathrm{n}\) Select the correct answer using the following codes: (a) 2 and 3 (b) 1,3 and 4 (c) 2,3 and 4 (d) 1,2 and 3

The rate law for the reaction \(\mathrm{RCl}+\mathrm{NaOH}(\mathrm{aq}) \longrightarrow \mathrm{ROH}+\mathrm{NaCl}\) is given by Rate \(=k[\mathrm{RCl}]\). The rate of the reaction will be (a) doubled on doubling the concentration of sodium hydroxide (b) halved on reducing the concentration of alkyl halide to one half (c) decreased on increasing the temperature of reaction (d) unaffected by increasing the temperature of the reaction.

A graph plotted between concentration of reactant, consumed at any time \((\mathrm{x})\) and time ' \(\mathrm{t}\) ' is found to be a straight line passing through the origin. The reaction is of (a) first-order (b) zero-order (c) third-order (d) second-order
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