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Chapter 11: Problem 71

For a first-order reaction, \(t_{0.75}=1386 \mathrm{~s}\). Its specific reaction rate is (a) \(10^{-3} \mathrm{~s}^{-1}\) (b) \(10^{-2} \mathrm{~s}^{-1}\) (c) \(10^{-4} \mathrm{~s}^{-1}\) (d) \(10^{-5} \mathrm{~s}^{-1}\)

Short Answer

Expert verified
\(k = 10^{-4} \mathrm{~s}^{-1}\)

Step by step solution

01

Understanding the Given Problem

We know that for a first-order reaction the time required to consume a certain fraction of reactant (in this case, 75%, which is the fraction left after the reaction) is related to the rate constant, k. The relationship is given by the formula: \( t = \frac{\ln(1/f)}{k} \), where \( f \) is the fraction of reactant remaining and \( t \) is the time. Here, \( t_{0.75} = 1386 \, s \) is the time at which 75% of the reactant remains unchanged.
02

Substitute the Known Values into the Equation

We substitute \( t_{0.75} = 1386 \, s \) and \( f = 0.75 \) into the equation to find \( k \): \( 1386 = \frac{\ln(1/0.75)}{k} \). We now need to solve for \( k \).
03

Calculate the Natural Logarithm Value

Calculate the value of \( \ln(1/0.75) = \ln(4/3) \) to find the numerator.
04

Solve for the Specific Reaction Rate, k

Rearrange the equation to solve for \( k \), which gives \( k = \frac{\ln(4/3)}{1386} \). Next, we calculate this value to find the rate constant.
05

Calculate the Specific Reaction Rate

Using a calculator, compute the value of \( k \) to determine the specific reaction rate. We calculate this to find which option (a, b, c, or d) corresponds with the calculated value.

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Key Concepts

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

Chemical Kinetics
Chemical kinetics is the branch of physical chemistry that deals with understanding the rates of chemical reactions. It's essential for predicting how fast reactions proceed, which is crucial for a wide range of scientific and industrial processes.

Within chemical kinetics, the term 'first-order reaction' refers to a reaction where the rate is directly proportional to the concentration of one of the reactants. This implies that if you were to double the concentration of the reactant, the rate of reaction would also double. A characteristic feature of first-order reactions is their unique half-life, which remains constant regardless of the initial concentration. This concept of half-life and reaction orders helps chemists to design and control chemical processes by predicting the timing and yields of the reactions involved.
Reaction Rate
The reaction rate is a measure of how fast a chemical reaction occurs. For a first-order reaction, the rate can be expressed in terms of the change in concentration of a reactant, A, over time using the equation: \[ -\frac{d[A]}{dt} = k[A] \]where \( [A] \) is the concentration of A, \( t \) is time, and \( k \) is the first-order rate constant unique to the reaction. The minus sign indicates that the concentration of A decreases with time.

In our original exercise, rather than directly measuring the change in concentration, the time taken for a certain percentage of the reactant to remain (here 75%) is used to deduce the rate constant, illustrating a practical approach to studying reaction kinetics without detailed concentration data.
Rate Constant Calculation
The rate constant, represented by the symbol \( k \), is a coefficient that quantifies the rate of a chemical reaction. For a first-order reaction, the time required for a certain fraction of the reactant to remain can be used to calculate \( k \). The equation for a first-order reaction is \[ t = \frac{\ln(1/f)}{k} \], where \( f \) is the fraction of the original concentration of reactant remaining after time \( t \).

In the step-by-step solution provided, the fraction remaining is 0.75, and the time is 1386 seconds. By rearranging the formula and inserting the given values, one can solve for \( k \) to find which of the provided options corresponds to the calculated rate constant. It's a straightforward yet powerful tool for scientists to predict the behavior of reactions under different conditions.

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Most popular questions from this chapter

Two substances, 'A' and 'B' are initially present as \(\left[A_{0}\right]=8\left[B_{0}\right]\) and \(t_{1 / 2}\) for the firstorder decomposition of 'A' and 'B' are 10 and \(20 \mathrm{~min}\), respectively. If they start decomposing at the same time, after how much time, the concentration of both of them would be same? (a) \(20 \mathrm{~min}\) (b) \(40 \mathrm{~min}\) (c) \(60 \mathrm{~min}\) (d) \(200 \mathrm{~min}\)

For the reaction: \(\mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g})\) \(\rightarrow 2 \mathrm{NH}_{3}(\mathrm{~g})\) under certain conditions of temperature and partial pressure of the reactants, the rate of formation of \(\mathrm{NH}_{3}\) is \(10^{-3} \mathrm{~kg} \mathrm{~h}^{-1}\). The rate of consumption of \(\mathrm{H}_{2}\) under same condition is (a) \(1.5 \times 10^{-3} \mathrm{~kg} \mathrm{~h}^{-1}\) (b) \(1.76 \times 10^{-4} \mathrm{~kg} \mathrm{~h}^{-1}\) (c) \(6.67 \times 10^{-4} \mathrm{~kg} \mathrm{~h}^{-1}\) (d) \(3 \times 10^{-3} \mathrm{~kg} \mathrm{~h}^{-1}\)

A substance 'A' decomposes in solution following first-order kinetics. Flask 1 contains 11 of \(1 \mathrm{M}\) solution of \(\mathrm{A}^{\prime}\) and flask 2 contains \(100 \mathrm{ml}\) of \(0.6 \mathrm{M}\) solution of 'A'. After \(8.0 \mathrm{~h}\), the concentration of 'A' in flask 1 becomes \(0.25 \mathrm{M}\). In what time, the concentration of 'A' in flask 2 becomes \(0.3 \mathrm{M}\) ? (a) \(8.0 \mathrm{~h}\) (b) \(3.2 \mathrm{~h}\) (c) \(4.0 \mathrm{~h}\) (d) \(9.6 \mathrm{~h}\)

In the gas phase, two butadiene molecules can dimerizes to give larger molecules according to the reaction: \(2 \mathrm{C}_{4} \mathrm{H}_{6}(\mathrm{~g})\) \(\rightarrow \mathrm{C}_{8} \mathrm{H}_{12}(\mathrm{~g})\). The rate law for this reac- tion is, \(r=K\left[\mathrm{C}_{4} \mathrm{H}_{6}\right]^{2}\) with \(K=6.1 \times 10^{-2}\) \(1 \mathrm{~mol}^{-1} \mathrm{~s}^{-1}\) at the temperature of reaction. The rate of formation of \(\mathrm{C}_{8} \mathrm{H}_{12}\), when the concentration of \(\mathrm{C}_{4} \mathrm{H}_{6}\) is \(0.02 \mathrm{M}\), is (a) \(2.44 \times 10^{-5} \mathrm{Ms}^{-1}\) (b) \(1.22 \times 10^{-5} \mathrm{Ms}^{-1}\) (c) \(1.22 \times 10^{-3} \mathrm{Ms}^{-1}\) (d) \(2.44 \times 10^{-6} \mathrm{Ms}^{-1}\)

Which of the following statements is not correct? (a) The efficiency of a solid catalyst depends upon its surface area. (b) Catalyst operates by providing alternate path for the reaction that involves lower activation energy. (c) Catalyst lowers the energy of activation of the forwards direction without affecting the energy of activation of the backward direction. (d) Catalyst does not affect the overall enthalpy change of the reaction.
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