General Chemistry
General Chemistry
7th Edition
ISBN: 9780073402758
Author: Chang, Raymond/ Goldsby
Publisher: McGraw-Hill College
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Chapter 14, Problem 14.70QP
Interpretation Introduction

Interpretation:

From the given information, the time taken for the concentration of A to be four times that of B has to be calculated.

Concept introduction:

The first order reaction:

It is a reaction whose rate depends on the reactant concentration whose power is raised to one in the rate expression.

Consider a reaction:

AProduct

In this reaction, the reactant A has been converted into the product.

The rate expression for the considered reaction can be given as follows:

Rate=-Δ[A]Δt

-Ve sign indicates the decrease in concentration of the reactant A.

[A]- Concentration of the reactant A in molarity.

Δt-Timeperiod

Δ- Difference between initial and final states.

The integrated form of the first order rate expression is:

[A]t=[A]oe-kAt

[A]t-Concentration at any time t or final Concentration of the reactant A.

[A]o- Initial Concentration of the reactant A.

e- Exponent of natural logarithm.

kA- Rate constant for the reaction with the reactant A.

t-Time period.

Half-life: The time required for half of a reactant to be consumed in a reaction is said to be half-life. It is represented by the symbol as t12

The expression for the half-life period in first order kinetics:

t12=0.693k

t12-Half-life period

k-Rate constant.

Expert Solution & Answer
Check Mark

Answer to Problem 14.70QP

The time taken for the concentration of A to be four times that of B is t=56.26min

Explanation of Solution

Given:

t12 of A is 50.0min

t12 of B is 18.0min

Calculating the rate constants from the given half-lives:

t12=0.693kk=0.693t12

Let the rate constant for the reaction with the reactant A be kA.

kA=0.693t12=0.69350.0min=0.01386min1

Therefore, kA=0.01386min1

Let the rate constant for the reaction with the reactant B be kB.

kB=0.693t12=0.69318.0min=0.0385min1

Therefore, kB=0.0385min1

The initial concentrations for A and B are equal. So,

[A]0 = [B]0[A]0[B]0=1

The final concentrations of A is four times of B are equal. So,

[A]t = 4[B]t[A]t [B]t=4

Using the integrated rate expression of first order kinetics, the initial and the final concentrations can be equated as follows:

[A]t[B]t[A]0e-kAt[B]0e-kBt4e-kAte-kBt[A]t = 4[B]tand[A]0 = [B]0=e(kB-kA)t=e(0.0385-0.01386)t

4=e(0.02464)tln4=(0.02464)tt=ln4(0.02464)=56.26min

Therefore, the time taken for the concentration of A to be four times of B is t=56.26min

Conclusion

The time taken for the concentration of A to be four times has been calculated.

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Chapter 14 Solutions

General Chemistry

Ch. 14.4 - Practice Exercise The first-order rate constant...Ch. 14.4 - Review of Concepts (a) What can you deduce about...Ch. 14.5 - Practice Exercise The reaction between NO2 and CO...Ch. 14.5 - Prob. 1RCCh. 14.6 - Prob. 1RCCh. 14 - Prob. 14.1QPCh. 14 - 15.2 Explain the difference between physical...Ch. 14 - Prob. 14.3QPCh. 14 - Prob. 14.4QPCh. 14 - Prob. 14.5QPCh. 14 - 14.6 Consider the reaction Suppose that at a...Ch. 14 - Prob. 14.7QPCh. 14 - 14.8 What are the units for the rate constants of...Ch. 14 - Prob. 14.9QPCh. 14 - Prob. 14.10QPCh. 14 - Prob. 14.11QPCh. 14 - Prob. 14.13QPCh. 14 - Prob. 14.14QPCh. 14 - Prob. 14.15QPCh. 14 - Prob. 14.16QPCh. 14 - Prob. 14.17QPCh. 14 - Prob. 14.18QPCh. 14 - Prob. 14.19QPCh. 14 - Prob. 14.20QPCh. 14 - 14.21 What is the half-life of a compound if 75...Ch. 14 - 14.22 The thermal decomposition of phosphine (PH3)...Ch. 14 - Prob. 14.23QPCh. 14 - Prob. 14.24QPCh. 14 - 14.25 Consider the first-order reaction A → B...Ch. 14 - Prob. 14.26QPCh. 14 - 14.27 Define activation energy. What role does...Ch. 14 - Prob. 14.28QPCh. 14 - Prob. 14.29QPCh. 14 - 14.30 As we know, methane burns readily in oxygen...Ch. 14 - Prob. 14.31QPCh. 14 - Prob. 14.32QPCh. 14 - Prob. 14.33QPCh. 14 - Prob. 14.34QPCh. 14 - Prob. 14.35QPCh. 14 - Prob. 14.36QPCh. 14 - Prob. 14.37QPCh. 14 - 14.38 The rate at which tree crickets chirp is 2.0...Ch. 14 - 14.39 The diagram here describes the initial state...Ch. 14 - Prob. 14.40QPCh. 14 - Prob. 14.41QPCh. 14 - Prob. 14.42QPCh. 14 - 14.43 Explain why termolecular reactions are...Ch. 14 - 14.44 What is the rate-determining step of a...Ch. 14 - Prob. 14.45QPCh. 14 - Prob. 14.46QPCh. 14 - Prob. 14.47QPCh. 14 - Prob. 14.48QPCh. 14 - Prob. 14.49QPCh. 14 - Prob. 14.50QPCh. 14 - Prob. 14.51QPCh. 14 - Prob. 14.52QPCh. 14 - Prob. 14.53QPCh. 14 - Prob. 14.54QPCh. 14 - Prob. 14.55QPCh. 14 - Prob. 14.56QPCh. 14 - Prob. 14.57QPCh. 14 - Prob. 14.58QPCh. 14 - Prob. 14.59QPCh. 14 - Prob. 14.60QPCh. 14 - Prob. 14.61QPCh. 14 - Prob. 14.62QPCh. 14 - Prob. 14.63QPCh. 14 - Prob. 14.64QPCh. 14 - Prob. 14.65QPCh. 14 - 14.66 The decomposition of N2O to N2 and O2 is a...Ch. 14 - Prob. 14.67QPCh. 14 - Prob. 14.68QPCh. 14 - 14.69 Consider the zero-order reaction a → B....Ch. 14 - Prob. 14.70QPCh. 14 - Prob. 14.72QPCh. 14 - Prob. 14.73QPCh. 14 - Prob. 14.74QPCh. 14 - Prob. 14.75QPCh. 14 - Prob. 14.76QPCh. 14 - Prob. 14.77QPCh. 14 - Prob. 14.78QPCh. 14 - Prob. 14.79QPCh. 14 - Prob. 14.80QPCh. 14 - Prob. 14.81QPCh. 14 - Prob. 14.82QPCh. 14 - 14.83 When a mixture of methane and bromine is...Ch. 14 - 14.84 Consider this elementary step: (a)...Ch. 14 - Prob. 14.85QPCh. 14 - Prob. 14.86QPCh. 14 - 14.87 In recent years ozone in the stratosphere...Ch. 14 - Prob. 14.88QPCh. 14 - Prob. 14.90QPCh. 14 - Prob. 14.91QPCh. 14 - Prob. 14.92QPCh. 14 - Prob. 14.93QPCh. 14 - Prob. 14.94QPCh. 14 - Prob. 14.95QPCh. 14 - Prob. 14.96QPCh. 14 - Prob. 14.97QPCh. 14 - Prob. 14.98QPCh. 14 - Prob. 14.100QPCh. 14 - Prob. 14.101QPCh. 14 - 14.102 Consider the potential energy profiles...Ch. 14 - Prob. 14.103QPCh. 14 - Prob. 14.104QPCh. 14 - 14.105 The activation energy for the...Ch. 14 - Prob. 14.106QPCh. 14 - Prob. 14.107SPCh. 14 - Prob. 14.108SPCh. 14 - Prob. 14.109SPCh. 14 - Prob. 14.110SPCh. 14 - Prob. 14.111SPCh. 14 - Prob. 14.112SPCh. 14 - Prob. 14.113SPCh. 14 - Prob. 14.114SPCh. 14 - Prob. 14.115SPCh. 14 - 14.116 To prevent brain damage, a drastic medical...Ch. 14 - Prob. 14.117SPCh. 14 - Prob. 14.118SP
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