General Chemistry
General Chemistry
7th Edition
ISBN: 9780073402758
Author: Chang, Raymond/ Goldsby
Publisher: McGraw-Hill College
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Chapter 14, Problem 14.26QP

(a)

Interpretation Introduction

Interpretation:

The relative rate of the reaction in the given three containers has to be determined.

Concept introduction:

Rate of the reaction is the change in the concentration of reactant or a product with time.

Rate equation for the general reaction A+BProduct is,

Rate=krateconstat[A][B]

(a)

Expert Solution
Check Mark

Explanation of Solution

The given first order reaction is,

XY

For first order reaction, Rate=k[X]

X molecules are placed in three equal-volume containers at the same temperature.

General Chemistry, Chapter 14, Problem 14.26QP , additional homework tip  1

Fig (1)

The relative rates of the reaction in the given three containers can be determined as follows,

For convenience, we can use the number of molecules to represent the concentration. Therefore, the relative rates of reaction for the three containers are,

(i)Rate=8k(ii)Rate=6k(iii)Rate=12k

To get the relative rates dividing each rate by 2k

Therefore

Relative rate of the reaction in three containers are 4:3:6

(b)

Interpretation Introduction

Interpretation:

To determine the relative rates be affected if the volume of the each container were doubled

Concept introduction:

Rate of the reaction is the change in the concentration of reactant or a product with time.

Rate equation for the general reaction A+BProduct is,

Rate=krateconstat[A][B]

(b)

Expert Solution
Check Mark

Explanation of Solution

The given first order reaction is,

XY

For first order reaction, Rate=k[X]

X molecules are placed in three equal-volume containers at the same temperature.

General Chemistry, Chapter 14, Problem 14.26QP , additional homework tip  2

Fig (1)

The relative rates of the reaction in the given three containers can be determined as follows,

For convenience, we can use the number of molecules to represent the concentration. Therefore, the relative rates of reaction for the three containers are,

(i)Rate=8k(ii)Rate=6k(iii)Rate=12k

To get the relative rates dividing each rate by 2k

Therefore

Relative rate of the reaction in three containers are 4:3:6

The relative rates would be unaffected if the volume of the each container were doubled. Therefore, the relative rates between the three containers would remain same and so the actual rate would decrease by 50%  .

(c)

Interpretation Introduction

Interpretation:

The relative half-life of the reactions in (i) to (iii) has to be determined.

Concept introduction:

Rate of the reaction is the change in the concentration of reactant or a product with time.

Rate equation for the general reaction A+BProduct is,

Rate=krateconstat[A][B]

Half-life is the time required for one half of a reactant to react.

For first order reaction kt=ln([A][A0])

[A] Is the concentration of reactant A at time t [A]0 is the initial concentration of reactant, k is the rate constant.

t12=ln(2)k

Half-life for a first order reaction is t12=0.693k ; half-life for a first order reaction is independent of initial concentration of reactant.

(c)

Expert Solution
Check Mark

Explanation of Solution

The given first order reaction is,

XY

For first order reaction, Rate=k[X]

X molecules are placed in three equal-volume containers at the same temperature.

General Chemistry, Chapter 14, Problem 14.26QP , additional homework tip  3

Fig (1)

We know that, the half-life of a first order reaction is independent on substrate (reactant) concentration; it does not depend on substrate concentration.

Therefore, the relative half-life of the reactions in (i) to (iii) will be same; 1:1:1

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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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