The rate constant for the first order reaction of the below described equation at 500 ° C is 5 .5 × 10 -4 s -1 . The half-life of cyclopropane at 500 ° C has to be calculated. Also, the concentration of cyclopropane that remains after 2 .0 hours has to be calculated, if the initial cyclopropane concentration is 1 .00 × 10 -3 M at 500 ° C . Concept Introduction: The half-life of first order reaction is calculated by the following equation, t 1/2 = 0 .693 k t 1/2 = half-life of the reaction k = Rate constant of the reaction The concentration of A that remains after time, t in a first order reaction can be calculated by the following equation: ln [ A ] [ A ] 0 = -kt [ A ] = concentration of A at time,t [ A ] 0 = initial concentration of A k = Rate constant t = time

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Answer 1

Question

Chapter 17, Problem 17.22P

Interpretation Introduction

Interpretation:

The rate constant for the first order reaction of the below described equation at $500°C$ is $5.5 × 10-4 s-1$ .

The half-life of cyclopropane at $500°C$ has to be calculated. Also, the concentration of cyclopropane that remains after $2.0 hours$ has to be calculated, if the initial cyclopropane concentration is $1.00 × 10-3 M$ at $500°C$ .

Concept Introduction:

The half-life of first order reaction is calculated by the following equation,

$t1/2 = 0.693kt1/2 = half-life of the reactionk = Rate constant of the reaction$

The concentration of $A$ that remains after time, $t$ in a first order reaction can be calculated by the following equation:

$ln [A][A]0 = -kt[A] = concentration of A at time,t[A]0 = initial concentration of Ak = Rate constantt = time$

Expert Solution & Answer

Explanation of Solution

Given,

$k = 5.5 × 10-4 s-1[cyclopropane]0 = 1.00 × 10-3 Mt = 2.0 hours (1 hour = 3600 s2.0 hours = 7200 s)$

The half-life of cyclopropane can be calculated as

$t1/2 = 0.693k = 0.6935.5 × 10-4 s-1 = 1260 s$

The concentration of cyclopropane that remains after $2.0 hours$ can be calculated as

$ln [cyclopropane][cyclopropane]0 = -ktln [cyclopropane][1.00 × 10-3 M] = - (5.5 × 10-4 s-1) (7200 s)ln [cyclopropane][1.00 × 10-3 M] = - 3.96[cyclopropane][1.00 × 10-3 M] = e-3.96[cyclopropane][1.00 × 10-3 M] = 0.019[cyclopropane] = 0.019 × (1.00 × 10-3 M) = 1.9 × 10-5 M$

The concentration of cyclopropane that remains after $2.0 hours$ is $1.9 × 10-5 M$ .

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Answer 2

Question

Chapter 17, Problem 17.22P

Interpretation Introduction

Interpretation:

The rate constant for the first order reaction of the below described equation at $500°C$ is $5.5 × 10-4 s-1$ .

The half-life of cyclopropane at $500°C$ has to be calculated. Also, the concentration of cyclopropane that remains after $2.0 hours$ has to be calculated, if the initial cyclopropane concentration is $1.00 × 10-3 M$ at $500°C$ .

Concept Introduction:

The half-life of first order reaction is calculated by the following equation,

$t1/2 = 0.693kt1/2 = half-life of the reactionk = Rate constant of the reaction$

The concentration of $A$ that remains after time, $t$ in a first order reaction can be calculated by the following equation:

$ln [A][A]0 = -kt[A] = concentration of A at time,t[A]0 = initial concentration of Ak = Rate constantt = time$

Expert Solution & Answer

Explanation of Solution

Given,

$k = 5.5 × 10-4 s-1[cyclopropane]0 = 1.00 × 10-3 Mt = 2.0 hours (1 hour = 3600 s2.0 hours = 7200 s)$

The half-life of cyclopropane can be calculated as

$t1/2 = 0.693k = 0.6935.5 × 10-4 s-1 = 1260 s$

The concentration of cyclopropane that remains after $2.0 hours$ can be calculated as

$ln [cyclopropane][cyclopropane]0 = -ktln [cyclopropane][1.00 × 10-3 M] = - (5.5 × 10-4 s-1) (7200 s)ln [cyclopropane][1.00 × 10-3 M] = - 3.96[cyclopropane][1.00 × 10-3 M] = e-3.96[cyclopropane][1.00 × 10-3 M] = 0.019[cyclopropane] = 0.019 × (1.00 × 10-3 M) = 1.9 × 10-5 M$

The concentration of cyclopropane that remains after $2.0 hours$ is $1.9 × 10-5 M$ .

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Answer 3

Question

Chapter 17, Problem 17.22P

Interpretation Introduction

Interpretation:

The rate constant for the first order reaction of the below described equation at $500°C$ is $5.5 × 10-4 s-1$ .

The half-life of cyclopropane at $500°C$ has to be calculated. Also, the concentration of cyclopropane that remains after $2.0 hours$ has to be calculated, if the initial cyclopropane concentration is $1.00 × 10-3 M$ at $500°C$ .

Concept Introduction:

The half-life of first order reaction is calculated by the following equation,

$t1/2 = 0.693kt1/2 = half-life of the reactionk = Rate constant of the reaction$

The concentration of $A$ that remains after time, $t$ in a first order reaction can be calculated by the following equation:

$ln [A][A]0 = -kt[A] = concentration of A at time,t[A]0 = initial concentration of Ak = Rate constantt = time$

Expert Solution & Answer

Explanation of Solution

Given,

$k = 5.5 × 10-4 s-1[cyclopropane]0 = 1.00 × 10-3 Mt = 2.0 hours (1 hour = 3600 s2.0 hours = 7200 s)$

The half-life of cyclopropane can be calculated as

$t1/2 = 0.693k = 0.6935.5 × 10-4 s-1 = 1260 s$

The concentration of cyclopropane that remains after $2.0 hours$ can be calculated as

$ln [cyclopropane][cyclopropane]0 = -ktln [cyclopropane][1.00 × 10-3 M] = - (5.5 × 10-4 s-1) (7200 s)ln [cyclopropane][1.00 × 10-3 M] = - 3.96[cyclopropane][1.00 × 10-3 M] = e-3.96[cyclopropane][1.00 × 10-3 M] = 0.019[cyclopropane] = 0.019 × (1.00 × 10-3 M) = 1.9 × 10-5 M$

The concentration of cyclopropane that remains after $2.0 hours$ is $1.9 × 10-5 M$ .

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Answer 4

Question

Chapter 17, Problem 17.22P

Interpretation Introduction

Interpretation:

The rate constant for the first order reaction of the below described equation at $500°C$ is $5.5 × 10-4 s-1$ .

The half-life of cyclopropane at $500°C$ has to be calculated. Also, the concentration of cyclopropane that remains after $2.0 hours$ has to be calculated, if the initial cyclopropane concentration is $1.00 × 10-3 M$ at $500°C$ .

Concept Introduction:

The half-life of first order reaction is calculated by the following equation,

$t1/2 = 0.693kt1/2 = half-life of the reactionk = Rate constant of the reaction$

The concentration of $A$ that remains after time, $t$ in a first order reaction can be calculated by the following equation:

$ln [A][A]0 = -kt[A] = concentration of A at time,t[A]0 = initial concentration of Ak = Rate constantt = time$

Expert Solution & Answer

Explanation of Solution

Given,

$k = 5.5 × 10-4 s-1[cyclopropane]0 = 1.00 × 10-3 Mt = 2.0 hours (1 hour = 3600 s2.0 hours = 7200 s)$

The half-life of cyclopropane can be calculated as

$t1/2 = 0.693k = 0.6935.5 × 10-4 s-1 = 1260 s$

The concentration of cyclopropane that remains after $2.0 hours$ can be calculated as

$ln [cyclopropane][cyclopropane]0 = -ktln [cyclopropane][1.00 × 10-3 M] = - (5.5 × 10-4 s-1) (7200 s)ln [cyclopropane][1.00 × 10-3 M] = - 3.96[cyclopropane][1.00 × 10-3 M] = e-3.96[cyclopropane][1.00 × 10-3 M] = 0.019[cyclopropane] = 0.019 × (1.00 × 10-3 M) = 1.9 × 10-5 M$

The concentration of cyclopropane that remains after $2.0 hours$ is $1.9 × 10-5 M$ .

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Answer 5

Chapter 17, Problem 17.22P

Interpretation Introduction

Interpretation:

The rate constant for the first order reaction of the below described equation at $500°C$ is $5.5 × 10-4 s-1$ .

The half-life of cyclopropane at $500°C$ has to be calculated. Also, the concentration of cyclopropane that remains after $2.0 hours$ has to be calculated, if the initial cyclopropane concentration is $1.00 × 10-3 M$ at $500°C$ .

Concept Introduction:

The half-life of first order reaction is calculated by the following equation,

$t1/2 = 0.693kt1/2 = half-life of the reactionk = Rate constant of the reaction$

The concentration of $A$ that remains after time, $t$ in a first order reaction can be calculated by the following equation:

$ln [A][A]0 = -kt[A] = concentration of A at time,t[A]0 = initial concentration of Ak = Rate constantt = time$

Expert Solution & Answer

Explanation of Solution

Given,

$k = 5.5 × 10-4 s-1[cyclopropane]0 = 1.00 × 10-3 Mt = 2.0 hours (1 hour = 3600 s2.0 hours = 7200 s)$

The half-life of cyclopropane can be calculated as

$t1/2 = 0.693k = 0.6935.5 × 10-4 s-1 = 1260 s$

The concentration of cyclopropane that remains after $2.0 hours$ can be calculated as

$ln [cyclopropane][cyclopropane]0 = -ktln [cyclopropane][1.00 × 10-3 M] = - (5.5 × 10-4 s-1) (7200 s)ln [cyclopropane][1.00 × 10-3 M] = - 3.96[cyclopropane][1.00 × 10-3 M] = e-3.96[cyclopropane][1.00 × 10-3 M] = 0.019[cyclopropane] = 0.019 × (1.00 × 10-3 M) = 1.9 × 10-5 M$

The concentration of cyclopropane that remains after $2.0 hours$ is $1.9 × 10-5 M$ .

Answer 6

Chapter 17, Problem 17.22P

Interpretation Introduction

Interpretation:

The rate constant for the first order reaction of the below described equation at $500°C$ is $5.5 × 10-4 s-1$ .

The half-life of cyclopropane at $500°C$ has to be calculated. Also, the concentration of cyclopropane that remains after $2.0 hours$ has to be calculated, if the initial cyclopropane concentration is $1.00 × 10-3 M$ at $500°C$ .

Concept Introduction:

The half-life of first order reaction is calculated by the following equation,

$t1/2 = 0.693kt1/2 = half-life of the reactionk = Rate constant of the reaction$

The concentration of $A$ that remains after time, $t$ in a first order reaction can be calculated by the following equation:

$ln [A][A]0 = -kt[A] = concentration of A at time,t[A]0 = initial concentration of Ak = Rate constantt = time$

Expert Solution & Answer

Explanation of Solution

Given,

$k = 5.5 × 10-4 s-1[cyclopropane]0 = 1.00 × 10-3 Mt = 2.0 hours (1 hour = 3600 s2.0 hours = 7200 s)$

The half-life of cyclopropane can be calculated as

$t1/2 = 0.693k = 0.6935.5 × 10-4 s-1 = 1260 s$

The concentration of cyclopropane that remains after $2.0 hours$ can be calculated as

$ln [cyclopropane][cyclopropane]0 = -ktln [cyclopropane][1.00 × 10-3 M] = - (5.5 × 10-4 s-1) (7200 s)ln [cyclopropane][1.00 × 10-3 M] = - 3.96[cyclopropane][1.00 × 10-3 M] = e-3.96[cyclopropane][1.00 × 10-3 M] = 0.019[cyclopropane] = 0.019 × (1.00 × 10-3 M) = 1.9 × 10-5 M$

The concentration of cyclopropane that remains after $2.0 hours$ is $1.9 × 10-5 M$ .

Answer 7

Chapter 17, Problem 17.22P

Interpretation Introduction

Interpretation:

The rate constant for the first order reaction of the below described equation at $500°C$ is $5.5 × 10-4 s-1$ .

The half-life of cyclopropane at $500°C$ has to be calculated. Also, the concentration of cyclopropane that remains after $2.0 hours$ has to be calculated, if the initial cyclopropane concentration is $1.00 × 10-3 M$ at $500°C$ .

Concept Introduction:

The half-life of first order reaction is calculated by the following equation,

$t1/2 = 0.693kt1/2 = half-life of the reactionk = Rate constant of the reaction$

The concentration of $A$ that remains after time, $t$ in a first order reaction can be calculated by the following equation:

$ln [A][A]0 = -kt[A] = concentration of A at time,t[A]0 = initial concentration of Ak = Rate constantt = time$

Answer 8

Expert Solution & Answer

Explanation of Solution

Given,

$k = 5.5 × 10-4 s-1[cyclopropane]0 = 1.00 × 10-3 Mt = 2.0 hours (1 hour = 3600 s2.0 hours = 7200 s)$

The half-life of cyclopropane can be calculated as

$t1/2 = 0.693k = 0.6935.5 × 10-4 s-1 = 1260 s$

The concentration of cyclopropane that remains after $2.0 hours$ can be calculated as

$ln [cyclopropane][cyclopropane]0 = -ktln [cyclopropane][1.00 × 10-3 M] = - (5.5 × 10-4 s-1) (7200 s)ln [cyclopropane][1.00 × 10-3 M] = - 3.96[cyclopropane][1.00 × 10-3 M] = e-3.96[cyclopropane][1.00 × 10-3 M] = 0.019[cyclopropane] = 0.019 × (1.00 × 10-3 M) = 1.9 × 10-5 M$