CHEMISTRY >CUSTOM<
CHEMISTRY >CUSTOM<
8th Edition
ISBN: 9781309097182
Author: SILBERBERG
Publisher: MCG/CREATE
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Chapter 16, Problem 16.20P

(a)

Interpretation Introduction

Interpretation:

Average rate over the entire experiment has to be determined.

Concept introduction:

Relative rates and stoichiometry: During a chemical reaction, amounts of reactant decrease with time and amounts of products increases.

Reaction Rate = - 1coefficientΔ[reactants]Δt = +1coefficient Δ[products]Δt

(a)

Expert Solution
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Explanation of Solution

The reaction given is:

Givenreaction: NOBr(g)NO(g)+12Br2(g)

The average rate of the chemical reaction:

Reaction Rate = - 1coefficientΔ[reactants]Δt = +1coefficient Δ[products]ΔtReaction Rate = -Δ[NOBr]Δt(1)

Δ[NOBr]=(0.0033)-(0.0100)=0.0067mol/L;Δt=(10.0sec)(0.00sec)=10.0sec

Substituting in the equation (1) as follows,

Reaction Rate = - Δ[NOBr]Δt=(-(0.0067)10.0)=0.00067=6.7×10-4mol/L.s.

Therefore, the average rate over the entire experiment was 6.7×10-4mol/L.s.

(b)

Interpretation Introduction

Interpretation:

Average rate between 2.0and4.0sec has to be determined.

Concept introduction:

Relative rates and stoichiometry: During a chemical reaction, amounts of reactant decrease with time and amounts of products increases.

Reaction Rate = - 1coefficientΔ[reactants]Δt = +1coefficient Δ[products]Δt

(b)

Expert Solution
Check Mark

Explanation of Solution

The reaction given is:

Givenreaction: NOBr(g)NO(g)+12Br2(g)

The average rate of the chemical reaction:

Reaction Rate = - 1coefficientΔ[reactants]Δt = +1coefficient Δ[products]ΔtReaction Rate = - Δ[NOBr]Δt(1)

Δ[NOBr]=(0.0055)-(0.0071)=0.0016mol/L;Δt=(4.0sec)(2.00sec)=2.0sec

Substituting in the equation (1) as follows,

Reaction Rate = -Δ[NOBr]Δt=(-(0.0016)2.0)=0.0008=8.0×10-4mol/L.s.

Therefore, the average rate between 2.0and4.0sec was 8.0×10-4mol/L.s.

(c)

Interpretation Introduction

Interpretation:

Initial reaction rate has to be calculated using graphical method.

Concept introduction:

Rate law or rate equation: The relationship between the reactant concentrations and reaction rate is expressed by an equation.

aA + bBxXRate of reaction = k [A]m[B]nTotalorderof reaction = (m + n).

Reaction Rate = k [A]m[B]n[C]p,where 'm, n and p' are orders of the reactants.

Order of a reaction: The order of a reaction with respect to a particular reactant is the exponent of its concentration term in the rate law expression, and the overall reaction order is the sum of the exponents on all concentration terms.

Rate constant, k: It is a proportionality constant that relates rate and concentration at a given temperature.

(c)

Expert Solution
Check Mark

Explanation of Solution

CHEMISTRY >CUSTOM<                     , Chapter 16, Problem 16.20P

Figure.1

Using the given values, the graph has been plotted as shown above.

The initial reaction rate is calculated as shown below,

Reaction Rate = ΔyΔx=-Δ[NOBr]Δt=((0.0040 - 0.0100)(4.00.0))=1.5×10-3mol/L.s.

Therefore, the initial reaction rate obtained as 1.5×10-3mol/L.s.

(d)

Interpretation Introduction

Interpretation:

Rate of reaction at 7.0sec has to be calculated using the graphical method.

Concept introduction:

Rate law or rate equation: The relationship between the reactant concentrations and reaction rate is expressed by an equation.

aA + bBxXRate of reaction = k [A]m[B]nTotalorderof reaction = (m + n).

Reaction Rate = k [A]m[B]n[C]p,where 'm, n and p' are orders of the reactants.

Order of a reaction: The order of a reaction with respect to a particular reactant is the exponent of its concentration term in the rate law expression, and the overall reaction order is the sum of the exponents on all concentration terms.

Rate constant, k: It is a proportionality constant that relates rate and concentration at a given temperature.

(d)

Expert Solution
Check Mark

Explanation of Solution

The reaction rate at 7.0sec is calculated as shown below,

Reaction Rate = ΔyΔx=-Δ[NOBr]Δt=((0.0030 - 0.0050)(11.04.0))=2.857×10-4mol/L.s.

Therefore, the reaction rate at 7.0sec obtained as 2.9×10-4mol/L.s.

(e)

Interpretation Introduction

Interpretation:

Particular time the instantaneous rate equal the average rate over the entire experiment has to be determined.

Concept introduction:

Rate law or rate equation: The relationship between the reactant concentrations and reaction rate is expressed by an equation.

aA + bBxXRate of reaction = k [A]m[B]nTotalorderof reaction = (m + n).

Reaction Rate = k [A]m[B]n[C]p,where 'm, n and p' are orders of the reactants.

Order of a reaction: The order of a reaction with respect to a particular reactant is the exponent of its concentration term in the rate law expression, and the overall reaction order is the sum of the exponents on all concentration terms.

Rate constant, k: It is a proportionality constant that relates rate and concentration at a given temperature.

(e)

Expert Solution
Check Mark

Explanation of Solution

The average between t=3.0secandt = 5 sec is calculated as shown below,

Reaction Rate = ΔyΔx=-Δ[NOBr]Δt=((0.0050 - 0.0063)(5.03.0))=6.5×10-4mol/L.s.

Rate of reaction at t=4 sec is approximately 6.5×10-4mol/L.s, hence the rates are equal at about t=4 sec

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Chapter 16 Solutions

CHEMISTRY >CUSTOM<

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