Chemistry: The Molecular Science
Chemistry: The Molecular Science
5th Edition
ISBN: 9781285199047
Author: John W. Moore, Conrad L. Stanitski
Publisher: Cengage Learning
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Chapter 11, Problem 13QRT

(a)

Interpretation Introduction

Interpretation:

Average rate of reaction in the interval 0.00to0.50h has to be calculated.

Concept Introduction:

Consider the following reaction.

  aA+bBcC

Where A and B are reactants and C is the product. a,bandc are stoichiometric co-efficient of A,BandC.

  rate=-1aΔ[A]Δt=-1bΔ[B]Δt=1cΔ[C]Δt

Where,

    Δ[A]Δt= rate of disappearance of A

    Δ[B]Δt= rate of disappearance of B

    Δ[C]Δt= rate of appearance of C

Rate of the reaction can be expressed in terms of change in concentration of reactant and products by multiplying the reciprocal of the corresponding stoichiometric co-efficient to that.

When the rate is expressed in terms of change in reactant concentration, a minus sign has to be given.  Since change in time will be a positive quantity and reactant concentration decreases with time change in concentration of reactant will be negative.  So in order to make the rate a positive quantity negative sign is given.

But if the rate is expressed in terms of change in concentration of products, which is a positive quantity, no need of negative sign in the rate expression.

(a)

Expert Solution
Check Mark

Answer to Problem 13QRT

Average rate is 0.23molL-1h-1.

Explanation of Solution

Given that the concentration of N2O5 at t=0.00h is 0.849mol/L and the concentration of N2O5 at t=0.50h is 0.733mol/L.

So the average rate can be calculated as follows,

  rate=changeinconcentrationchangeintime=[0.733mol/L0.849mol/L][0.50h0.00h]=0.23molL-1h-1.

Average rate is 0.23molL-1h-1.

(b)

Interpretation Introduction

Interpretation:

Average rate of reaction in the interval 0.50to1.0h has to be calculated.

Concept Introduction:

Refer to part (a).

(b)

Expert Solution
Check Mark

Answer to Problem 13QRT

Average rate is 0.20molL-1h-1.

Explanation of Solution

Given that the concentration of N2O5 at t=0.50h is 0.733mol/L and the concentration of N2O5 at t=1.0h is 0.633mol/L.

So the average rate can be calculated as follows,

  rate=changeinconcentrationchangeintime=[0.633mol/L0.733mol/L][1.0h0.50h]=0.20molL-1h-1.

Average rate is 0.20molL-1h-1.

(c)

Interpretation Introduction

Interpretation:

Average rate of reaction in the interval 1to2h has to be calculated.

Concept Introduction:

Refer to part (a).

(c)

Expert Solution
Check Mark

Answer to Problem 13QRT

Average rate is 0.161molL-1h-1.

Explanation of Solution

Given that the concentration of N2O5 at t=1h is 0.633mol/L and the concentration of N2O5 at t=2.0h is 0.472mol/L.

So the average rate can be calculated as follows,

  rate=changeinconcentrationchangeintime=[0.472mol/L0.633mol/L][2h1h]=0.161molL-1h-1.

Average rate is 0.161molL-1h-1.

(d)

Interpretation Introduction

Interpretation:

Average rate of reaction in the interval 2to3h has to be calculated.

Concept Introduction:

Refer to part (a).

(d)

Expert Solution
Check Mark

Answer to Problem 13QRT

Average rate is 0.120molL-1h-1.

Explanation of Solution

Given that the concentration of N2O5 at t=2h is 0.473mol/L and the concentration of N2O5 at t=3.0h is 0.352mol/L.

So the average rate can be calculated as follows,

  rate=changeinconcentrationchangeintime=[0.352mol/L0.473mol/L][3h2h]=0.120molL-1h-1.

Average rate is 0.120molL-1h-1.

(e)

Interpretation Introduction

Interpretation:

Average rate of reaction in the interval 3to4h has to be calculated.

Concept Introduction:

Refer to part (a).

(e)

Expert Solution
Check Mark

Answer to Problem 13QRT

Average rate is 0.090molL-1h-1.

Explanation of Solution

Given that the concentration of N2O5 at t=3h is 0.352mol/L and the concentration of N2O5 at t=4h is 0.262mol/L.

So the average rate can be calculated as follows,

  rate=changeinconcentrationchangeintime=[0.262mol/L0.352mol/L][4h3h]=0.090molL-1h-1.

Average rate is 0.090molL-1h-1.

(f)

Interpretation Introduction

Interpretation:

Average rate of reaction in the interval 4to5h has to be calculated.

Concept Introduction:

Refer to part (a).

(f)

Expert Solution
Check Mark

Answer to Problem 13QRT

Average rate is 0.066molL-1h-1.

Explanation of Solution

Given that the concentration of N2O5 at t=4h is 0.262mol/L and the concentration of N2O5 at t=5h is 0.196mol/L.

So the average rate can be calculated as follows,

  rate=changeinconcentrationchangeintime=[0.196mol/L0.262mol/L][5h4h]=0.066molL-1h-1.

Average rate is 0.066molL-1h-1.

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Chapter 11 Solutions

Chemistry: The Molecular Science

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