Average rate of reaction in the interval 0.00 t o 0.50 h has to be calculated. Concept Introduction: Consider the following reaction. a A + b B → c C Where A and B are reactants and C is the product. a , b a n d c are stoichiometric co-efficient of A , B a n d C . r a t e = - 1 a Δ [ A ] Δ t = - 1 b Δ [ B ] Δ t = 1 c Δ [ C ] Δ t Where, − Δ [ A ] Δ t = rate of disappearance of A − Δ [ B ] Δ t = rate of disappearance of B Δ [ C ] Δ t = 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.

Question

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Answer 1

Question

Chapter 11, Problem 13QRT

(a)

Interpretation Introduction

Interpretation:

Average rate of reaction in the interval 0.00 to 0.50 h has to be calculated.

Concept Introduction:

Consider the following reaction.

a A+b B→c C

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

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

Answer to Problem 13QRT

Average rate is 0.23 mol L-1 h-1.

Explanation of Solution

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

So the average rate can be calculated as follows,

rate=change in concentrationchange in time =−[0.733 mol/L−0.849 mol/L][ 0.50 h− 0.00 h] = 0.23 mol L-1 h-1.

Average rate is 0.23 mol L-1 h-1.

(b)

Interpretation Introduction

Interpretation:

Average rate of reaction in the interval 0.50 to 1.0 h has to be calculated.

Concept Introduction:

Refer to part (a).

(b)

Expert Solution

Answer to Problem 13QRT

Average rate is 0.20 mol L-1 h-1.

Explanation of Solution

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

So the average rate can be calculated as follows,

rate=change in concentrationchange in time =−[0.633 mol/L−0.733 mol/L][ 1.0 h− 0.50 h] = 0.20 mol L-1 h-1.

Average rate is 0.20 mol L-1 h-1.

(c)

Interpretation Introduction

Interpretation:

Average rate of reaction in the interval 1 to 2 h has to be calculated.

Concept Introduction:

Refer to part (a).

(c)

Expert Solution

Answer to Problem 13QRT

Average rate is 0.161 mol L-1 h-1.

Explanation of Solution

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

So the average rate can be calculated as follows,

rate=change in concentrationchange in time =−[0.472 mol/L−0.633 mol/L][2 h − 1 h] = 0.161 mol L-1 h-1.

Average rate is 0.161 mol L-1 h-1.

(d)

Interpretation Introduction

Interpretation:

Average rate of reaction in the interval 2 to 3 h has to be calculated.

Concept Introduction:

Refer to part (a).

(d)

Expert Solution

Answer to Problem 13QRT

Average rate is 0.120 mol L-1 h-1.

Explanation of Solution

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

So the average rate can be calculated as follows,

rate=change in concentrationchange in time =−[0.352 mol/L− 0.473 mol/L][3 h − 2 h] = 0.120 mol L-1 h-1.

Average rate is 0.120 mol L-1 h-1.

(e)

Interpretation Introduction

Interpretation:

Average rate of reaction in the interval 3 to 4 h has to be calculated.

Concept Introduction:

Refer to part (a).

(e)

Expert Solution

Answer to Problem 13QRT

Average rate is 0.090 mol L-1 h-1.

Explanation of Solution

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

So the average rate can be calculated as follows,

rate=change in concentrationchange in time =−[0.262 mol/L−0.352 mol/L][4 h − 3 h] = 0.090 mol L-1 h-1.

Average rate is 0.090 mol L-1 h-1.

(f)

Interpretation Introduction

Interpretation:

Average rate of reaction in the interval 4 to 5 h has to be calculated.

Concept Introduction:

Refer to part (a).

(f)

Expert Solution

Answer to Problem 13QRT

Average rate is 0.066 mol L-1 h-1.

Explanation of Solution

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

So the average rate can be calculated as follows,

rate=change in concentrationchange in time =−[0.196 mol/L−0.262 mol/L][5 h − 4 h] = 0.066 mol L-1 h-1.

Average rate is 0.066 mol L-1 h-1.

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Please correct answer and don't used hand raiting

this is an organic chemistry question please answer accordindly!! please post the solution in your hand writing not an AI generated answer please draw the figures and structures if needed to support your explanation hand drawn only!!!! answer the question in a very simple and straight forward manner thanks!!!!! im reposting this please solve all parts and draw it not just word explanations!!

2B: The retrosynthetic cut below provides two options for a Suzuki coupling, provide the identities of A, B, C and D then identify which pairing is better and justify your choice. O₂N. Retro-Suzuki NO2 MeO OMe A + B OR C + D

Answer 2

Question

Chapter 11, Problem 13QRT

(a)

Interpretation Introduction

Interpretation:

Average rate of reaction in the interval 0.00 to 0.50 h has to be calculated.

Concept Introduction:

Consider the following reaction.

a A+b B→c C

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

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

Answer to Problem 13QRT

Average rate is 0.23 mol L-1 h-1.

Explanation of Solution

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

So the average rate can be calculated as follows,

rate=change in concentrationchange in time =−[0.733 mol/L−0.849 mol/L][ 0.50 h− 0.00 h] = 0.23 mol L-1 h-1.

Average rate is 0.23 mol L-1 h-1.

(b)

Interpretation Introduction

Interpretation:

Average rate of reaction in the interval 0.50 to 1.0 h has to be calculated.

Concept Introduction:

Refer to part (a).

(b)

Expert Solution

Answer to Problem 13QRT

Average rate is 0.20 mol L-1 h-1.

Explanation of Solution

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

So the average rate can be calculated as follows,

rate=change in concentrationchange in time =−[0.633 mol/L−0.733 mol/L][ 1.0 h− 0.50 h] = 0.20 mol L-1 h-1.

Average rate is 0.20 mol L-1 h-1.

(c)

Interpretation Introduction

Interpretation:

Average rate of reaction in the interval 1 to 2 h has to be calculated.

Concept Introduction:

Refer to part (a).

(c)

Expert Solution

Answer to Problem 13QRT

Average rate is 0.161 mol L-1 h-1.

Explanation of Solution

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

So the average rate can be calculated as follows,

rate=change in concentrationchange in time =−[0.472 mol/L−0.633 mol/L][2 h − 1 h] = 0.161 mol L-1 h-1.

Average rate is 0.161 mol L-1 h-1.

(d)

Interpretation Introduction

Interpretation:

Average rate of reaction in the interval 2 to 3 h has to be calculated.

Concept Introduction:

Refer to part (a).

(d)

Expert Solution

Answer to Problem 13QRT

Average rate is 0.120 mol L-1 h-1.

Explanation of Solution

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

So the average rate can be calculated as follows,

rate=change in concentrationchange in time =−[0.352 mol/L− 0.473 mol/L][3 h − 2 h] = 0.120 mol L-1 h-1.

Average rate is 0.120 mol L-1 h-1.

(e)

Interpretation Introduction

Interpretation:

Average rate of reaction in the interval 3 to 4 h has to be calculated.

Concept Introduction:

Refer to part (a).

(e)

Expert Solution

Answer to Problem 13QRT

Average rate is 0.090 mol L-1 h-1.

Explanation of Solution

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

So the average rate can be calculated as follows,

rate=change in concentrationchange in time =−[0.262 mol/L−0.352 mol/L][4 h − 3 h] = 0.090 mol L-1 h-1.

Average rate is 0.090 mol L-1 h-1.

(f)

Interpretation Introduction

Interpretation:

Average rate of reaction in the interval 4 to 5 h has to be calculated.

Concept Introduction:

Refer to part (a).

(f)

Expert Solution

Answer to Problem 13QRT

Average rate is 0.066 mol L-1 h-1.

Explanation of Solution

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

So the average rate can be calculated as follows,

rate=change in concentrationchange in time =−[0.196 mol/L−0.262 mol/L][5 h − 4 h] = 0.066 mol L-1 h-1.

Average rate is 0.066 mol L-1 h-1.

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Answer 3

Question

Chapter 11, Problem 13QRT

(a)

Interpretation Introduction

Interpretation:

Average rate of reaction in the interval 0.00 to 0.50 h has to be calculated.

Concept Introduction:

Consider the following reaction.

a A+b B→c C

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

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

Answer to Problem 13QRT

Average rate is 0.23 mol L-1 h-1.

Explanation of Solution

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

So the average rate can be calculated as follows,

rate=change in concentrationchange in time =−[0.733 mol/L−0.849 mol/L][ 0.50 h− 0.00 h] = 0.23 mol L-1 h-1.

Average rate is 0.23 mol L-1 h-1.

(b)

Interpretation Introduction

Interpretation:

Average rate of reaction in the interval 0.50 to 1.0 h has to be calculated.

Concept Introduction:

Refer to part (a).

(b)

Expert Solution

Answer to Problem 13QRT

Average rate is 0.20 mol L-1 h-1.

Explanation of Solution

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

So the average rate can be calculated as follows,

rate=change in concentrationchange in time =−[0.633 mol/L−0.733 mol/L][ 1.0 h− 0.50 h] = 0.20 mol L-1 h-1.

Average rate is 0.20 mol L-1 h-1.

(c)

Interpretation Introduction

Interpretation:

Average rate of reaction in the interval 1 to 2 h has to be calculated.

Concept Introduction:

Refer to part (a).

(c)

Expert Solution

Answer to Problem 13QRT

Average rate is 0.161 mol L-1 h-1.

Explanation of Solution

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

So the average rate can be calculated as follows,

rate=change in concentrationchange in time =−[0.472 mol/L−0.633 mol/L][2 h − 1 h] = 0.161 mol L-1 h-1.

Average rate is 0.161 mol L-1 h-1.

(d)

Interpretation Introduction

Interpretation:

Average rate of reaction in the interval 2 to 3 h has to be calculated.

Concept Introduction:

Refer to part (a).

(d)

Expert Solution

Answer to Problem 13QRT

Average rate is 0.120 mol L-1 h-1.

Explanation of Solution

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

So the average rate can be calculated as follows,

rate=change in concentrationchange in time =−[0.352 mol/L− 0.473 mol/L][3 h − 2 h] = 0.120 mol L-1 h-1.

Average rate is 0.120 mol L-1 h-1.

(e)

Interpretation Introduction

Interpretation:

Average rate of reaction in the interval 3 to 4 h has to be calculated.

Concept Introduction:

Refer to part (a).

(e)

Expert Solution

Answer to Problem 13QRT

Average rate is 0.090 mol L-1 h-1.

Explanation of Solution

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

So the average rate can be calculated as follows,

rate=change in concentrationchange in time =−[0.262 mol/L−0.352 mol/L][4 h − 3 h] = 0.090 mol L-1 h-1.

Average rate is 0.090 mol L-1 h-1.

(f)

Interpretation Introduction

Interpretation:

Average rate of reaction in the interval 4 to 5 h has to be calculated.

Concept Introduction:

Refer to part (a).

(f)

Expert Solution

Answer to Problem 13QRT

Average rate is 0.066 mol L-1 h-1.

Explanation of Solution

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

So the average rate can be calculated as follows,

rate=change in concentrationchange in time =−[0.196 mol/L−0.262 mol/L][5 h − 4 h] = 0.066 mol L-1 h-1.

Average rate is 0.066 mol L-1 h-1.

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Answer 4

Question

Chapter 11, Problem 13QRT

(a)

Interpretation Introduction

Interpretation:

Average rate of reaction in the interval 0.00 to 0.50 h has to be calculated.

Concept Introduction:

Consider the following reaction.

a A+b B→c C

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

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

Answer to Problem 13QRT

Average rate is 0.23 mol L-1 h-1.

Explanation of Solution

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

So the average rate can be calculated as follows,

rate=change in concentrationchange in time =−[0.733 mol/L−0.849 mol/L][ 0.50 h− 0.00 h] = 0.23 mol L-1 h-1.

Average rate is 0.23 mol L-1 h-1.

(b)

Interpretation Introduction

Interpretation:

Average rate of reaction in the interval 0.50 to 1.0 h has to be calculated.

Concept Introduction:

Refer to part (a).

(b)

Expert Solution

Answer to Problem 13QRT

Average rate is 0.20 mol L-1 h-1.

Explanation of Solution

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

So the average rate can be calculated as follows,

rate=change in concentrationchange in time =−[0.633 mol/L−0.733 mol/L][ 1.0 h− 0.50 h] = 0.20 mol L-1 h-1.

Average rate is 0.20 mol L-1 h-1.

(c)

Interpretation Introduction

Interpretation:

Average rate of reaction in the interval 1 to 2 h has to be calculated.

Concept Introduction:

Refer to part (a).

(c)

Expert Solution

Answer to Problem 13QRT

Average rate is 0.161 mol L-1 h-1.

Explanation of Solution

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

So the average rate can be calculated as follows,

rate=change in concentrationchange in time =−[0.472 mol/L−0.633 mol/L][2 h − 1 h] = 0.161 mol L-1 h-1.

Average rate is 0.161 mol L-1 h-1.

(d)

Interpretation Introduction

Interpretation:

Average rate of reaction in the interval 2 to 3 h has to be calculated.

Concept Introduction:

Refer to part (a).

(d)

Expert Solution

Answer to Problem 13QRT

Average rate is 0.120 mol L-1 h-1.

Explanation of Solution

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

So the average rate can be calculated as follows,

rate=change in concentrationchange in time =−[0.352 mol/L− 0.473 mol/L][3 h − 2 h] = 0.120 mol L-1 h-1.

Average rate is 0.120 mol L-1 h-1.

(e)

Interpretation Introduction

Interpretation:

Average rate of reaction in the interval 3 to 4 h has to be calculated.

Concept Introduction:

Refer to part (a).

(e)

Expert Solution

Answer to Problem 13QRT

Average rate is 0.090 mol L-1 h-1.

Explanation of Solution

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

So the average rate can be calculated as follows,

rate=change in concentrationchange in time =−[0.262 mol/L−0.352 mol/L][4 h − 3 h] = 0.090 mol L-1 h-1.

Average rate is 0.090 mol L-1 h-1.

(f)

Interpretation Introduction

Interpretation:

Average rate of reaction in the interval 4 to 5 h has to be calculated.

Concept Introduction:

Refer to part (a).

(f)

Expert Solution

Answer to Problem 13QRT

Average rate is 0.066 mol L-1 h-1.

Explanation of Solution

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

So the average rate can be calculated as follows,

rate=change in concentrationchange in time =−[0.196 mol/L−0.262 mol/L][5 h − 4 h] = 0.066 mol L-1 h-1.

Average rate is 0.066 mol L-1 h-1.

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Answer 5

Chapter 11, Problem 13QRT

(a)

Interpretation Introduction

Interpretation:

Average rate of reaction in the interval 0.00 to 0.50 h has to be calculated.

Concept Introduction:

Consider the following reaction.

a A+b B→c C

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

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

Answer to Problem 13QRT

Average rate is 0.23 mol L-1 h-1.

Explanation of Solution

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

So the average rate can be calculated as follows,

rate=change in concentrationchange in time =−[0.733 mol/L−0.849 mol/L][ 0.50 h− 0.00 h] = 0.23 mol L-1 h-1.

Average rate is 0.23 mol L-1 h-1.

(b)

Interpretation Introduction

Interpretation:

Average rate of reaction in the interval 0.50 to 1.0 h has to be calculated.

Concept Introduction:

Refer to part (a).

(b)

Expert Solution

Answer to Problem 13QRT

Average rate is 0.20 mol L-1 h-1.

Explanation of Solution

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

So the average rate can be calculated as follows,

rate=change in concentrationchange in time =−[0.633 mol/L−0.733 mol/L][ 1.0 h− 0.50 h] = 0.20 mol L-1 h-1.

Average rate is 0.20 mol L-1 h-1.

(c)

Interpretation Introduction

Interpretation:

Average rate of reaction in the interval 1 to 2 h has to be calculated.

Concept Introduction:

Refer to part (a).

(c)

Expert Solution

Answer to Problem 13QRT

Average rate is 0.161 mol L-1 h-1.

Explanation of Solution

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

So the average rate can be calculated as follows,

rate=change in concentrationchange in time =−[0.472 mol/L−0.633 mol/L][2 h − 1 h] = 0.161 mol L-1 h-1.

Average rate is 0.161 mol L-1 h-1.

(d)

Interpretation Introduction

Interpretation:

Average rate of reaction in the interval 2 to 3 h has to be calculated.

Concept Introduction:

Refer to part (a).

(d)

Expert Solution

Answer to Problem 13QRT

Average rate is 0.120 mol L-1 h-1.

Explanation of Solution

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

So the average rate can be calculated as follows,

rate=change in concentrationchange in time =−[0.352 mol/L− 0.473 mol/L][3 h − 2 h] = 0.120 mol L-1 h-1.

Average rate is 0.120 mol L-1 h-1.

(e)

Interpretation Introduction

Interpretation:

Average rate of reaction in the interval 3 to 4 h has to be calculated.

Concept Introduction:

Refer to part (a).

(e)

Expert Solution

Answer to Problem 13QRT

Average rate is 0.090 mol L-1 h-1.

Explanation of Solution

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

So the average rate can be calculated as follows,

rate=change in concentrationchange in time =−[0.262 mol/L−0.352 mol/L][4 h − 3 h] = 0.090 mol L-1 h-1.

Average rate is 0.090 mol L-1 h-1.

(f)

Interpretation Introduction

Interpretation:

Average rate of reaction in the interval 4 to 5 h has to be calculated.

Concept Introduction:

Refer to part (a).

(f)

Expert Solution

Answer to Problem 13QRT

Average rate is 0.066 mol L-1 h-1.

Explanation of Solution

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

So the average rate can be calculated as follows,

rate=change in concentrationchange in time =−[0.196 mol/L−0.262 mol/L][5 h − 4 h] = 0.066 mol L-1 h-1.

Average rate is 0.066 mol L-1 h-1.

Answer 6

Chapter 11, Problem 13QRT

(a)

Interpretation Introduction

Interpretation:

Average rate of reaction in the interval 0.00 to 0.50 h has to be calculated.

Concept Introduction:

Consider the following reaction.

a A+b B→c C

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

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

Answer to Problem 13QRT

Average rate is 0.23 mol L-1 h-1.

Explanation of Solution

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

So the average rate can be calculated as follows,

rate=change in concentrationchange in time =−[0.733 mol/L−0.849 mol/L][ 0.50 h− 0.00 h] = 0.23 mol L-1 h-1.

Average rate is 0.23 mol L-1 h-1.

Answer 7

Chapter 11, Problem 13QRT

(a)

Interpretation Introduction

Interpretation:

Average rate of reaction in the interval 0.00 to 0.50 h has to be calculated.

Concept Introduction:

Consider the following reaction.

a A+b B→c C

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

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.

Answer 8

(a)

Expert Solution

Answer to Problem 13QRT

Average rate is 0.23 mol L-1 h-1.

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

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

So the average rate can be calculated as follows,

rate=change in concentrationchange in time =−[0.733 mol/L−0.849 mol/L][ 0.50 h− 0.00 h] = 0.23 mol L-1 h-1.

Average rate is 0.23 mol L-1 h-1.

Answer 13

(b)

Interpretation Introduction

Interpretation:

Average rate of reaction in the interval 0.50 to 1.0 h has to be calculated.

Concept Introduction:

Refer to part (a).

(b)

Expert Solution

Answer to Problem 13QRT

Average rate is 0.20 mol L-1 h-1.

Explanation of Solution

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

So the average rate can be calculated as follows,

rate=change in concentrationchange in time =−[0.633 mol/L−0.733 mol/L][ 1.0 h− 0.50 h] = 0.20 mol L-1 h-1.

Average rate is 0.20 mol L-1 h-1.

Answer 14

(b)

Interpretation Introduction

Interpretation:

Average rate of reaction in the interval 0.50 to 1.0 h has to be calculated.

Concept Introduction:

Refer to part (a).

Answer 15

(b)

Expert Solution

Answer to Problem 13QRT

Average rate is 0.20 mol L-1 h-1.

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

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

So the average rate can be calculated as follows,

rate=change in concentrationchange in time =−[0.633 mol/L−0.733 mol/L][ 1.0 h− 0.50 h] = 0.20 mol L-1 h-1.

Average rate is 0.20 mol L-1 h-1.

Answer 20

(c)

Interpretation Introduction

Interpretation:

Average rate of reaction in the interval 1 to 2 h has to be calculated.

Concept Introduction:

Refer to part (a).

(c)

Expert Solution

Answer to Problem 13QRT

Average rate is 0.161 mol L-1 h-1.

Explanation of Solution

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

So the average rate can be calculated as follows,

rate=change in concentrationchange in time =−[0.472 mol/L−0.633 mol/L][2 h − 1 h] = 0.161 mol L-1 h-1.

Average rate is 0.161 mol L-1 h-1.

Answer 21

(c)

Interpretation Introduction

Interpretation:

Average rate of reaction in the interval 1 to 2 h has to be calculated.

Concept Introduction:

Refer to part (a).

Answer 22

(c)

Expert Solution

Answer to Problem 13QRT

Average rate is 0.161 mol L-1 h-1.

Answer 23

(c)

Expert Solution

Answer 24

(c)

Expert Solution

Answer 25

Expert Solution

Answer 26

Explanation of Solution

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

So the average rate can be calculated as follows,

rate=change in concentrationchange in time =−[0.472 mol/L−0.633 mol/L][2 h − 1 h] = 0.161 mol L-1 h-1.

Average rate is 0.161 mol L-1 h-1.

Answer 27

(d)

Interpretation Introduction

Interpretation:

Average rate of reaction in the interval 2 to 3 h has to be calculated.

Concept Introduction:

Refer to part (a).

(d)

Expert Solution

Answer to Problem 13QRT

Average rate is 0.120 mol L-1 h-1.

Explanation of Solution

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

So the average rate can be calculated as follows,

rate=change in concentrationchange in time =−[0.352 mol/L− 0.473 mol/L][3 h − 2 h] = 0.120 mol L-1 h-1.

Average rate is 0.120 mol L-1 h-1.

Answer 28

(d)

Interpretation Introduction

Interpretation:

Average rate of reaction in the interval 2 to 3 h has to be calculated.

Concept Introduction:

Refer to part (a).

Answer 29

(d)

Expert Solution

Answer to Problem 13QRT

Average rate is 0.120 mol L-1 h-1.

Answer 30

(d)

Expert Solution

Answer 31

(d)

Expert Solution

Answer 32

Expert Solution

Answer 33

Explanation of Solution

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

So the average rate can be calculated as follows,

rate=change in concentrationchange in time =−[0.352 mol/L− 0.473 mol/L][3 h − 2 h] = 0.120 mol L-1 h-1.

Average rate is 0.120 mol L-1 h-1.

Answer 34

(e)

Interpretation Introduction

Interpretation:

Average rate of reaction in the interval 3 to 4 h has to be calculated.

Concept Introduction:

Refer to part (a).

(e)

Expert Solution

Answer to Problem 13QRT

Average rate is 0.090 mol L-1 h-1.

Explanation of Solution

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

So the average rate can be calculated as follows,

rate=change in concentrationchange in time =−[0.262 mol/L−0.352 mol/L][4 h − 3 h] = 0.090 mol L-1 h-1.

Average rate is 0.090 mol L-1 h-1.

Answer 35

(e)

Interpretation Introduction

Interpretation:

Average rate of reaction in the interval 3 to 4 h has to be calculated.

Concept Introduction:

Refer to part (a).

Answer 36

(e)

Expert Solution

Answer to Problem 13QRT

Average rate is 0.090 mol L-1 h-1.

Answer 37

(e)

Expert Solution

Answer 38

(e)

Expert Solution

Answer 39

Expert Solution

Answer 40

Explanation of Solution

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

So the average rate can be calculated as follows,

rate=change in concentrationchange in time =−[0.262 mol/L−0.352 mol/L][4 h − 3 h] = 0.090 mol L-1 h-1.

Average rate is 0.090 mol L-1 h-1.

Answer 41

(f)

Interpretation Introduction

Interpretation:

Average rate of reaction in the interval 4 to 5 h has to be calculated.

Concept Introduction:

Refer to part (a).

(f)

Expert Solution

Answer to Problem 13QRT

Average rate is 0.066 mol L-1 h-1.

Explanation of Solution

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

So the average rate can be calculated as follows,

rate=change in concentrationchange in time =−[0.196 mol/L−0.262 mol/L][5 h − 4 h] = 0.066 mol L-1 h-1.

Average rate is 0.066 mol L-1 h-1.

Answer 42

(f)

Interpretation Introduction

Interpretation:

Average rate of reaction in the interval 4 to 5 h has to be calculated.

Concept Introduction:

Refer to part (a).

Answer 43

(f)

Expert Solution

Answer to Problem 13QRT

Average rate is 0.066 mol L-1 h-1.

Answer 44

(f)

Expert Solution

Answer 45

(f)

Expert Solution

Answer 46

Expert Solution

Answer 47

Explanation of Solution

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

So the average rate can be calculated as follows,

rate=change in concentrationchange in time =−[0.196 mol/L−0.262 mol/L][5 h − 4 h] = 0.066 mol L-1 h-1.

Average rate is 0.066 mol L-1 h-1.

Answer 48

Ch. 11.1 - For the reaction of crystal violet with NaOH(aq),...Ch. 11.1 - (a) From data in Table 11.1, calculate the rate of...Ch. 11.1 - For the reaction 4NO2(g)+O2(g)2N2O5(g) (a) express...Ch. 11.1 - Instantaneous rates for the reaction of hydroxide...Ch. 11.1 - Prob. 11.3CE Ch. 11.2 - Prob. 11.4E Ch. 11.2 - Prob. 11.3PSP Ch. 11.2 - Prob. 11.5E Ch. 11.3 - Prob. 11.4PSP Ch. 11.3 - Prob. 11.5PSP

Videos

Answer to Problem 13QRT

Explanation of Solution

Interpretation:

Average rate of reaction in the interval 0.50 to 1.0 h has to be calculated.

Concept Introduction:

Refer to part (a).

Answer to Problem 13QRT

Explanation of Solution

Interpretation:

Average rate of reaction in the interval 1 to 2 h has to be calculated.

Concept Introduction:

Refer to part (a).

Answer to Problem 13QRT

Explanation of Solution

Interpretation:

Average rate of reaction in the interval 2 to 3 h has to be calculated.

Concept Introduction:

Refer to part (a).

Answer to Problem 13QRT

Explanation of Solution

Interpretation:

Average rate of reaction in the interval 3 to 4 h has to be calculated.

Concept Introduction:

Refer to part (a).

Answer to Problem 13QRT

Explanation of Solution

Interpretation:

Average rate of reaction in the interval 4 to 5 h has to be calculated.

Concept Introduction:

Refer to part (a).

Answer to Problem 13QRT

Explanation of Solution

Want to see more full solutions like this?

Chapter 11 Solutions

Videos

Answer to Problem 13QRT

Explanation of Solution

Interpretation: Average rate of reaction in the interval 0.50 to 1.0 h has to be calculated. Concept Introduction: Refer to part (a).

Answer to Problem 13QRT

Explanation of Solution

Interpretation: Average rate of reaction in the interval 1 to 2 h has to be calculated. Concept Introduction: Refer to part (a).

Answer to Problem 13QRT

Explanation of Solution

Interpretation: Average rate of reaction in the interval 2 to 3 h has to be calculated. Concept Introduction: Refer to part (a).

Answer to Problem 13QRT

Explanation of Solution

Interpretation: Average rate of reaction in the interval 3 to 4 h has to be calculated. Concept Introduction: Refer to part (a).

Answer to Problem 13QRT

Explanation of Solution

Interpretation: Average rate of reaction in the interval 4 to 5 h has to be calculated. Concept Introduction: Refer to part (a).

Answer to Problem 13QRT

Explanation of Solution

Want to see more full solutions like this?

Chapter 11 Solutions

Interpretation:

Average rate of reaction in the interval 0.50 to 1.0 h has to be calculated.

Concept Introduction:

Refer to part (a).

Interpretation:

Average rate of reaction in the interval 1 to 2 h has to be calculated.

Concept Introduction:

Refer to part (a).

Interpretation:

Average rate of reaction in the interval 2 to 3 h has to be calculated.

Concept Introduction:

Refer to part (a).

Interpretation:

Average rate of reaction in the interval 3 to 4 h has to be calculated.

Concept Introduction:

Refer to part (a).

Interpretation:

Average rate of reaction in the interval 4 to 5 h has to be calculated.

Concept Introduction:

Refer to part (a).