It has to be shown that the reaction obeys the rate law as given below. N 2 O 5 → 2 N O 2 + 1 2 O 2 R a t e = − Δ [ N 2 O 5 ] Δ t = k [ N 2 O 5 ] Concept Introduction: In a first-order reaction, the graph of concentration vs. rate is linear.

Question

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Answer 1

Question

Chapter 11, Problem 14QRT

(a)

Interpretation Introduction

Interpretation:

It has to be shown that the reaction obeys the rate law as given below.

N2O5→2NO2+12O2Rate=−Δ[N2O5]Δt=k[N2O5]

Concept Introduction:

In a first-order reaction, the graph of concentration vs. rate is linear.

(a)

Expert Solution

Explanation of Solution

Average concentrations can be calculated as given below.

From 0 to 0.50 h:

Average concentration=(0.849 mol/L)+(0.733 mol/L)2=0.791 mol/L

From 0.50 to 1.00 h:

Average concentration=(0.733 mol/L)+( 0.633mol/L)2=0.683 mol/L

From 1.00 to 2.00 h:

Average concentration=(0.633 mol/L)+( 0.472 mol/L)2=0.553 mol/L

From 2.00 to 3.00 h:

Average concentration=(0.472 mol/L)+( 0.352 mol/L)2=0.412 mol/L

From 3.00 to 4.00 h:

Average concentration=(0.352 mol/L)+( 0.262 mol/L)2=0.307 mol/L

From 4.00 to 5.00 h:

Average concentration=(0.262 mol/L)+( 0.196 mol/L)2=0.229 mol/L

The average rate and concentration is given in the table below.

Average concentration (mol/L)	Average rate (mol L−1h−1)
0.791	0.23
0.683	0.20
0.553	0.161
0.412	0.120
0.307	0.090
0.229	0.066

A graph of concentration vs. rate is given below.

OWLV2 FOR MOORE/STANITSKI'S CHEMISTRY:, Chapter 11, Problem 14QRT

Figure 1

The above graph is linear which satisfies the rate law.

(b)

Interpretation Introduction

Interpretation:

The rate constant k has to be evaluated as an average of the values obtained for the six intervals.

(b)

Expert Solution

Answer to Problem 14QRT

The average value of k is 0.29 h-1.

Explanation of Solution

The rate constant k as an average of the values obtained for the six intervals can be calculated as given below.

From 0 to 0.50 h:

Average rate is 0.23 mol L−1h−1.

Average concentration is 0.791 mol L−1.

Then, the rate constant can be calculated as given below.

k=Rate[N2O5]=0.23 mol L−1h−10.791 mol L−1=0.29 h−1.

From 0.50 to 1.00 h:

Average rate is 0.20 mol L−1h−1.

Average concentration is 0.683 mol L−1.

Then, the rate constant can be calculated as given below.

k=Rate[N2O5]=0.20 mol L−1h−10.683 mol L−1=0.29 h−1.

From 1.00 to 2.00 h:

Average rate is 0.161 mol L−1h−1.

Average concentration is 0.553 mol L−1.

Then, the rate constant can be calculated as given below.

k=Rate[N2O5]=0.161 mol L−1h−10.553 mol L−1=0.291 h−1.

From 2.00 to 3.00 h:

Average rate is 0.120 mol L−1h−1.

Average concentration is 0.412 mol L−1.

Then, the rate constant can be calculated as given below.

k=Rate[N2O5]=0.120 mol L−1h−10.412 mol L−1=0.291 h−1.

From 3.00 to 4.00 h:

Average rate is 0.090 mol L−1h−1.

Average concentration is 0.307 mol L−1.

Then, the rate constant can be calculated as given below.

k=Rate[N2O5]=0.090 mol L−1h−10.307 mol L−1=0.29 h−1.

From 4.00 to 5.00 h:

Average rate is 0.066 mol L−1h−1.

Average concentration is 0.229 mol L−1.

Then, the rate constant can be calculated as given below.

k=Rate[N2O5]=0.066 mol L−1h−10.229 mol L−1=0.29 h−1.

The average value of k is 0.29 h-1.

(c)

Interpretation Introduction

Interpretation:

The reaction rate exactly 2.5 h from the start has to be calculated.

(c)

Expert Solution

Answer to Problem 14QRT

The reaction rate exactly 2.5 h from the start is 0.12 mol L-1h-1.

Explanation of Solution

The initial concentration of N2O5 is 0.849 mol/L. The rate constant is 0.29 h−1.

The rate law describes that the reaction is a first-order reaction. The integrated first-order rate law is given below.

ln[N2O5]t=−kt+ln[N2O5]0=−(0.29 h−1)×(2.50 h)+ln(0.849)ln[N2O5]t=−0.889[N2O5]t=0.411 mol/L.

Now, the rate of the reaction at 2.5 h can be calculated as follows,

Rate=k[N2O5]=(0.29 h−1)×(0.411 mol/L)=0.12 mol L-1h-1.

Therefore, the reaction rate exactly 2.5 h from the start is 0.12 mol L-1h-1.

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Students have asked these similar questions

Consider the reaction A + B ¡ C + D. Is each of the following statements true or false? (a) The rate law for the reaction must be Rate = k3A43B4. (b) If the reaction is an elementary reaction, the rate law is second order. (c) If the reaction is an elementary reaction, the rate law of the reverse reaction is first order. (d) The activation energy for the reverse reaction must be greater than that for the forward reaction.

The following kinetic data are collected for the initial rates of a reaction 2 X + Z→ products: Experiment [X ]o(M) [Z]o(M) Rate (M/s) 0.25 0.25 4.0 x 10! 0.50 0.50 3.2 x 102 0.50 0.75 7.2 x 102 (a) What is the rate law for this reaction? (b) What is the value of the rate constant with proper units? (c) What is the reaction rate when the initial concentration of X is 0.75 M and that of Z is 1.25 M? 2. 3.

Consider the following reaction:2 NO(g) + 2 H2(g)--->N2(g) + 2 H2O(g)(a) The rate law for this reaction is first order in H2 andsecond order in NO. Write the rate law. (b) If the rateconstant for this reaction at 1000 K is 6.0 x 104 M-2 s-1,what is the reaction rate when [NO] = 0.035 M and[H2] = 0.015 M? (c) What is the reaction rate at 1000 Kwhen the concentration of NO is increased to 0.10 M,while the concentration of H2 is 0.010 M? (d) What is thereaction rate at 1000 K if [NO] is decreased to 0.010 M and[H2] is increased to 0.030 M?

Answer 2

Question

Chapter 11, Problem 14QRT

(a)

Interpretation Introduction

Interpretation:

It has to be shown that the reaction obeys the rate law as given below.

N2O5→2NO2+12O2Rate=−Δ[N2O5]Δt=k[N2O5]

Concept Introduction:

In a first-order reaction, the graph of concentration vs. rate is linear.

(a)

Expert Solution

Explanation of Solution

Average concentrations can be calculated as given below.

From 0 to 0.50 h:

Average concentration=(0.849 mol/L)+(0.733 mol/L)2=0.791 mol/L

From 0.50 to 1.00 h:

Average concentration=(0.733 mol/L)+( 0.633mol/L)2=0.683 mol/L

From 1.00 to 2.00 h:

Average concentration=(0.633 mol/L)+( 0.472 mol/L)2=0.553 mol/L

From 2.00 to 3.00 h:

Average concentration=(0.472 mol/L)+( 0.352 mol/L)2=0.412 mol/L

From 3.00 to 4.00 h:

Average concentration=(0.352 mol/L)+( 0.262 mol/L)2=0.307 mol/L

From 4.00 to 5.00 h:

Average concentration=(0.262 mol/L)+( 0.196 mol/L)2=0.229 mol/L

The average rate and concentration is given in the table below.

Average concentration (mol/L)	Average rate (mol L−1h−1)
0.791	0.23
0.683	0.20
0.553	0.161
0.412	0.120
0.307	0.090
0.229	0.066

A graph of concentration vs. rate is given below.

OWLV2 FOR MOORE/STANITSKI'S CHEMISTRY:, Chapter 11, Problem 14QRT

Figure 1

The above graph is linear which satisfies the rate law.

(b)

Interpretation Introduction

Interpretation:

The rate constant k has to be evaluated as an average of the values obtained for the six intervals.

(b)

Expert Solution

Answer to Problem 14QRT

The average value of k is 0.29 h-1.

Explanation of Solution

The rate constant k as an average of the values obtained for the six intervals can be calculated as given below.

From 0 to 0.50 h:

Average rate is 0.23 mol L−1h−1.

Average concentration is 0.791 mol L−1.

Then, the rate constant can be calculated as given below.

k=Rate[N2O5]=0.23 mol L−1h−10.791 mol L−1=0.29 h−1.

From 0.50 to 1.00 h:

Average rate is 0.20 mol L−1h−1.

Average concentration is 0.683 mol L−1.

Then, the rate constant can be calculated as given below.

k=Rate[N2O5]=0.20 mol L−1h−10.683 mol L−1=0.29 h−1.

From 1.00 to 2.00 h:

Average rate is 0.161 mol L−1h−1.

Average concentration is 0.553 mol L−1.

Then, the rate constant can be calculated as given below.

k=Rate[N2O5]=0.161 mol L−1h−10.553 mol L−1=0.291 h−1.

From 2.00 to 3.00 h:

Average rate is 0.120 mol L−1h−1.

Average concentration is 0.412 mol L−1.

Then, the rate constant can be calculated as given below.

k=Rate[N2O5]=0.120 mol L−1h−10.412 mol L−1=0.291 h−1.

From 3.00 to 4.00 h:

Average rate is 0.090 mol L−1h−1.

Average concentration is 0.307 mol L−1.

Then, the rate constant can be calculated as given below.

k=Rate[N2O5]=0.090 mol L−1h−10.307 mol L−1=0.29 h−1.

From 4.00 to 5.00 h:

Average rate is 0.066 mol L−1h−1.

Average concentration is 0.229 mol L−1.

Then, the rate constant can be calculated as given below.

k=Rate[N2O5]=0.066 mol L−1h−10.229 mol L−1=0.29 h−1.

The average value of k is 0.29 h-1.

(c)

Interpretation Introduction

Interpretation:

The reaction rate exactly 2.5 h from the start has to be calculated.

(c)

Expert Solution

Answer to Problem 14QRT

The reaction rate exactly 2.5 h from the start is 0.12 mol L-1h-1.

Explanation of Solution

The initial concentration of N2O5 is 0.849 mol/L. The rate constant is 0.29 h−1.

The rate law describes that the reaction is a first-order reaction. The integrated first-order rate law is given below.

ln[N2O5]t=−kt+ln[N2O5]0=−(0.29 h−1)×(2.50 h)+ln(0.849)ln[N2O5]t=−0.889[N2O5]t=0.411 mol/L.

Now, the rate of the reaction at 2.5 h can be calculated as follows,

Rate=k[N2O5]=(0.29 h−1)×(0.411 mol/L)=0.12 mol L-1h-1.

Therefore, the reaction rate exactly 2.5 h from the start is 0.12 mol L-1h-1.

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Answer 3

Question

Chapter 11, Problem 14QRT

(a)

Interpretation Introduction

Interpretation:

It has to be shown that the reaction obeys the rate law as given below.

N2O5→2NO2+12O2Rate=−Δ[N2O5]Δt=k[N2O5]

Concept Introduction:

In a first-order reaction, the graph of concentration vs. rate is linear.

(a)

Expert Solution

Explanation of Solution

Average concentrations can be calculated as given below.

From 0 to 0.50 h:

Average concentration=(0.849 mol/L)+(0.733 mol/L)2=0.791 mol/L

From 0.50 to 1.00 h:

Average concentration=(0.733 mol/L)+( 0.633mol/L)2=0.683 mol/L

From 1.00 to 2.00 h:

Average concentration=(0.633 mol/L)+( 0.472 mol/L)2=0.553 mol/L

From 2.00 to 3.00 h:

Average concentration=(0.472 mol/L)+( 0.352 mol/L)2=0.412 mol/L

From 3.00 to 4.00 h:

Average concentration=(0.352 mol/L)+( 0.262 mol/L)2=0.307 mol/L

From 4.00 to 5.00 h:

Average concentration=(0.262 mol/L)+( 0.196 mol/L)2=0.229 mol/L

The average rate and concentration is given in the table below.

Average concentration (mol/L)	Average rate (mol L−1h−1)
0.791	0.23
0.683	0.20
0.553	0.161
0.412	0.120
0.307	0.090
0.229	0.066

A graph of concentration vs. rate is given below.

OWLV2 FOR MOORE/STANITSKI'S CHEMISTRY:, Chapter 11, Problem 14QRT

Figure 1

The above graph is linear which satisfies the rate law.

(b)

Interpretation Introduction

Interpretation:

The rate constant k has to be evaluated as an average of the values obtained for the six intervals.

(b)

Expert Solution

Answer to Problem 14QRT

The average value of k is 0.29 h-1.

Explanation of Solution

The rate constant k as an average of the values obtained for the six intervals can be calculated as given below.

From 0 to 0.50 h:

Average rate is 0.23 mol L−1h−1.

Average concentration is 0.791 mol L−1.

Then, the rate constant can be calculated as given below.

k=Rate[N2O5]=0.23 mol L−1h−10.791 mol L−1=0.29 h−1.

From 0.50 to 1.00 h:

Average rate is 0.20 mol L−1h−1.

Average concentration is 0.683 mol L−1.

Then, the rate constant can be calculated as given below.

k=Rate[N2O5]=0.20 mol L−1h−10.683 mol L−1=0.29 h−1.

From 1.00 to 2.00 h:

Average rate is 0.161 mol L−1h−1.

Average concentration is 0.553 mol L−1.

Then, the rate constant can be calculated as given below.

k=Rate[N2O5]=0.161 mol L−1h−10.553 mol L−1=0.291 h−1.

From 2.00 to 3.00 h:

Average rate is 0.120 mol L−1h−1.

Average concentration is 0.412 mol L−1.

Then, the rate constant can be calculated as given below.

k=Rate[N2O5]=0.120 mol L−1h−10.412 mol L−1=0.291 h−1.

From 3.00 to 4.00 h:

Average rate is 0.090 mol L−1h−1.

Average concentration is 0.307 mol L−1.

Then, the rate constant can be calculated as given below.

k=Rate[N2O5]=0.090 mol L−1h−10.307 mol L−1=0.29 h−1.

From 4.00 to 5.00 h:

Average rate is 0.066 mol L−1h−1.

Average concentration is 0.229 mol L−1.

Then, the rate constant can be calculated as given below.

k=Rate[N2O5]=0.066 mol L−1h−10.229 mol L−1=0.29 h−1.

The average value of k is 0.29 h-1.

(c)

Interpretation Introduction

Interpretation:

The reaction rate exactly 2.5 h from the start has to be calculated.

(c)

Expert Solution

Answer to Problem 14QRT

The reaction rate exactly 2.5 h from the start is 0.12 mol L-1h-1.

Explanation of Solution

The initial concentration of N2O5 is 0.849 mol/L. The rate constant is 0.29 h−1.

The rate law describes that the reaction is a first-order reaction. The integrated first-order rate law is given below.

ln[N2O5]t=−kt+ln[N2O5]0=−(0.29 h−1)×(2.50 h)+ln(0.849)ln[N2O5]t=−0.889[N2O5]t=0.411 mol/L.

Now, the rate of the reaction at 2.5 h can be calculated as follows,

Rate=k[N2O5]=(0.29 h−1)×(0.411 mol/L)=0.12 mol L-1h-1.

Therefore, the reaction rate exactly 2.5 h from the start is 0.12 mol L-1h-1.

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Answer 4

Question

Chapter 11, Problem 14QRT

(a)

Interpretation Introduction

Interpretation:

It has to be shown that the reaction obeys the rate law as given below.

N2O5→2NO2+12O2Rate=−Δ[N2O5]Δt=k[N2O5]

Concept Introduction:

In a first-order reaction, the graph of concentration vs. rate is linear.

(a)

Expert Solution

Explanation of Solution

Average concentrations can be calculated as given below.

From 0 to 0.50 h:

Average concentration=(0.849 mol/L)+(0.733 mol/L)2=0.791 mol/L

From 0.50 to 1.00 h:

Average concentration=(0.733 mol/L)+( 0.633mol/L)2=0.683 mol/L

From 1.00 to 2.00 h:

Average concentration=(0.633 mol/L)+( 0.472 mol/L)2=0.553 mol/L

From 2.00 to 3.00 h:

Average concentration=(0.472 mol/L)+( 0.352 mol/L)2=0.412 mol/L

From 3.00 to 4.00 h:

Average concentration=(0.352 mol/L)+( 0.262 mol/L)2=0.307 mol/L

From 4.00 to 5.00 h:

Average concentration=(0.262 mol/L)+( 0.196 mol/L)2=0.229 mol/L

The average rate and concentration is given in the table below.

Average concentration (mol/L)	Average rate (mol L−1h−1)
0.791	0.23
0.683	0.20
0.553	0.161
0.412	0.120
0.307	0.090
0.229	0.066

A graph of concentration vs. rate is given below.

OWLV2 FOR MOORE/STANITSKI'S CHEMISTRY:, Chapter 11, Problem 14QRT

Figure 1

The above graph is linear which satisfies the rate law.

(b)

Interpretation Introduction

Interpretation:

The rate constant k has to be evaluated as an average of the values obtained for the six intervals.

(b)

Expert Solution

Answer to Problem 14QRT

The average value of k is 0.29 h-1.

Explanation of Solution

The rate constant k as an average of the values obtained for the six intervals can be calculated as given below.

From 0 to 0.50 h:

Average rate is 0.23 mol L−1h−1.

Average concentration is 0.791 mol L−1.

Then, the rate constant can be calculated as given below.

k=Rate[N2O5]=0.23 mol L−1h−10.791 mol L−1=0.29 h−1.

From 0.50 to 1.00 h:

Average rate is 0.20 mol L−1h−1.

Average concentration is 0.683 mol L−1.

Then, the rate constant can be calculated as given below.

k=Rate[N2O5]=0.20 mol L−1h−10.683 mol L−1=0.29 h−1.

From 1.00 to 2.00 h:

Average rate is 0.161 mol L−1h−1.

Average concentration is 0.553 mol L−1.

Then, the rate constant can be calculated as given below.

k=Rate[N2O5]=0.161 mol L−1h−10.553 mol L−1=0.291 h−1.

From 2.00 to 3.00 h:

Average rate is 0.120 mol L−1h−1.

Average concentration is 0.412 mol L−1.

Then, the rate constant can be calculated as given below.

k=Rate[N2O5]=0.120 mol L−1h−10.412 mol L−1=0.291 h−1.

From 3.00 to 4.00 h:

Average rate is 0.090 mol L−1h−1.

Average concentration is 0.307 mol L−1.

Then, the rate constant can be calculated as given below.

k=Rate[N2O5]=0.090 mol L−1h−10.307 mol L−1=0.29 h−1.

From 4.00 to 5.00 h:

Average rate is 0.066 mol L−1h−1.

Average concentration is 0.229 mol L−1.

Then, the rate constant can be calculated as given below.

k=Rate[N2O5]=0.066 mol L−1h−10.229 mol L−1=0.29 h−1.

The average value of k is 0.29 h-1.

(c)

Interpretation Introduction

Interpretation:

The reaction rate exactly 2.5 h from the start has to be calculated.

(c)

Expert Solution

Answer to Problem 14QRT

The reaction rate exactly 2.5 h from the start is 0.12 mol L-1h-1.

Explanation of Solution

The initial concentration of N2O5 is 0.849 mol/L. The rate constant is 0.29 h−1.

The rate law describes that the reaction is a first-order reaction. The integrated first-order rate law is given below.

ln[N2O5]t=−kt+ln[N2O5]0=−(0.29 h−1)×(2.50 h)+ln(0.849)ln[N2O5]t=−0.889[N2O5]t=0.411 mol/L.

Now, the rate of the reaction at 2.5 h can be calculated as follows,

Rate=k[N2O5]=(0.29 h−1)×(0.411 mol/L)=0.12 mol L-1h-1.

Therefore, the reaction rate exactly 2.5 h from the start is 0.12 mol L-1h-1.

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Answer 5

Chapter 11, Problem 14QRT

(a)

Interpretation Introduction

Interpretation:

It has to be shown that the reaction obeys the rate law as given below.

N2O5→2NO2+12O2Rate=−Δ[N2O5]Δt=k[N2O5]

Concept Introduction:

In a first-order reaction, the graph of concentration vs. rate is linear.

(a)

Expert Solution

Explanation of Solution

Average concentrations can be calculated as given below.

From 0 to 0.50 h:

Average concentration=(0.849 mol/L)+(0.733 mol/L)2=0.791 mol/L

From 0.50 to 1.00 h:

Average concentration=(0.733 mol/L)+( 0.633mol/L)2=0.683 mol/L

From 1.00 to 2.00 h:

Average concentration=(0.633 mol/L)+( 0.472 mol/L)2=0.553 mol/L

From 2.00 to 3.00 h:

Average concentration=(0.472 mol/L)+( 0.352 mol/L)2=0.412 mol/L

From 3.00 to 4.00 h:

Average concentration=(0.352 mol/L)+( 0.262 mol/L)2=0.307 mol/L

From 4.00 to 5.00 h:

Average concentration=(0.262 mol/L)+( 0.196 mol/L)2=0.229 mol/L

The average rate and concentration is given in the table below.

Average concentration (mol/L)	Average rate (mol L−1h−1)
0.791	0.23
0.683	0.20
0.553	0.161
0.412	0.120
0.307	0.090
0.229	0.066

A graph of concentration vs. rate is given below.

OWLV2 FOR MOORE/STANITSKI'S CHEMISTRY:, Chapter 11, Problem 14QRT

Figure 1

The above graph is linear which satisfies the rate law.

(b)

Interpretation Introduction

Interpretation:

The rate constant k has to be evaluated as an average of the values obtained for the six intervals.

(b)

Expert Solution

Answer to Problem 14QRT

The average value of k is 0.29 h-1.

Explanation of Solution

The rate constant k as an average of the values obtained for the six intervals can be calculated as given below.

From 0 to 0.50 h:

Average rate is 0.23 mol L−1h−1.

Average concentration is 0.791 mol L−1.

Then, the rate constant can be calculated as given below.

k=Rate[N2O5]=0.23 mol L−1h−10.791 mol L−1=0.29 h−1.

From 0.50 to 1.00 h:

Average rate is 0.20 mol L−1h−1.

Average concentration is 0.683 mol L−1.

Then, the rate constant can be calculated as given below.

k=Rate[N2O5]=0.20 mol L−1h−10.683 mol L−1=0.29 h−1.

From 1.00 to 2.00 h:

Average rate is 0.161 mol L−1h−1.

Average concentration is 0.553 mol L−1.

Then, the rate constant can be calculated as given below.

k=Rate[N2O5]=0.161 mol L−1h−10.553 mol L−1=0.291 h−1.

From 2.00 to 3.00 h:

Average rate is 0.120 mol L−1h−1.

Average concentration is 0.412 mol L−1.

Then, the rate constant can be calculated as given below.

k=Rate[N2O5]=0.120 mol L−1h−10.412 mol L−1=0.291 h−1.

From 3.00 to 4.00 h:

Average rate is 0.090 mol L−1h−1.

Average concentration is 0.307 mol L−1.

Then, the rate constant can be calculated as given below.

k=Rate[N2O5]=0.090 mol L−1h−10.307 mol L−1=0.29 h−1.

From 4.00 to 5.00 h:

Average rate is 0.066 mol L−1h−1.

Average concentration is 0.229 mol L−1.

Then, the rate constant can be calculated as given below.

k=Rate[N2O5]=0.066 mol L−1h−10.229 mol L−1=0.29 h−1.

The average value of k is 0.29 h-1.

(c)

Interpretation Introduction

Interpretation:

The reaction rate exactly 2.5 h from the start has to be calculated.

(c)

Expert Solution

Answer to Problem 14QRT

The reaction rate exactly 2.5 h from the start is 0.12 mol L-1h-1.

Explanation of Solution

The initial concentration of N2O5 is 0.849 mol/L. The rate constant is 0.29 h−1.

The rate law describes that the reaction is a first-order reaction. The integrated first-order rate law is given below.

ln[N2O5]t=−kt+ln[N2O5]0=−(0.29 h−1)×(2.50 h)+ln(0.849)ln[N2O5]t=−0.889[N2O5]t=0.411 mol/L.

Now, the rate of the reaction at 2.5 h can be calculated as follows,

Rate=k[N2O5]=(0.29 h−1)×(0.411 mol/L)=0.12 mol L-1h-1.

Therefore, the reaction rate exactly 2.5 h from the start is 0.12 mol L-1h-1.

Answer 6

Chapter 11, Problem 14QRT

(a)

Interpretation Introduction

Interpretation:

It has to be shown that the reaction obeys the rate law as given below.

N2O5→2NO2+12O2Rate=−Δ[N2O5]Δt=k[N2O5]

Concept Introduction:

In a first-order reaction, the graph of concentration vs. rate is linear.

(a)

Expert Solution

Explanation of Solution

Average concentrations can be calculated as given below.

From 0 to 0.50 h:

Average concentration=(0.849 mol/L)+(0.733 mol/L)2=0.791 mol/L

From 0.50 to 1.00 h:

Average concentration=(0.733 mol/L)+( 0.633mol/L)2=0.683 mol/L

From 1.00 to 2.00 h:

Average concentration=(0.633 mol/L)+( 0.472 mol/L)2=0.553 mol/L

From 2.00 to 3.00 h:

Average concentration=(0.472 mol/L)+( 0.352 mol/L)2=0.412 mol/L

From 3.00 to 4.00 h:

Average concentration=(0.352 mol/L)+( 0.262 mol/L)2=0.307 mol/L

From 4.00 to 5.00 h:

Average concentration=(0.262 mol/L)+( 0.196 mol/L)2=0.229 mol/L

The average rate and concentration is given in the table below.

Average concentration (mol/L)	Average rate (mol L−1h−1)
0.791	0.23
0.683	0.20
0.553	0.161
0.412	0.120
0.307	0.090
0.229	0.066

A graph of concentration vs. rate is given below.

OWLV2 FOR MOORE/STANITSKI'S CHEMISTRY:, Chapter 11, Problem 14QRT

Figure 1

The above graph is linear which satisfies the rate law.

Answer 7

Chapter 11, Problem 14QRT

(a)

Interpretation Introduction

Interpretation:

It has to be shown that the reaction obeys the rate law as given below.

N2O5→2NO2+12O2Rate=−Δ[N2O5]Δt=k[N2O5]

Concept Introduction:

In a first-order reaction, the graph of concentration vs. rate is linear.

Answer 8

(a)

Expert Solution

Explanation of Solution

Average concentrations can be calculated as given below.

From 0 to 0.50 h:

Average concentration=(0.849 mol/L)+(0.733 mol/L)2=0.791 mol/L

From 0.50 to 1.00 h:

Average concentration=(0.733 mol/L)+( 0.633mol/L)2=0.683 mol/L

From 1.00 to 2.00 h:

Average concentration=(0.633 mol/L)+( 0.472 mol/L)2=0.553 mol/L

From 2.00 to 3.00 h:

Average concentration=(0.472 mol/L)+( 0.352 mol/L)2=0.412 mol/L

From 3.00 to 4.00 h:

Average concentration=(0.352 mol/L)+( 0.262 mol/L)2=0.307 mol/L

From 4.00 to 5.00 h:

Average concentration=(0.262 mol/L)+( 0.196 mol/L)2=0.229 mol/L

The average rate and concentration is given in the table below.

Average concentration (mol/L)	Average rate (mol L−1h−1)
0.791	0.23
0.683	0.20
0.553	0.161
0.412	0.120
0.307	0.090
0.229	0.066

A graph of concentration vs. rate is given below.

OWLV2 FOR MOORE/STANITSKI'S CHEMISTRY:, Chapter 11, Problem 14QRT

Figure 1

The above graph is linear which satisfies the rate law.

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

(b)

Interpretation Introduction

Interpretation:

The rate constant k has to be evaluated as an average of the values obtained for the six intervals.

(b)

Expert Solution

Answer to Problem 14QRT

The average value of k is 0.29 h-1.

Explanation of Solution

The rate constant k as an average of the values obtained for the six intervals can be calculated as given below.

From 0 to 0.50 h:

Average rate is 0.23 mol L−1h−1.

Average concentration is 0.791 mol L−1.

Then, the rate constant can be calculated as given below.

k=Rate[N2O5]=0.23 mol L−1h−10.791 mol L−1=0.29 h−1.

From 0.50 to 1.00 h:

Average rate is 0.20 mol L−1h−1.

Average concentration is 0.683 mol L−1.

Then, the rate constant can be calculated as given below.

k=Rate[N2O5]=0.20 mol L−1h−10.683 mol L−1=0.29 h−1.

From 1.00 to 2.00 h:

Average rate is 0.161 mol L−1h−1.

Average concentration is 0.553 mol L−1.

Then, the rate constant can be calculated as given below.

k=Rate[N2O5]=0.161 mol L−1h−10.553 mol L−1=0.291 h−1.

From 2.00 to 3.00 h:

Average rate is 0.120 mol L−1h−1.

Average concentration is 0.412 mol L−1.

Then, the rate constant can be calculated as given below.

k=Rate[N2O5]=0.120 mol L−1h−10.412 mol L−1=0.291 h−1.

From 3.00 to 4.00 h:

Average rate is 0.090 mol L−1h−1.

Average concentration is 0.307 mol L−1.

Then, the rate constant can be calculated as given below.

k=Rate[N2O5]=0.090 mol L−1h−10.307 mol L−1=0.29 h−1.

From 4.00 to 5.00 h:

Average rate is 0.066 mol L−1h−1.

Average concentration is 0.229 mol L−1.

Then, the rate constant can be calculated as given below.

k=Rate[N2O5]=0.066 mol L−1h−10.229 mol L−1=0.29 h−1.

The average value of k is 0.29 h-1.

Answer 13

(b)

Interpretation Introduction

Interpretation:

The rate constant k has to be evaluated as an average of the values obtained for the six intervals.

Answer 14

(b)

Expert Solution

Answer to Problem 14QRT

The average value of k is 0.29 h-1.

Answer 15

(b)

Expert Solution

Answer 16

(b)

Expert Solution

Answer 17

Expert Solution

Answer 18

Explanation of Solution

The rate constant k as an average of the values obtained for the six intervals can be calculated as given below.

From 0 to 0.50 h:

Average rate is 0.23 mol L−1h−1.

Average concentration is 0.791 mol L−1.

Then, the rate constant can be calculated as given below.

k=Rate[N2O5]=0.23 mol L−1h−10.791 mol L−1=0.29 h−1.

From 0.50 to 1.00 h:

Average rate is 0.20 mol L−1h−1.

Average concentration is 0.683 mol L−1.

Then, the rate constant can be calculated as given below.

k=Rate[N2O5]=0.20 mol L−1h−10.683 mol L−1=0.29 h−1.

From 1.00 to 2.00 h:

Average rate is 0.161 mol L−1h−1.

Average concentration is 0.553 mol L−1.

Then, the rate constant can be calculated as given below.

k=Rate[N2O5]=0.161 mol L−1h−10.553 mol L−1=0.291 h−1.

From 2.00 to 3.00 h:

Average rate is 0.120 mol L−1h−1.

Average concentration is 0.412 mol L−1.

Then, the rate constant can be calculated as given below.

k=Rate[N2O5]=0.120 mol L−1h−10.412 mol L−1=0.291 h−1.

From 3.00 to 4.00 h:

Average rate is 0.090 mol L−1h−1.

Average concentration is 0.307 mol L−1.

Then, the rate constant can be calculated as given below.

k=Rate[N2O5]=0.090 mol L−1h−10.307 mol L−1=0.29 h−1.

From 4.00 to 5.00 h:

Average rate is 0.066 mol L−1h−1.

Average concentration is 0.229 mol L−1.

Then, the rate constant can be calculated as given below.

k=Rate[N2O5]=0.066 mol L−1h−10.229 mol L−1=0.29 h−1.

The average value of k is 0.29 h-1.

Answer 19

(c)

Interpretation Introduction

Interpretation:

The reaction rate exactly 2.5 h from the start has to be calculated.

(c)

Expert Solution

Answer to Problem 14QRT

The reaction rate exactly 2.5 h from the start is 0.12 mol L-1h-1.

Explanation of Solution

The initial concentration of N2O5 is 0.849 mol/L. The rate constant is 0.29 h−1.

The rate law describes that the reaction is a first-order reaction. The integrated first-order rate law is given below.

ln[N2O5]t=−kt+ln[N2O5]0=−(0.29 h−1)×(2.50 h)+ln(0.849)ln[N2O5]t=−0.889[N2O5]t=0.411 mol/L.

Now, the rate of the reaction at 2.5 h can be calculated as follows,

Rate=k[N2O5]=(0.29 h−1)×(0.411 mol/L)=0.12 mol L-1h-1.

Therefore, the reaction rate exactly 2.5 h from the start is 0.12 mol L-1h-1.

Answer 20

(c)

Interpretation Introduction

Interpretation:

The reaction rate exactly 2.5 h from the start has to be calculated.

Answer 21

(c)

Expert Solution

Answer to Problem 14QRT

The reaction rate exactly 2.5 h from the start is 0.12 mol L-1h-1.

Answer 22

(c)

Expert Solution

Answer 23

(c)

Expert Solution

Answer 24

Expert Solution

Answer 25

Explanation of Solution

The initial concentration of N2O5 is 0.849 mol/L. The rate constant is 0.29 h−1.

The rate law describes that the reaction is a first-order reaction. The integrated first-order rate law is given below.

ln[N2O5]t=−kt+ln[N2O5]0=−(0.29 h−1)×(2.50 h)+ln(0.849)ln[N2O5]t=−0.889[N2O5]t=0.411 mol/L.

Now, the rate of the reaction at 2.5 h can be calculated as follows,

Rate=k[N2O5]=(0.29 h−1)×(0.411 mol/L)=0.12 mol L-1h-1.

Therefore, the reaction rate exactly 2.5 h from the start is 0.12 mol L-1h-1.

Answer 26

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

It has to be shown that the reaction obeys the rate law as given below. N 2 O 5 → 2 N O 2 + 1 2 O 2 R a t e = − Δ [ N 2 O 5 ] Δ t = k [ N 2 O 5 ] Concept Introduction: In a first-order reaction, the graph of concentration vs. rate is linear.

Videos

Interpretation:

It has to be shown that the reaction obeys the rate law as given below.

N2O5→2NO2+12O2Rate=−Δ[N2O5]Δt=k[N2O5]

Concept Introduction:

In a first-order reaction, the graph of concentration vs. rate is linear.

Explanation of Solution

Interpretation:

The rate constant k has to be evaluated as an average of the values obtained for the six intervals.

Answer to Problem 14QRT

Explanation of Solution

Interpretation:

The reaction rate exactly 2.5 h from the start has to be calculated.

Answer to Problem 14QRT

Explanation of Solution

Want to see more full solutions like this?

Chapter 11 Solutions

It has to be shown that the reaction obeys the rate law as given below. N 2 O 5 → 2 N O 2 + 1 2 O 2 R a t e = − Δ [ N 2 O 5 ] Δ t = k [ N 2 O 5 ] Concept Introduction: In a first-order reaction, the graph of concentration vs. rate is linear.

Videos

Interpretation: It has to be shown that the reaction obeys the rate law as given below. N2O5→2NO2+12O2Rate=−Δ[N2O5]Δt=k[N2O5] Concept Introduction: In a first-order reaction, the graph of concentration vs. rate is linear.

Explanation of Solution

Interpretation: The rate constant k has to be evaluated as an average of the values obtained for the six intervals.

Answer to Problem 14QRT

Explanation of Solution

Interpretation: The reaction rate exactly 2.5 h from the start has to be calculated.

Answer to Problem 14QRT

Explanation of Solution

Want to see more full solutions like this?

Chapter 11 Solutions

Interpretation:

It has to be shown that the reaction obeys the rate law as given below.

N2O5→2NO2+12O2Rate=−Δ[N2O5]Δt=k[N2O5]

Concept Introduction:

In a first-order reaction, the graph of concentration vs. rate is linear.

Interpretation:

The rate constant k has to be evaluated as an average of the values obtained for the six intervals.

Interpretation:

The reaction rate exactly 2.5 h from the start has to be calculated.