At 40°C, H 2 O 2 ( aq ) will decompose according to the following reaction: 2H 2 O 2 ( a q ) → 2H 2 O ( l ) + O 2 ( g ) The following data were collected for the concentration of H 2 O 2 at various times. Times(s) [H 2 O 2 ](mol/L) 0 1.000 2.16 × 10 4 0.500 4.32 × 10 4 0.250 a. Calculate the average rate of decomposition of H 2 O 2 between 0 and 2.16 × 10 4 s. Use this rate to calculate the average rate of production of O 2 ( g ) over the same time period. b. What are these rates for the time period 2.16 × 10 4 s to 4.32 × 10 4 s?

Question

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Answer 1

Textbook Question

Chapter 11, Problem 25E

At 40°C, H₂O₂ (aq) will decompose according to the following reaction:

$2H 2 O 2 ( a q ) → 2H 2 O ( l ) + O 2 ( g )$

The following data were collected for the concentration of H₂O₂ at various times.

Times(s)	[H₂O₂](mol/L)
0	1.000
2.16 × 10⁴	0.500
4.32 × 10⁴	0.250

a. Calculate the average rate of decomposition of H₂O₂ between 0 and 2.16 × 10⁴ s. Use this rate to calculate the average rate of production of O₂(g) over the same time period.

b. What are these rates for the time period 2.16 × 10⁴ s to 4.32 × 10⁴ s?

(a)

Expert Solution

Interpretation Introduction

Interpretation: The decomposition reaction of $H2O2$ is given. The answers for each option using appropriate data from the given table are to be stated.

Concept introduction: The change observed in the concentration of a reactant or a product per unit time is known as the rate of the particular reaction. The differential rate law provides the rate of a reaction at specific reaction concentrations.

Answer to Problem 25E

Answer

The average rate of decomposition of $H2O2$ between $0$ and $2.16×104 s$ is $−2.31×10−5 mol/L⋅s_$ . The average rate of production of $O2(g)$ between $0$ and $2.16×104 s$ is $1.16×10−5 mol/L⋅s_$ .

Explanation of Solution

Explanation

Given

Concentration of $H2O2$ at $0 s$ is $1.000 mol/L$ .

Concentration of $H2O2$ at $2.16×104 s$ is $0.500 mol/L$ .

The reaction that takes place is,

$2H2O2(aq)→2H2O(l)+O2(g)$

The rate law for the above equation is written as,

$Rate=−12Δ[H2O2]Δt=12Δ[H2O]Δt=12Δ[O2]Δt$ (1)

Where,

$Δ[H2O2]$ is the average rate of consumption of $H2O2$ .
$Δ[H2O]$ is the average rate of production of $H2O$ .
$Δ[O2]$ is the average rate of production of $O2$ .
$Δt$ is the average time.

The average rate of decomposition of $H2O2$ between $0$ and $2.16×104 s$ is calculated as,

$Average Rate=Final concentration−Initial concentrationFinal time−Initial time$

Substitute the values of concentration of $H2O2$ at $0$ and $2.16×104 s$ in the above equation.

$Average Rate=Final concentration−Initial concentrationFinal time−Initial time=(0.500−1.00) mol/L(2.16×104−0) s=−2.31×10−5 mol/L⋅s_$

The average rate of decomposition of $H2O2$ between $0$ and $2.16×104 s$ is $−2.31×10−5 mol/L⋅s$ .

Substitute the above value in equation (1).

$−12Δ[H2O2]Δt=12Δ[O2]ΔtΔ[O2]Δt=−12×(−2.31×10−5 mol/L⋅s)=1.16×10−5 mol/L⋅s_$

Conclusion

Conclusion

The calculated average rate of decomposition of $H2O2$ between $0$ and $2.16×104 s$ is $−2.31×10−5 mol/L⋅s_$ . The average rate of production of $O2(g)$ between $0$ and $2.16×104 s$ is $1.16×10−5 mol/L⋅s_$ .

(b)

Expert Solution

Interpretation Introduction

Interpretation: The decomposition reaction of $H2O2$ is given. The answers for each option using appropriate data from the given table are to be stated.

Concept introduction: The change observed in the concentration of a reactant or a product per unit time is known as the rate of the particular reaction. The differential rate law provides the rate of a reaction at specific reaction concentrations.

Answer to Problem 25E

Answer

The average rate of decomposition of $H2O2$ between $2.16×104 s$ and $4.32×104 s$ is $−1.15×10−5 mol/L⋅s_$ . The average rate of production of $O2(g)$ between $2.16×104 s$ and $4.32×104 s$ is $5.78×10−6 mol/L⋅s_$ .

Explanation of Solution

Given:

Concentration of $H2O2$ at $2.16×104 s$ is $0.500 mol/L$ .

Concentration of $H2O2$ at $4.32×104 s$ is $0.250 mol/L$ .

The reaction that takes place is,

$2H2O2(aq)→2H2O(l)+O2(g)$

The rate law for the above equation is written as,

$Rate=−12Δ[H2O2]Δt=12Δ[H2O]Δt=12Δ[O2]Δt$ (2)

Where,

$Δ[H2O2]$ is the average rate of consumption of $H2O2$ .
$Δ[H2O]$ is the average rate of production of $H2O$ .
$Δ[O2]$ is the average rate of production of $O2$ .
$Δt$ is the average time.

The average rate of decomposition of $H2O2$ between $0$ and $2.16×104 s$ is calculated as,

$Average Rate=Final concentration−Initial concentrationFinal time−Initial time$

Substitute the values of concentration of $H2O2$ at $4.32×104 s$ and $2.16×104 s$ in the above equation.

$Average Rate=Final concentration−Initial concentrationFinal time−Initial time=(0.25−0.500) mol/L(4.32×104 −2.16×104) s=−1.15×10−5 mol/L⋅s_$

The average rate of production of $O2(g)$ between $2.16×104 s$ and $4.32×104 s$ is $5.78×10−6 mol/L⋅s_$ .

The average rate of decomposition of $H2O2$ between $2.16×104 s$ and $4.32×104 s$ is $−1.15×10−5 mol/L⋅s$ .

Substitute the above value in equation (2).

$−12Δ[H2O2]Δt=12Δ[O2]ΔtΔ[O2]Δt=−12×(−1.15×10−5 mol/L⋅s)=5.78×10−6 mol/L⋅s_$

Conclusion

Conclusion

The calculated average rate of decomposition of $H2O2$ between $2.16×104 s$ and $4.32×104 s$ is $−1.15×10−5 mol/L⋅s_$ . The average rate of production of $O2(g)$ between $2.16×104 s$ and $4.32×104 s$ is $5.78×10−6 mol/L⋅s_$ .

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Answer 2

Textbook Question

Chapter 11, Problem 25E

At 40°C, H₂O₂ (aq) will decompose according to the following reaction:

$2H 2 O 2 ( a q ) → 2H 2 O ( l ) + O 2 ( g )$

The following data were collected for the concentration of H₂O₂ at various times.

Times(s)	[H₂O₂](mol/L)
0	1.000
2.16 × 10⁴	0.500
4.32 × 10⁴	0.250

a. Calculate the average rate of decomposition of H₂O₂ between 0 and 2.16 × 10⁴ s. Use this rate to calculate the average rate of production of O₂(g) over the same time period.

b. What are these rates for the time period 2.16 × 10⁴ s to 4.32 × 10⁴ s?

(a)

Expert Solution

Interpretation Introduction

Interpretation: The decomposition reaction of $H2O2$ is given. The answers for each option using appropriate data from the given table are to be stated.

Concept introduction: The change observed in the concentration of a reactant or a product per unit time is known as the rate of the particular reaction. The differential rate law provides the rate of a reaction at specific reaction concentrations.

Answer to Problem 25E

Answer

The average rate of decomposition of $H2O2$ between $0$ and $2.16×104 s$ is $−2.31×10−5 mol/L⋅s_$ . The average rate of production of $O2(g)$ between $0$ and $2.16×104 s$ is $1.16×10−5 mol/L⋅s_$ .

Explanation of Solution

Explanation

Given

Concentration of $H2O2$ at $0 s$ is $1.000 mol/L$ .

Concentration of $H2O2$ at $2.16×104 s$ is $0.500 mol/L$ .

The reaction that takes place is,

$2H2O2(aq)→2H2O(l)+O2(g)$

The rate law for the above equation is written as,

$Rate=−12Δ[H2O2]Δt=12Δ[H2O]Δt=12Δ[O2]Δt$ (1)

Where,

$Δ[H2O2]$ is the average rate of consumption of $H2O2$ .
$Δ[H2O]$ is the average rate of production of $H2O$ .
$Δ[O2]$ is the average rate of production of $O2$ .
$Δt$ is the average time.

The average rate of decomposition of $H2O2$ between $0$ and $2.16×104 s$ is calculated as,

$Average Rate=Final concentration−Initial concentrationFinal time−Initial time$

Substitute the values of concentration of $H2O2$ at $0$ and $2.16×104 s$ in the above equation.

$Average Rate=Final concentration−Initial concentrationFinal time−Initial time=(0.500−1.00) mol/L(2.16×104−0) s=−2.31×10−5 mol/L⋅s_$

The average rate of decomposition of $H2O2$ between $0$ and $2.16×104 s$ is $−2.31×10−5 mol/L⋅s$ .

Substitute the above value in equation (1).

$−12Δ[H2O2]Δt=12Δ[O2]ΔtΔ[O2]Δt=−12×(−2.31×10−5 mol/L⋅s)=1.16×10−5 mol/L⋅s_$

Conclusion

Conclusion

The calculated average rate of decomposition of $H2O2$ between $0$ and $2.16×104 s$ is $−2.31×10−5 mol/L⋅s_$ . The average rate of production of $O2(g)$ between $0$ and $2.16×104 s$ is $1.16×10−5 mol/L⋅s_$ .

(b)

Expert Solution

Interpretation Introduction

Interpretation: The decomposition reaction of $H2O2$ is given. The answers for each option using appropriate data from the given table are to be stated.

Concept introduction: The change observed in the concentration of a reactant or a product per unit time is known as the rate of the particular reaction. The differential rate law provides the rate of a reaction at specific reaction concentrations.

Answer to Problem 25E

Answer

The average rate of decomposition of $H2O2$ between $2.16×104 s$ and $4.32×104 s$ is $−1.15×10−5 mol/L⋅s_$ . The average rate of production of $O2(g)$ between $2.16×104 s$ and $4.32×104 s$ is $5.78×10−6 mol/L⋅s_$ .

Explanation of Solution

Given:

Concentration of $H2O2$ at $2.16×104 s$ is $0.500 mol/L$ .

Concentration of $H2O2$ at $4.32×104 s$ is $0.250 mol/L$ .

The reaction that takes place is,

$2H2O2(aq)→2H2O(l)+O2(g)$

The rate law for the above equation is written as,

$Rate=−12Δ[H2O2]Δt=12Δ[H2O]Δt=12Δ[O2]Δt$ (2)

Where,

$Δ[H2O2]$ is the average rate of consumption of $H2O2$ .
$Δ[H2O]$ is the average rate of production of $H2O$ .
$Δ[O2]$ is the average rate of production of $O2$ .
$Δt$ is the average time.

The average rate of decomposition of $H2O2$ between $0$ and $2.16×104 s$ is calculated as,

$Average Rate=Final concentration−Initial concentrationFinal time−Initial time$

Substitute the values of concentration of $H2O2$ at $4.32×104 s$ and $2.16×104 s$ in the above equation.

$Average Rate=Final concentration−Initial concentrationFinal time−Initial time=(0.25−0.500) mol/L(4.32×104 −2.16×104) s=−1.15×10−5 mol/L⋅s_$

The average rate of production of $O2(g)$ between $2.16×104 s$ and $4.32×104 s$ is $5.78×10−6 mol/L⋅s_$ .

The average rate of decomposition of $H2O2$ between $2.16×104 s$ and $4.32×104 s$ is $−1.15×10−5 mol/L⋅s$ .

Substitute the above value in equation (2).

$−12Δ[H2O2]Δt=12Δ[O2]ΔtΔ[O2]Δt=−12×(−1.15×10−5 mol/L⋅s)=5.78×10−6 mol/L⋅s_$

Conclusion

Conclusion

The calculated average rate of decomposition of $H2O2$ between $2.16×104 s$ and $4.32×104 s$ is $−1.15×10−5 mol/L⋅s_$ . The average rate of production of $O2(g)$ between $2.16×104 s$ and $4.32×104 s$ is $5.78×10−6 mol/L⋅s_$ .

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Answer 3

Textbook Question

Chapter 11, Problem 25E

At 40°C, H₂O₂ (aq) will decompose according to the following reaction:

$2H 2 O 2 ( a q ) → 2H 2 O ( l ) + O 2 ( g )$

The following data were collected for the concentration of H₂O₂ at various times.

Times(s)	[H₂O₂](mol/L)
0	1.000
2.16 × 10⁴	0.500
4.32 × 10⁴	0.250

a. Calculate the average rate of decomposition of H₂O₂ between 0 and 2.16 × 10⁴ s. Use this rate to calculate the average rate of production of O₂(g) over the same time period.

b. What are these rates for the time period 2.16 × 10⁴ s to 4.32 × 10⁴ s?

(a)

Expert Solution

Interpretation Introduction

Interpretation: The decomposition reaction of $H2O2$ is given. The answers for each option using appropriate data from the given table are to be stated.

Concept introduction: The change observed in the concentration of a reactant or a product per unit time is known as the rate of the particular reaction. The differential rate law provides the rate of a reaction at specific reaction concentrations.

Answer to Problem 25E

Answer

The average rate of decomposition of $H2O2$ between $0$ and $2.16×104 s$ is $−2.31×10−5 mol/L⋅s_$ . The average rate of production of $O2(g)$ between $0$ and $2.16×104 s$ is $1.16×10−5 mol/L⋅s_$ .

Explanation of Solution

Explanation

Given

Concentration of $H2O2$ at $0 s$ is $1.000 mol/L$ .

Concentration of $H2O2$ at $2.16×104 s$ is $0.500 mol/L$ .

The reaction that takes place is,

$2H2O2(aq)→2H2O(l)+O2(g)$

The rate law for the above equation is written as,

$Rate=−12Δ[H2O2]Δt=12Δ[H2O]Δt=12Δ[O2]Δt$ (1)

Where,

$Δ[H2O2]$ is the average rate of consumption of $H2O2$ .
$Δ[H2O]$ is the average rate of production of $H2O$ .
$Δ[O2]$ is the average rate of production of $O2$ .
$Δt$ is the average time.

The average rate of decomposition of $H2O2$ between $0$ and $2.16×104 s$ is calculated as,

$Average Rate=Final concentration−Initial concentrationFinal time−Initial time$

Substitute the values of concentration of $H2O2$ at $0$ and $2.16×104 s$ in the above equation.

$Average Rate=Final concentration−Initial concentrationFinal time−Initial time=(0.500−1.00) mol/L(2.16×104−0) s=−2.31×10−5 mol/L⋅s_$

The average rate of decomposition of $H2O2$ between $0$ and $2.16×104 s$ is $−2.31×10−5 mol/L⋅s$ .

Substitute the above value in equation (1).

$−12Δ[H2O2]Δt=12Δ[O2]ΔtΔ[O2]Δt=−12×(−2.31×10−5 mol/L⋅s)=1.16×10−5 mol/L⋅s_$

Conclusion

Conclusion

The calculated average rate of decomposition of $H2O2$ between $0$ and $2.16×104 s$ is $−2.31×10−5 mol/L⋅s_$ . The average rate of production of $O2(g)$ between $0$ and $2.16×104 s$ is $1.16×10−5 mol/L⋅s_$ .

(b)

Expert Solution

Interpretation Introduction

Interpretation: The decomposition reaction of $H2O2$ is given. The answers for each option using appropriate data from the given table are to be stated.

Concept introduction: The change observed in the concentration of a reactant or a product per unit time is known as the rate of the particular reaction. The differential rate law provides the rate of a reaction at specific reaction concentrations.

Answer to Problem 25E

Answer

The average rate of decomposition of $H2O2$ between $2.16×104 s$ and $4.32×104 s$ is $−1.15×10−5 mol/L⋅s_$ . The average rate of production of $O2(g)$ between $2.16×104 s$ and $4.32×104 s$ is $5.78×10−6 mol/L⋅s_$ .

Explanation of Solution

Given:

Concentration of $H2O2$ at $2.16×104 s$ is $0.500 mol/L$ .

Concentration of $H2O2$ at $4.32×104 s$ is $0.250 mol/L$ .

The reaction that takes place is,

$2H2O2(aq)→2H2O(l)+O2(g)$

The rate law for the above equation is written as,

$Rate=−12Δ[H2O2]Δt=12Δ[H2O]Δt=12Δ[O2]Δt$ (2)

Where,

$Δ[H2O2]$ is the average rate of consumption of $H2O2$ .
$Δ[H2O]$ is the average rate of production of $H2O$ .
$Δ[O2]$ is the average rate of production of $O2$ .
$Δt$ is the average time.

The average rate of decomposition of $H2O2$ between $0$ and $2.16×104 s$ is calculated as,

$Average Rate=Final concentration−Initial concentrationFinal time−Initial time$

Substitute the values of concentration of $H2O2$ at $4.32×104 s$ and $2.16×104 s$ in the above equation.

$Average Rate=Final concentration−Initial concentrationFinal time−Initial time=(0.25−0.500) mol/L(4.32×104 −2.16×104) s=−1.15×10−5 mol/L⋅s_$

The average rate of production of $O2(g)$ between $2.16×104 s$ and $4.32×104 s$ is $5.78×10−6 mol/L⋅s_$ .

The average rate of decomposition of $H2O2$ between $2.16×104 s$ and $4.32×104 s$ is $−1.15×10−5 mol/L⋅s$ .

Substitute the above value in equation (2).

$−12Δ[H2O2]Δt=12Δ[O2]ΔtΔ[O2]Δt=−12×(−1.15×10−5 mol/L⋅s)=5.78×10−6 mol/L⋅s_$

Conclusion

Conclusion

The calculated average rate of decomposition of $H2O2$ between $2.16×104 s$ and $4.32×104 s$ is $−1.15×10−5 mol/L⋅s_$ . The average rate of production of $O2(g)$ between $2.16×104 s$ and $4.32×104 s$ is $5.78×10−6 mol/L⋅s_$ .

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Answer 4

Textbook Question

Chapter 11, Problem 25E

At 40°C, H₂O₂ (aq) will decompose according to the following reaction:

$2H 2 O 2 ( a q ) → 2H 2 O ( l ) + O 2 ( g )$

The following data were collected for the concentration of H₂O₂ at various times.

Times(s)	[H₂O₂](mol/L)
0	1.000
2.16 × 10⁴	0.500
4.32 × 10⁴	0.250

a. Calculate the average rate of decomposition of H₂O₂ between 0 and 2.16 × 10⁴ s. Use this rate to calculate the average rate of production of O₂(g) over the same time period.

b. What are these rates for the time period 2.16 × 10⁴ s to 4.32 × 10⁴ s?

(a)

Expert Solution

Interpretation Introduction

Interpretation: The decomposition reaction of $H2O2$ is given. The answers for each option using appropriate data from the given table are to be stated.

Concept introduction: The change observed in the concentration of a reactant or a product per unit time is known as the rate of the particular reaction. The differential rate law provides the rate of a reaction at specific reaction concentrations.

Answer to Problem 25E

Answer

The average rate of decomposition of $H2O2$ between $0$ and $2.16×104 s$ is $−2.31×10−5 mol/L⋅s_$ . The average rate of production of $O2(g)$ between $0$ and $2.16×104 s$ is $1.16×10−5 mol/L⋅s_$ .

Explanation of Solution

Explanation

Given

Concentration of $H2O2$ at $0 s$ is $1.000 mol/L$ .

Concentration of $H2O2$ at $2.16×104 s$ is $0.500 mol/L$ .

The reaction that takes place is,

$2H2O2(aq)→2H2O(l)+O2(g)$

The rate law for the above equation is written as,

$Rate=−12Δ[H2O2]Δt=12Δ[H2O]Δt=12Δ[O2]Δt$ (1)

Where,

$Δ[H2O2]$ is the average rate of consumption of $H2O2$ .
$Δ[H2O]$ is the average rate of production of $H2O$ .
$Δ[O2]$ is the average rate of production of $O2$ .
$Δt$ is the average time.

The average rate of decomposition of $H2O2$ between $0$ and $2.16×104 s$ is calculated as,

$Average Rate=Final concentration−Initial concentrationFinal time−Initial time$

Substitute the values of concentration of $H2O2$ at $0$ and $2.16×104 s$ in the above equation.

$Average Rate=Final concentration−Initial concentrationFinal time−Initial time=(0.500−1.00) mol/L(2.16×104−0) s=−2.31×10−5 mol/L⋅s_$

The average rate of decomposition of $H2O2$ between $0$ and $2.16×104 s$ is $−2.31×10−5 mol/L⋅s$ .

Substitute the above value in equation (1).

$−12Δ[H2O2]Δt=12Δ[O2]ΔtΔ[O2]Δt=−12×(−2.31×10−5 mol/L⋅s)=1.16×10−5 mol/L⋅s_$

Conclusion

Conclusion

The calculated average rate of decomposition of $H2O2$ between $0$ and $2.16×104 s$ is $−2.31×10−5 mol/L⋅s_$ . The average rate of production of $O2(g)$ between $0$ and $2.16×104 s$ is $1.16×10−5 mol/L⋅s_$ .

(b)

Expert Solution

Interpretation Introduction

Interpretation: The decomposition reaction of $H2O2$ is given. The answers for each option using appropriate data from the given table are to be stated.

Concept introduction: The change observed in the concentration of a reactant or a product per unit time is known as the rate of the particular reaction. The differential rate law provides the rate of a reaction at specific reaction concentrations.

Answer to Problem 25E

Answer

The average rate of decomposition of $H2O2$ between $2.16×104 s$ and $4.32×104 s$ is $−1.15×10−5 mol/L⋅s_$ . The average rate of production of $O2(g)$ between $2.16×104 s$ and $4.32×104 s$ is $5.78×10−6 mol/L⋅s_$ .

Explanation of Solution

Given:

Concentration of $H2O2$ at $2.16×104 s$ is $0.500 mol/L$ .

Concentration of $H2O2$ at $4.32×104 s$ is $0.250 mol/L$ .

The reaction that takes place is,

$2H2O2(aq)→2H2O(l)+O2(g)$

The rate law for the above equation is written as,

$Rate=−12Δ[H2O2]Δt=12Δ[H2O]Δt=12Δ[O2]Δt$ (2)

Where,

$Δ[H2O2]$ is the average rate of consumption of $H2O2$ .
$Δ[H2O]$ is the average rate of production of $H2O$ .
$Δ[O2]$ is the average rate of production of $O2$ .
$Δt$ is the average time.

The average rate of decomposition of $H2O2$ between $0$ and $2.16×104 s$ is calculated as,

$Average Rate=Final concentration−Initial concentrationFinal time−Initial time$

Substitute the values of concentration of $H2O2$ at $4.32×104 s$ and $2.16×104 s$ in the above equation.

$Average Rate=Final concentration−Initial concentrationFinal time−Initial time=(0.25−0.500) mol/L(4.32×104 −2.16×104) s=−1.15×10−5 mol/L⋅s_$

The average rate of production of $O2(g)$ between $2.16×104 s$ and $4.32×104 s$ is $5.78×10−6 mol/L⋅s_$ .

The average rate of decomposition of $H2O2$ between $2.16×104 s$ and $4.32×104 s$ is $−1.15×10−5 mol/L⋅s$ .

Substitute the above value in equation (2).

$−12Δ[H2O2]Δt=12Δ[O2]ΔtΔ[O2]Δt=−12×(−1.15×10−5 mol/L⋅s)=5.78×10−6 mol/L⋅s_$

Conclusion

Conclusion

The calculated average rate of decomposition of $H2O2$ between $2.16×104 s$ and $4.32×104 s$ is $−1.15×10−5 mol/L⋅s_$ . The average rate of production of $O2(g)$ between $2.16×104 s$ and $4.32×104 s$ is $5.78×10−6 mol/L⋅s_$ .

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Answer 5

Textbook Question

Answer 6

Textbook Question

Answer 7

(a)

Expert Solution

Interpretation Introduction

Interpretation: The decomposition reaction of $H2O2$ is given. The answers for each option using appropriate data from the given table are to be stated.

Concept introduction: The change observed in the concentration of a reactant or a product per unit time is known as the rate of the particular reaction. The differential rate law provides the rate of a reaction at specific reaction concentrations.

Answer to Problem 25E

Answer

The average rate of decomposition of $H2O2$ between $0$ and $2.16×104 s$ is $−2.31×10−5 mol/L⋅s_$ . The average rate of production of $O2(g)$ between $0$ and $2.16×104 s$ is $1.16×10−5 mol/L⋅s_$ .

Explanation of Solution

Explanation

Given

Concentration of $H2O2$ at $0 s$ is $1.000 mol/L$ .

Concentration of $H2O2$ at $2.16×104 s$ is $0.500 mol/L$ .

The reaction that takes place is,

$2H2O2(aq)→2H2O(l)+O2(g)$

The rate law for the above equation is written as,

$Rate=−12Δ[H2O2]Δt=12Δ[H2O]Δt=12Δ[O2]Δt$ (1)

Where,

$Δ[H2O2]$ is the average rate of consumption of $H2O2$ .
$Δ[H2O]$ is the average rate of production of $H2O$ .
$Δ[O2]$ is the average rate of production of $O2$ .
$Δt$ is the average time.

The average rate of decomposition of $H2O2$ between $0$ and $2.16×104 s$ is calculated as,

$Average Rate=Final concentration−Initial concentrationFinal time−Initial time$

Substitute the values of concentration of $H2O2$ at $0$ and $2.16×104 s$ in the above equation.

$Average Rate=Final concentration−Initial concentrationFinal time−Initial time=(0.500−1.00) mol/L(2.16×104−0) s=−2.31×10−5 mol/L⋅s_$

The average rate of decomposition of $H2O2$ between $0$ and $2.16×104 s$ is $−2.31×10−5 mol/L⋅s$ .

Substitute the above value in equation (1).

$−12Δ[H2O2]Δt=12Δ[O2]ΔtΔ[O2]Δt=−12×(−2.31×10−5 mol/L⋅s)=1.16×10−5 mol/L⋅s_$

Conclusion

Conclusion

The calculated average rate of decomposition of $H2O2$ between $0$ and $2.16×104 s$ is $−2.31×10−5 mol/L⋅s_$ . The average rate of production of $O2(g)$ between $0$ and $2.16×104 s$ is $1.16×10−5 mol/L⋅s_$ .

(b)

Expert Solution

Interpretation Introduction

Interpretation: The decomposition reaction of $H2O2$ is given. The answers for each option using appropriate data from the given table are to be stated.

Concept introduction: The change observed in the concentration of a reactant or a product per unit time is known as the rate of the particular reaction. The differential rate law provides the rate of a reaction at specific reaction concentrations.

Answer to Problem 25E

Answer

The average rate of decomposition of $H2O2$ between $2.16×104 s$ and $4.32×104 s$ is $−1.15×10−5 mol/L⋅s_$ . The average rate of production of $O2(g)$ between $2.16×104 s$ and $4.32×104 s$ is $5.78×10−6 mol/L⋅s_$ .

Explanation of Solution

Given:

Concentration of $H2O2$ at $2.16×104 s$ is $0.500 mol/L$ .

Concentration of $H2O2$ at $4.32×104 s$ is $0.250 mol/L$ .

The reaction that takes place is,

$2H2O2(aq)→2H2O(l)+O2(g)$

The rate law for the above equation is written as,

$Rate=−12Δ[H2O2]Δt=12Δ[H2O]Δt=12Δ[O2]Δt$ (2)

Where,

$Δ[H2O2]$ is the average rate of consumption of $H2O2$ .
$Δ[H2O]$ is the average rate of production of $H2O$ .
$Δ[O2]$ is the average rate of production of $O2$ .
$Δt$ is the average time.

The average rate of decomposition of $H2O2$ between $0$ and $2.16×104 s$ is calculated as,

$Average Rate=Final concentration−Initial concentrationFinal time−Initial time$

Substitute the values of concentration of $H2O2$ at $4.32×104 s$ and $2.16×104 s$ in the above equation.

$Average Rate=Final concentration−Initial concentrationFinal time−Initial time=(0.25−0.500) mol/L(4.32×104 −2.16×104) s=−1.15×10−5 mol/L⋅s_$

The average rate of production of $O2(g)$ between $2.16×104 s$ and $4.32×104 s$ is $5.78×10−6 mol/L⋅s_$ .

The average rate of decomposition of $H2O2$ between $2.16×104 s$ and $4.32×104 s$ is $−1.15×10−5 mol/L⋅s$ .

Substitute the above value in equation (2).

$−12Δ[H2O2]Δt=12Δ[O2]ΔtΔ[O2]Δt=−12×(−1.15×10−5 mol/L⋅s)=5.78×10−6 mol/L⋅s_$

Conclusion

Conclusion

The calculated average rate of decomposition of $H2O2$ between $2.16×104 s$ and $4.32×104 s$ is $−1.15×10−5 mol/L⋅s_$ . The average rate of production of $O2(g)$ between $2.16×104 s$ and $4.32×104 s$ is $5.78×10−6 mol/L⋅s_$ .

Answer 8

(a)

Expert Solution

Interpretation Introduction

Interpretation: The decomposition reaction of $H2O2$ is given. The answers for each option using appropriate data from the given table are to be stated.

Concept introduction: The change observed in the concentration of a reactant or a product per unit time is known as the rate of the particular reaction. The differential rate law provides the rate of a reaction at specific reaction concentrations.

Answer to Problem 25E

Answer

The average rate of decomposition of $H2O2$ between $0$ and $2.16×104 s$ is $−2.31×10−5 mol/L⋅s_$ . The average rate of production of $O2(g)$ between $0$ and $2.16×104 s$ is $1.16×10−5 mol/L⋅s_$ .

Explanation of Solution

Explanation

Given

Concentration of $H2O2$ at $0 s$ is $1.000 mol/L$ .

Concentration of $H2O2$ at $2.16×104 s$ is $0.500 mol/L$ .

The reaction that takes place is,

$2H2O2(aq)→2H2O(l)+O2(g)$

The rate law for the above equation is written as,

$Rate=−12Δ[H2O2]Δt=12Δ[H2O]Δt=12Δ[O2]Δt$ (1)

Where,

$Δ[H2O2]$ is the average rate of consumption of $H2O2$ .
$Δ[H2O]$ is the average rate of production of $H2O$ .
$Δ[O2]$ is the average rate of production of $O2$ .
$Δt$ is the average time.

The average rate of decomposition of $H2O2$ between $0$ and $2.16×104 s$ is calculated as,

$Average Rate=Final concentration−Initial concentrationFinal time−Initial time$

Substitute the values of concentration of $H2O2$ at $0$ and $2.16×104 s$ in the above equation.

$Average Rate=Final concentration−Initial concentrationFinal time−Initial time=(0.500−1.00) mol/L(2.16×104−0) s=−2.31×10−5 mol/L⋅s_$

The average rate of decomposition of $H2O2$ between $0$ and $2.16×104 s$ is $−2.31×10−5 mol/L⋅s$ .

Substitute the above value in equation (1).

$−12Δ[H2O2]Δt=12Δ[O2]ΔtΔ[O2]Δt=−12×(−2.31×10−5 mol/L⋅s)=1.16×10−5 mol/L⋅s_$

Conclusion

Conclusion

The calculated average rate of decomposition of $H2O2$ between $0$ and $2.16×104 s$ is $−2.31×10−5 mol/L⋅s_$ . The average rate of production of $O2(g)$ between $0$ and $2.16×104 s$ is $1.16×10−5 mol/L⋅s_$ .

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Interpretation Introduction

Interpretation: The decomposition reaction of $H2O2$ is given. The answers for each option using appropriate data from the given table are to be stated.

Concept introduction: The change observed in the concentration of a reactant or a product per unit time is known as the rate of the particular reaction. The differential rate law provides the rate of a reaction at specific reaction concentrations.

Answer 13

Answer to Problem 25E

Answer

The average rate of decomposition of $H2O2$ between $0$ and $2.16×104 s$ is $−2.31×10−5 mol/L⋅s_$ . The average rate of production of $O2(g)$ between $0$ and $2.16×104 s$ is $1.16×10−5 mol/L⋅s_$ .

Answer 14

Explanation of Solution

Explanation

Given

Concentration of $H2O2$ at $0 s$ is $1.000 mol/L$ .

Concentration of $H2O2$ at $2.16×104 s$ is $0.500 mol/L$ .

The reaction that takes place is,

$2H2O2(aq)→2H2O(l)+O2(g)$

The rate law for the above equation is written as,

$Rate=−12Δ[H2O2]Δt=12Δ[H2O]Δt=12Δ[O2]Δt$ (1)

Where,

$Δ[H2O2]$ is the average rate of consumption of $H2O2$ .
$Δ[H2O]$ is the average rate of production of $H2O$ .
$Δ[O2]$ is the average rate of production of $O2$ .
$Δt$ is the average time.

The average rate of decomposition of $H2O2$ between $0$ and $2.16×104 s$ is calculated as,

$Average Rate=Final concentration−Initial concentrationFinal time−Initial time$

Substitute the values of concentration of $H2O2$ at $0$ and $2.16×104 s$ in the above equation.

$Average Rate=Final concentration−Initial concentrationFinal time−Initial time=(0.500−1.00) mol/L(2.16×104−0) s=−2.31×10−5 mol/L⋅s_$

The average rate of decomposition of $H2O2$ between $0$ and $2.16×104 s$ is $−2.31×10−5 mol/L⋅s$ .

Substitute the above value in equation (1).

$−12Δ[H2O2]Δt=12Δ[O2]ΔtΔ[O2]Δt=−12×(−2.31×10−5 mol/L⋅s)=1.16×10−5 mol/L⋅s_$

Answer 15

Conclusion

Conclusion

The calculated average rate of decomposition of $H2O2$ between $0$ and $2.16×104 s$ is $−2.31×10−5 mol/L⋅s_$ . The average rate of production of $O2(g)$ between $0$ and $2.16×104 s$ is $1.16×10−5 mol/L⋅s_$ .

Answer 16

(b)

Expert Solution

Interpretation Introduction

Interpretation: The decomposition reaction of $H2O2$ is given. The answers for each option using appropriate data from the given table are to be stated.

Concept introduction: The change observed in the concentration of a reactant or a product per unit time is known as the rate of the particular reaction. The differential rate law provides the rate of a reaction at specific reaction concentrations.

Answer to Problem 25E

Answer

The average rate of decomposition of $H2O2$ between $2.16×104 s$ and $4.32×104 s$ is $−1.15×10−5 mol/L⋅s_$ . The average rate of production of $O2(g)$ between $2.16×104 s$ and $4.32×104 s$ is $5.78×10−6 mol/L⋅s_$ .

Explanation of Solution

Given:

Concentration of $H2O2$ at $2.16×104 s$ is $0.500 mol/L$ .

Concentration of $H2O2$ at $4.32×104 s$ is $0.250 mol/L$ .

The reaction that takes place is,

$2H2O2(aq)→2H2O(l)+O2(g)$

The rate law for the above equation is written as,

$Rate=−12Δ[H2O2]Δt=12Δ[H2O]Δt=12Δ[O2]Δt$ (2)

Where,

$Δ[H2O2]$ is the average rate of consumption of $H2O2$ .
$Δ[H2O]$ is the average rate of production of $H2O$ .
$Δ[O2]$ is the average rate of production of $O2$ .
$Δt$ is the average time.

The average rate of decomposition of $H2O2$ between $0$ and $2.16×104 s$ is calculated as,

$Average Rate=Final concentration−Initial concentrationFinal time−Initial time$

Substitute the values of concentration of $H2O2$ at $4.32×104 s$ and $2.16×104 s$ in the above equation.

$Average Rate=Final concentration−Initial concentrationFinal time−Initial time=(0.25−0.500) mol/L(4.32×104 −2.16×104) s=−1.15×10−5 mol/L⋅s_$

The average rate of production of $O2(g)$ between $2.16×104 s$ and $4.32×104 s$ is $5.78×10−6 mol/L⋅s_$ .

The average rate of decomposition of $H2O2$ between $2.16×104 s$ and $4.32×104 s$ is $−1.15×10−5 mol/L⋅s$ .

Substitute the above value in equation (2).

$−12Δ[H2O2]Δt=12Δ[O2]ΔtΔ[O2]Δt=−12×(−1.15×10−5 mol/L⋅s)=5.78×10−6 mol/L⋅s_$

Conclusion

Conclusion

The calculated average rate of decomposition of $H2O2$ between $2.16×104 s$ and $4.32×104 s$ is $−1.15×10−5 mol/L⋅s_$ . The average rate of production of $O2(g)$ between $2.16×104 s$ and $4.32×104 s$ is $5.78×10−6 mol/L⋅s_$ .

Answer 17

(b)

Expert Solution

Answer 18

(b)

Expert Solution

Answer 19

Expert Solution

Answer 20

Interpretation Introduction

Interpretation: The decomposition reaction of $H2O2$ is given. The answers for each option using appropriate data from the given table are to be stated.

Concept introduction: The change observed in the concentration of a reactant or a product per unit time is known as the rate of the particular reaction. The differential rate law provides the rate of a reaction at specific reaction concentrations.

Answer 21

Answer to Problem 25E

Answer

The average rate of decomposition of $H2O2$ between $2.16×104 s$ and $4.32×104 s$ is $−1.15×10−5 mol/L⋅s_$ . The average rate of production of $O2(g)$ between $2.16×104 s$ and $4.32×104 s$ is $5.78×10−6 mol/L⋅s_$ .

Answer 22

Explanation of Solution

Given:

Concentration of $H2O2$ at $2.16×104 s$ is $0.500 mol/L$ .

Concentration of $H2O2$ at $4.32×104 s$ is $0.250 mol/L$ .

The reaction that takes place is,

$2H2O2(aq)→2H2O(l)+O2(g)$

The rate law for the above equation is written as,

$Rate=−12Δ[H2O2]Δt=12Δ[H2O]Δt=12Δ[O2]Δt$ (2)

Where,

$Δ[H2O2]$ is the average rate of consumption of $H2O2$ .
$Δ[H2O]$ is the average rate of production of $H2O$ .
$Δ[O2]$ is the average rate of production of $O2$ .
$Δt$ is the average time.

The average rate of decomposition of $H2O2$ between $0$ and $2.16×104 s$ is calculated as,

$Average Rate=Final concentration−Initial concentrationFinal time−Initial time$

Substitute the values of concentration of $H2O2$ at $4.32×104 s$ and $2.16×104 s$ in the above equation.

$Average Rate=Final concentration−Initial concentrationFinal time−Initial time=(0.25−0.500) mol/L(4.32×104 −2.16×104) s=−1.15×10−5 mol/L⋅s_$

The average rate of production of $O2(g)$ between $2.16×104 s$ and $4.32×104 s$ is $5.78×10−6 mol/L⋅s_$ .

The average rate of decomposition of $H2O2$ between $2.16×104 s$ and $4.32×104 s$ is $−1.15×10−5 mol/L⋅s$ .

Substitute the above value in equation (2).

$−12Δ[H2O2]Δt=12Δ[O2]ΔtΔ[O2]Δt=−12×(−1.15×10−5 mol/L⋅s)=5.78×10−6 mol/L⋅s_$

Answer 23

Conclusion

Conclusion

The calculated average rate of decomposition of $H2O2$ between $2.16×104 s$ and $4.32×104 s$ is $−1.15×10−5 mol/L⋅s_$ . The average rate of production of $O2(g)$ between $2.16×104 s$ and $4.32×104 s$ is $5.78×10−6 mol/L⋅s_$ .

Answer 24

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Answer to Problem 25E

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Answer to Problem 25E

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