The initial rate of the given reaction at the given temperature has to be calculated. Concept introduction: Rate of the reaction is the change in the concentration of reactant or a product with time. It can be varied in accordance with temperature, pressure, concentration, presence of catalyst, surface area Rate equation for the general reaction A+B → Product is, R a t e = k r a t e c o n s t a t [ A ] [ B ] Rate constants are independent of concentration but depend on other factors, most notably temperature. The reaction with the faster rate will have the larger rate constant. Order of a reaction: The sum of exponents of the concentrations in the rate law for the reaction is said to be order of a reaction. The partial pressure of a gas in a mixture can be expressed as, P A = X A P Total P A → Partial pressure of gas A X A → Mole fraction of gas A P Total → Total pressure of mixture The ideal gas Law equation is, PV = nRT P → Pressure V → Volume n → Number of moles R → Ideal gas constant T → Temperature

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Answer 1

Question

Chapter 14, Problem 14.115SP

Interpretation Introduction

Interpretation:

The initial rate of the given reaction at the given temperature has to be calculated.

Concept introduction:

Rate of the reaction is the change in the concentration of reactant or a product with time. It can be varied in accordance with temperature, pressure, concentration, presence of catalyst, surface area

Rate equation for the general reaction $A+B → Product$ is,

$Rate = k rate constat [A] [B]$

Rate constants are independent of concentration but depend on other factors, most notably temperature.

The reaction with the faster rate will have the larger rate constant.

Order of a reaction: The sum of exponents of the concentrations in the rate law for the reaction is said to be order of a reaction.

The partial pressure of a gas in a mixture can be expressed as,

$PA = XA PTotalPA →Partial pressure of gas AXA →Mole fraction of gas APTotal → Total pressure of mixture$

The ideal gas Law equation is,

$PV = nRTP → PressureV → Volumen → Number of molesR → Ideal gas constantT → Temperature$

Expert Solution & Answer

Answer to Problem 14.115SP

Initial rate of the given reaction is $2.6 × 10-4 M/s$

Explanation of Solution

Given,

Gas mixture containing $CH3$ fragments, $C2H6$ molecules, and an inert gas $(He)$ was prepared at $600 K$ with a total pressure of $5.42 atm$ .

The elementary reaction is,

$CH3 +C2H6 → CH4 +C2H5$

This reaction follows second order kinetics, with a rate constant of $3.0 × 104 / M.s$ .

Given mole fractions of $CH3$ and $C2H6$ are $0.00093 and 0.00077$ respectively,

The initial rate of the reaction at $600 K$ temperature can be determined as follows,

Rate law for the given reaction is,

$Rate = k [CH3][C2H6]$

Rate constant value for the given reaction is $3.0 × 104 / M.s$ . If the concentration of reactants were known it is easy to find out the initial rate of the reaction.

Using mole fraction value and total partial pressure, partial pressure of each reactant in the reaction can be calculated as follows,

$PCH3 = XCH3 PT = (0.00093)(5.42 atm) = 0.0050 atmPC2H6 = XC2H6 PT = (0.00077)(5.42 atm) = 0.0042 atm$

Molar concentration of reactants can be determined with the help of ideal gas equation,

$nV = PRTnV → molar concentration$

$MCH3 = PCH3RT = 0.00050 atm 0.0821 L.atm/mol.K × 600 K = 1.0 × 10−4 MMC2H6 = PC2H6RT = 0.00042 atm 0.0821 L.atm/mol.K × 600 K = 8.5 × 10−5 M$

Substitute the concentration and the rate constant into the rate law to find the initial rate of the reaction,

$Rate = k [CH3][C2H6]Rate = (3.0 ×104 M-1s-1) × (1.0 × 10-4 M) × (8.5 × 10-5 M )Rate = 2.6 × 10-4 M/s$

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Answer 2

Question

Chapter 14, Problem 14.115SP

Interpretation Introduction

Interpretation:

The initial rate of the given reaction at the given temperature has to be calculated.

Concept introduction:

Rate of the reaction is the change in the concentration of reactant or a product with time. It can be varied in accordance with temperature, pressure, concentration, presence of catalyst, surface area

Rate equation for the general reaction $A+B → Product$ is,

$Rate = k rate constat [A] [B]$

Rate constants are independent of concentration but depend on other factors, most notably temperature.

The reaction with the faster rate will have the larger rate constant.

Order of a reaction: The sum of exponents of the concentrations in the rate law for the reaction is said to be order of a reaction.

The partial pressure of a gas in a mixture can be expressed as,

$PA = XA PTotalPA →Partial pressure of gas AXA →Mole fraction of gas APTotal → Total pressure of mixture$

The ideal gas Law equation is,

$PV = nRTP → PressureV → Volumen → Number of molesR → Ideal gas constantT → Temperature$

Expert Solution & Answer

Answer to Problem 14.115SP

Initial rate of the given reaction is $2.6 × 10-4 M/s$

Explanation of Solution

Given,

Gas mixture containing $CH3$ fragments, $C2H6$ molecules, and an inert gas $(He)$ was prepared at $600 K$ with a total pressure of $5.42 atm$ .

The elementary reaction is,

$CH3 +C2H6 → CH4 +C2H5$

This reaction follows second order kinetics, with a rate constant of $3.0 × 104 / M.s$ .

Given mole fractions of $CH3$ and $C2H6$ are $0.00093 and 0.00077$ respectively,

The initial rate of the reaction at $600 K$ temperature can be determined as follows,

Rate law for the given reaction is,

$Rate = k [CH3][C2H6]$

Rate constant value for the given reaction is $3.0 × 104 / M.s$ . If the concentration of reactants were known it is easy to find out the initial rate of the reaction.

Using mole fraction value and total partial pressure, partial pressure of each reactant in the reaction can be calculated as follows,

$PCH3 = XCH3 PT = (0.00093)(5.42 atm) = 0.0050 atmPC2H6 = XC2H6 PT = (0.00077)(5.42 atm) = 0.0042 atm$

Molar concentration of reactants can be determined with the help of ideal gas equation,

$nV = PRTnV → molar concentration$

$MCH3 = PCH3RT = 0.00050 atm 0.0821 L.atm/mol.K × 600 K = 1.0 × 10−4 MMC2H6 = PC2H6RT = 0.00042 atm 0.0821 L.atm/mol.K × 600 K = 8.5 × 10−5 M$

Substitute the concentration and the rate constant into the rate law to find the initial rate of the reaction,

$Rate = k [CH3][C2H6]Rate = (3.0 ×104 M-1s-1) × (1.0 × 10-4 M) × (8.5 × 10-5 M )Rate = 2.6 × 10-4 M/s$

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Answer 3

Question

Chapter 14, Problem 14.115SP

Interpretation Introduction

Interpretation:

The initial rate of the given reaction at the given temperature has to be calculated.

Concept introduction:

Rate of the reaction is the change in the concentration of reactant or a product with time. It can be varied in accordance with temperature, pressure, concentration, presence of catalyst, surface area

Rate equation for the general reaction $A+B → Product$ is,

$Rate = k rate constat [A] [B]$

Rate constants are independent of concentration but depend on other factors, most notably temperature.

The reaction with the faster rate will have the larger rate constant.

Order of a reaction: The sum of exponents of the concentrations in the rate law for the reaction is said to be order of a reaction.

The partial pressure of a gas in a mixture can be expressed as,

$PA = XA PTotalPA →Partial pressure of gas AXA →Mole fraction of gas APTotal → Total pressure of mixture$

The ideal gas Law equation is,

$PV = nRTP → PressureV → Volumen → Number of molesR → Ideal gas constantT → Temperature$

Expert Solution & Answer

Answer to Problem 14.115SP

Initial rate of the given reaction is $2.6 × 10-4 M/s$

Explanation of Solution

Given,

Gas mixture containing $CH3$ fragments, $C2H6$ molecules, and an inert gas $(He)$ was prepared at $600 K$ with a total pressure of $5.42 atm$ .

The elementary reaction is,

$CH3 +C2H6 → CH4 +C2H5$

This reaction follows second order kinetics, with a rate constant of $3.0 × 104 / M.s$ .

Given mole fractions of $CH3$ and $C2H6$ are $0.00093 and 0.00077$ respectively,

The initial rate of the reaction at $600 K$ temperature can be determined as follows,

Rate law for the given reaction is,

$Rate = k [CH3][C2H6]$

Rate constant value for the given reaction is $3.0 × 104 / M.s$ . If the concentration of reactants were known it is easy to find out the initial rate of the reaction.

Using mole fraction value and total partial pressure, partial pressure of each reactant in the reaction can be calculated as follows,

$PCH3 = XCH3 PT = (0.00093)(5.42 atm) = 0.0050 atmPC2H6 = XC2H6 PT = (0.00077)(5.42 atm) = 0.0042 atm$

Molar concentration of reactants can be determined with the help of ideal gas equation,

$nV = PRTnV → molar concentration$

$MCH3 = PCH3RT = 0.00050 atm 0.0821 L.atm/mol.K × 600 K = 1.0 × 10−4 MMC2H6 = PC2H6RT = 0.00042 atm 0.0821 L.atm/mol.K × 600 K = 8.5 × 10−5 M$

Substitute the concentration and the rate constant into the rate law to find the initial rate of the reaction,

$Rate = k [CH3][C2H6]Rate = (3.0 ×104 M-1s-1) × (1.0 × 10-4 M) × (8.5 × 10-5 M )Rate = 2.6 × 10-4 M/s$

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Answer 4

Question

Chapter 14, Problem 14.115SP

Interpretation Introduction

Interpretation:

The initial rate of the given reaction at the given temperature has to be calculated.

Concept introduction:

Rate of the reaction is the change in the concentration of reactant or a product with time. It can be varied in accordance with temperature, pressure, concentration, presence of catalyst, surface area

Rate equation for the general reaction $A+B → Product$ is,

$Rate = k rate constat [A] [B]$

Rate constants are independent of concentration but depend on other factors, most notably temperature.

The reaction with the faster rate will have the larger rate constant.

Order of a reaction: The sum of exponents of the concentrations in the rate law for the reaction is said to be order of a reaction.

The partial pressure of a gas in a mixture can be expressed as,

$PA = XA PTotalPA →Partial pressure of gas AXA →Mole fraction of gas APTotal → Total pressure of mixture$

The ideal gas Law equation is,

$PV = nRTP → PressureV → Volumen → Number of molesR → Ideal gas constantT → Temperature$

Expert Solution & Answer

Answer to Problem 14.115SP

Initial rate of the given reaction is $2.6 × 10-4 M/s$

Explanation of Solution

Given,

Gas mixture containing $CH3$ fragments, $C2H6$ molecules, and an inert gas $(He)$ was prepared at $600 K$ with a total pressure of $5.42 atm$ .

The elementary reaction is,

$CH3 +C2H6 → CH4 +C2H5$

This reaction follows second order kinetics, with a rate constant of $3.0 × 104 / M.s$ .

Given mole fractions of $CH3$ and $C2H6$ are $0.00093 and 0.00077$ respectively,

The initial rate of the reaction at $600 K$ temperature can be determined as follows,

Rate law for the given reaction is,

$Rate = k [CH3][C2H6]$

Rate constant value for the given reaction is $3.0 × 104 / M.s$ . If the concentration of reactants were known it is easy to find out the initial rate of the reaction.

Using mole fraction value and total partial pressure, partial pressure of each reactant in the reaction can be calculated as follows,

$PCH3 = XCH3 PT = (0.00093)(5.42 atm) = 0.0050 atmPC2H6 = XC2H6 PT = (0.00077)(5.42 atm) = 0.0042 atm$

Molar concentration of reactants can be determined with the help of ideal gas equation,

$nV = PRTnV → molar concentration$

$MCH3 = PCH3RT = 0.00050 atm 0.0821 L.atm/mol.K × 600 K = 1.0 × 10−4 MMC2H6 = PC2H6RT = 0.00042 atm 0.0821 L.atm/mol.K × 600 K = 8.5 × 10−5 M$

Substitute the concentration and the rate constant into the rate law to find the initial rate of the reaction,

$Rate = k [CH3][C2H6]Rate = (3.0 ×104 M-1s-1) × (1.0 × 10-4 M) × (8.5 × 10-5 M )Rate = 2.6 × 10-4 M/s$

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Answer 5

Chapter 14, Problem 14.115SP

Interpretation Introduction

Interpretation:

The initial rate of the given reaction at the given temperature has to be calculated.

Concept introduction:

Rate of the reaction is the change in the concentration of reactant or a product with time. It can be varied in accordance with temperature, pressure, concentration, presence of catalyst, surface area

Rate equation for the general reaction $A+B → Product$ is,

$Rate = k rate constat [A] [B]$

Rate constants are independent of concentration but depend on other factors, most notably temperature.

The reaction with the faster rate will have the larger rate constant.

Order of a reaction: The sum of exponents of the concentrations in the rate law for the reaction is said to be order of a reaction.

The partial pressure of a gas in a mixture can be expressed as,

$PA = XA PTotalPA →Partial pressure of gas AXA →Mole fraction of gas APTotal → Total pressure of mixture$

The ideal gas Law equation is,

$PV = nRTP → PressureV → Volumen → Number of molesR → Ideal gas constantT → Temperature$

Expert Solution & Answer

Answer to Problem 14.115SP

Initial rate of the given reaction is $2.6 × 10-4 M/s$

Explanation of Solution

Given,

Gas mixture containing $CH3$ fragments, $C2H6$ molecules, and an inert gas $(He)$ was prepared at $600 K$ with a total pressure of $5.42 atm$ .

The elementary reaction is,

$CH3 +C2H6 → CH4 +C2H5$

This reaction follows second order kinetics, with a rate constant of $3.0 × 104 / M.s$ .

Given mole fractions of $CH3$ and $C2H6$ are $0.00093 and 0.00077$ respectively,

The initial rate of the reaction at $600 K$ temperature can be determined as follows,

Rate law for the given reaction is,

$Rate = k [CH3][C2H6]$

Rate constant value for the given reaction is $3.0 × 104 / M.s$ . If the concentration of reactants were known it is easy to find out the initial rate of the reaction.

Using mole fraction value and total partial pressure, partial pressure of each reactant in the reaction can be calculated as follows,

$PCH3 = XCH3 PT = (0.00093)(5.42 atm) = 0.0050 atmPC2H6 = XC2H6 PT = (0.00077)(5.42 atm) = 0.0042 atm$

Molar concentration of reactants can be determined with the help of ideal gas equation,

$nV = PRTnV → molar concentration$

$MCH3 = PCH3RT = 0.00050 atm 0.0821 L.atm/mol.K × 600 K = 1.0 × 10−4 MMC2H6 = PC2H6RT = 0.00042 atm 0.0821 L.atm/mol.K × 600 K = 8.5 × 10−5 M$

Substitute the concentration and the rate constant into the rate law to find the initial rate of the reaction,

$Rate = k [CH3][C2H6]Rate = (3.0 ×104 M-1s-1) × (1.0 × 10-4 M) × (8.5 × 10-5 M )Rate = 2.6 × 10-4 M/s$

Answer 6

Chapter 14, Problem 14.115SP

Interpretation Introduction

Interpretation:

The initial rate of the given reaction at the given temperature has to be calculated.

Concept introduction:

Rate of the reaction is the change in the concentration of reactant or a product with time. It can be varied in accordance with temperature, pressure, concentration, presence of catalyst, surface area

Rate equation for the general reaction $A+B → Product$ is,

$Rate = k rate constat [A] [B]$

Rate constants are independent of concentration but depend on other factors, most notably temperature.

The reaction with the faster rate will have the larger rate constant.

Order of a reaction: The sum of exponents of the concentrations in the rate law for the reaction is said to be order of a reaction.

The partial pressure of a gas in a mixture can be expressed as,

$PA = XA PTotalPA →Partial pressure of gas AXA →Mole fraction of gas APTotal → Total pressure of mixture$

The ideal gas Law equation is,

$PV = nRTP → PressureV → Volumen → Number of molesR → Ideal gas constantT → Temperature$

Expert Solution & Answer

Answer to Problem 14.115SP

Initial rate of the given reaction is $2.6 × 10-4 M/s$

Explanation of Solution

Given,

Gas mixture containing $CH3$ fragments, $C2H6$ molecules, and an inert gas $(He)$ was prepared at $600 K$ with a total pressure of $5.42 atm$ .

The elementary reaction is,

$CH3 +C2H6 → CH4 +C2H5$

This reaction follows second order kinetics, with a rate constant of $3.0 × 104 / M.s$ .

Given mole fractions of $CH3$ and $C2H6$ are $0.00093 and 0.00077$ respectively,

The initial rate of the reaction at $600 K$ temperature can be determined as follows,

Rate law for the given reaction is,

$Rate = k [CH3][C2H6]$

Rate constant value for the given reaction is $3.0 × 104 / M.s$ . If the concentration of reactants were known it is easy to find out the initial rate of the reaction.

Using mole fraction value and total partial pressure, partial pressure of each reactant in the reaction can be calculated as follows,

$PCH3 = XCH3 PT = (0.00093)(5.42 atm) = 0.0050 atmPC2H6 = XC2H6 PT = (0.00077)(5.42 atm) = 0.0042 atm$

Molar concentration of reactants can be determined with the help of ideal gas equation,

$nV = PRTnV → molar concentration$

$MCH3 = PCH3RT = 0.00050 atm 0.0821 L.atm/mol.K × 600 K = 1.0 × 10−4 MMC2H6 = PC2H6RT = 0.00042 atm 0.0821 L.atm/mol.K × 600 K = 8.5 × 10−5 M$

Substitute the concentration and the rate constant into the rate law to find the initial rate of the reaction,

$Rate = k [CH3][C2H6]Rate = (3.0 ×104 M-1s-1) × (1.0 × 10-4 M) × (8.5 × 10-5 M )Rate = 2.6 × 10-4 M/s$

Answer 7

Chapter 14, Problem 14.115SP

Interpretation Introduction

Interpretation:

The initial rate of the given reaction at the given temperature has to be calculated.

Concept introduction:

Rate of the reaction is the change in the concentration of reactant or a product with time. It can be varied in accordance with temperature, pressure, concentration, presence of catalyst, surface area

Rate equation for the general reaction $A+B → Product$ is,

$Rate = k rate constat [A] [B]$

Rate constants are independent of concentration but depend on other factors, most notably temperature.

The reaction with the faster rate will have the larger rate constant.

Order of a reaction: The sum of exponents of the concentrations in the rate law for the reaction is said to be order of a reaction.

The partial pressure of a gas in a mixture can be expressed as,

$PA = XA PTotalPA →Partial pressure of gas AXA →Mole fraction of gas APTotal → Total pressure of mixture$

The ideal gas Law equation is,

$PV = nRTP → PressureV → Volumen → Number of molesR → Ideal gas constantT → Temperature$

Answer 8

Expert Solution & Answer

Answer to Problem 14.115SP

Initial rate of the given reaction is $2.6 × 10-4 M/s$

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Explanation of Solution

Given,

Gas mixture containing $CH3$ fragments, $C2H6$ molecules, and an inert gas $(He)$ was prepared at $600 K$ with a total pressure of $5.42 atm$ .

The elementary reaction is,

$CH3 +C2H6 → CH4 +C2H5$

This reaction follows second order kinetics, with a rate constant of $3.0 × 104 / M.s$ .

Given mole fractions of $CH3$ and $C2H6$ are $0.00093 and 0.00077$ respectively,

The initial rate of the reaction at $600 K$ temperature can be determined as follows,

Rate law for the given reaction is,

$Rate = k [CH3][C2H6]$

Rate constant value for the given reaction is $3.0 × 104 / M.s$ . If the concentration of reactants were known it is easy to find out the initial rate of the reaction.

Using mole fraction value and total partial pressure, partial pressure of each reactant in the reaction can be calculated as follows,

$PCH3 = XCH3 PT = (0.00093)(5.42 atm) = 0.0050 atmPC2H6 = XC2H6 PT = (0.00077)(5.42 atm) = 0.0042 atm$

Molar concentration of reactants can be determined with the help of ideal gas equation,

$nV = PRTnV → molar concentration$

$MCH3 = PCH3RT = 0.00050 atm 0.0821 L.atm/mol.K × 600 K = 1.0 × 10−4 MMC2H6 = PC2H6RT = 0.00042 atm 0.0821 L.atm/mol.K × 600 K = 8.5 × 10−5 M$

Substitute the concentration and the rate constant into the rate law to find the initial rate of the reaction,

$Rate = k [CH3][C2H6]Rate = (3.0 ×104 M-1s-1) × (1.0 × 10-4 M) × (8.5 × 10-5 M )Rate = 2.6 × 10-4 M/s$

Answer 12

Ch. 14.1 - Practice Exercise Write the rate expressions for...Ch. 14.1 - Practice Exercise Consider the reaction Suppose...Ch. 14.1 - Review of Concepts Write a balanced equation for a...Ch. 14.2 - Practice Exercise The reaction of peroxydisulfate...Ch. 14.2 - Prob. 1RC Ch. 14.3 - Practice Exercise The reaction is first order in...Ch. 14.3 - Practice Exercise Calculate the half-life of the...Ch. 14.3 - Review of Concepts Consider the first-order...Ch. 14.3 - Practice Exercise The reaction is second order...Ch. 14.4 - Practice Exercise The second-order rate constant...

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Answer to Problem 14.115SP

Explanation of Solution

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