By what factor does the rate of the reaction with the followingrate law change i

Question

By what factor does the rate of the reaction with the followingrate law change if the concentrations of NO2 andCl2 are doubled. Rate =k[NO2]2[Cl2]. Express answer aswhole number.

In a first-order decomposition reaction, 50.0% of a compounddecomposes in 10.5 min. What is the rate constant of the reaction,in min-1? (Use 3 sig figs)

A decomposition reaction has a rate constant of 0.0012yr-1. How many years does it take for the reactantconcentration to reach 12.5% of its original value? (Use 4 sigfigs)

In a first-order decomposition reaction, 50.0% of a compounddecomposes in 10.5 min. What is the rate constant of the reaction,in min-1? (Use 3 sig figs)

A decomposition reaction has a rate constant of 0.0012yr-1. How many years does it take for the reactantconcentration to reach 12.5% of its original value? (Use 4 sigfigs)

Explanation / Answer

By what factor does the rate of the reaction with thefollowing rate law change if the concentrations of NO2and Cl2 are doubled. Rate =k[NO2]2[Cl2]. Express answer aswhole number. Rate , r =k[NO2]2[Cl2] new rate , r' =k[2NO2]2[2Cl2]                  = 2 2 * 2 1 *k[2NO2]2[2Cl2]                 = 8 k[2NO2]2[2Cl2]                 = 8 r So , r' / r = 8 In a first-order decomposition reaction, 50.0% of a compounddecomposes in 10.5 min. What is the rate constant of the reaction,in min-1 The rate constant for the first order reaction is givenby , K = ( 2.303 / t ) * log ( a / ( a-x)) Where t = time taken = 10.5 min a = initial concentration = 100 a-x = concentration remaining after time t = 50 Plug the values weget         K = (2.303 / t ) * log ( a / ( a-x))                                             = ( 2.303 / 10.5 ) * log ( 100 / 50 )                                             = 660 * 10^-4 min ^-1 A decomposition reaction has a rate constant of 0.0012yr-1. How many years does it take for the reactantconcentration to reach 12.5% of its original value We know that N = No e-t      -- ( 1) Where N0 = initial amount of the sample = 100 N = amount of the sampled remained = 12.5 = decay constant = ? t = time taken Plug the values in Eq ( 1 ) we get t = ( 1/ ) ln ( No/ N )                                                     = 752.5 years The rate constant for the first order reaction is givenby , K = ( 2.303 / t ) * log ( a / ( a-x)) Where t = time taken = 10.5 min a = initial concentration = 100 a-x = concentration remaining after time t = 50 Plug the values weget         K = (2.303 / t ) * log ( a / ( a-x))                                             = ( 2.303 / 10.5 ) * log ( 100 / 50 )                                             = 660 * 10^-4 min ^-1 A decomposition reaction has a rate constant of 0.0012yr-1. How many years does it take for the reactantconcentration to reach 12.5% of its original value We know that N = No e-t      -- ( 1) Where N0 = initial amount of the sample = 100 N = amount of the sampled remained = 12.5 = decay constant = ? t = time taken Plug the values in Eq ( 1 ) we get t = ( 1/ ) ln ( No/ N )                                                     = 752.5 years The rate constant for the first order reaction is givenby , K = ( 2.303 / t ) * log ( a / ( a-x)) Where t = time taken = 10.5 min a = initial concentration = 100 a-x = concentration remaining after time t = 50 Plug the values weget         K = (2.303 / t ) * log ( a / ( a-x))                                             = ( 2.303 / 10.5 ) * log ( 100 / 50 )                                             = 660 * 10^-4 min ^-1 A decomposition reaction has a rate constant of 0.0012yr-1. How many years does it take for the reactantconcentration to reach 12.5% of its original value We know that N = No e-t      -- ( 1) Where N0 = initial amount of the sample = 100 N = amount of the sampled remained = 12.5 = decay constant = ? t = time taken Plug the values in Eq ( 1 ) we get t = ( 1/ ) ln ( No/ N )                                                     = 752.5 years                                             = ( 2.303 / 10.5 ) * log ( 100 / 50 )                                             = 660 * 10^-4 min ^-1 A decomposition reaction has a rate constant of 0.0012yr-1. How many years does it take for the reactantconcentration to reach 12.5% of its original value We know that N = No e-t      -- ( 1) Where N0 = initial amount of the sample = 100 N = amount of the sampled remained = 12.5 = decay constant = ? t = time taken Plug the values in Eq ( 1 ) we get t = ( 1/ ) ln ( No/ N )                                                     = 752.5 years We know that N = No e-t      -- ( 1) Where N0 = initial amount of the sample = 100 N = amount of the sampled remained = 12.5 = decay constant = ? t = time taken Plug the values in Eq ( 1 ) we get t = ( 1/ ) ln ( No/ N )                                                     = 752.5 years Where N0 = initial amount of the sample = 100 N = amount of the sampled remained = 12.5 = decay constant = ? t = time taken Plug the values in Eq ( 1 ) we get t = ( 1/ ) ln ( No/ N )                                                     = 752.5 years

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