At 407 K, the rate contant of a chemical reaction is `9.5xx10^(-5) s^(-1)` and at 420 K the rate constant is `1.9xx10^(-4)s^(-1)`. Calculate the frequ

Question

At 407 K, the rate contant of a chemical reaction is `9.5xx10^(-5) s^(-1)` and at 420 K the rate constant is `1.9xx10^(-4)s^(-1)`. Calculate the frequ

At 407 K, the rate contant of a chemical reaction is `9.5xx10^(-5) s^(-1)` and at 420 K the rate constant is `1.9xx10^(-4)s^(-1)`. Calculate the frequency equation is

Monjur Alahi

4 views

2025-01-04 04:42

1 Answers

Related Questions

Rate of reaction is given by the equation Rate = `k[A]^(2)[B]` What are the units for the rate and rate constant for the section?

2 Answers 8 Views

The rate of reaction: `2NO + Cl_(2) rarr 2NOCl` is given by the rate, equation rate `= k[NO]_(2)[Cl_(2)]`. The value of the rate constant can be incre

2 Answers 5 Views

Activation energy `(E_(a))` and rate constant `(K_(1)` and `(K_(2))` for a chemical reaction at two different temperatures `T_(1)` and `T_(2)` are rel

2 Answers 5 Views

For a reaction taking place in three steps, the rate constant are `k_(1), k_(2)` and `k_(3)` and overall rate constant is `k=(k_(1)k_(3))/(k_(2)`. If

2 Answers 4 Views

The rate law expresses the relationship of the rate of a reaction to the rate constant and the concentration of the reactants raised to some powers fo

2 Answers 4 Views

Rate constant for the reaction `NO_2+COtoNO+CO_2` are `1.3M^(-1)S^(-1)` at 700 K and `23 M^(-1)S^(-1)` at 800 K. What will be the rate constant at 750

2 Answers 7 Views

A reaction takes place in three steps with individual rate constant and activation energy, `{:(,"Rate constant","Activation energy"),("Step 1",k_(1),E

2 Answers 5 Views

A system has two charges `q_(A)=2.5xx10^(-7) C` and `q_(B)=-2.5xx10^(-7) C` located at points A, `(0, 0, -0.15 m)` and B, `(0, 0, +0.15 m)` respective

2 Answers 4 Views

सिलिकॉन परमाणुओं की संख्या `5xx10^(28)` प्रति `m^(3)` है। यह साथ ही साथ आर्सेनिक के `5xx10^(22)`परमाणु प्रति `m^(3)` और इडियम के `5xx10^(20)` परमाणु प

2 Answers 4 Views

If the rate of a chemical reaction is expressed in the following manner : Rate = K[A]^2[B], then the order of this reaction would be–

2 Answers 4 Views

Salim Ahmed Kawsar · Accepted Answer · 2023-02-05T15:29:48+06:00

`"log"_(10) =(E_(2))/(2.303xxR)[(T_(2)-T_(1))/(T_(1)T_(2))]`
Given `k_(1)=9.5xx10^(-5)s^(-1),k_(2)=1.9xx10^(-4)s^(-1),`
R=8.314J `"mol"^(-1)K^(-1),`
`T_(1)-407 K and T_(2)=420K`
Substituting the values in Arrhenius equation
`"log"_(10)(1.9xx10^(-4))/(9.5xx10^(-5))=(E_(a))/(2.303xx8.314)[(420-407)/(420xx407)]`
Applying now log `k_(1)` = log `A-(E_(a))/(2.3030RT_(1))`
log `9.5xx10^(-5)="log"A-(75782.3)/(2.303xx8.314xx407)`
or log `(A)/(9.5xx10 ^(-5))=(757.82.3)/(2.3030xx8.314xx407)=9.7246`
`A=5.04xx10^(5)s^(-1)`