If the rate constant of a reaction is 3.0 mol `L^(-1)s^(-1)` at 700 K and 30 mol `L^(-1)s^(-1)` at 800 K, what is the energy of activation for this re

Question

If the rate constant of a reaction is 3.0 mol `L^(-1)s^(-1)` at 700 K and 30 mol `L^(-1)s^(-1)` at 800 K, what is the energy of activation for this re

If the rate constant of a reaction is 3.0 mol `L^(-1)s^(-1)` at 700 K and 30 mol `L^(-1)s^(-1)` at 800 K, what is the energy of activation for this reaction? `R = 8.314 JK^(-1) mol^(-1)`

Monjur Alahi

6 views

2025-01-04 04:42

1 Answers

Related Questions

For a reaction, the activation energy is zero. What is the value of rate constant at `300 K` if `k=1.6 xx 10^(6)s^(-1)` at 280 K `(R= 8.31 JK^(-1)mol^

2 Answers 4 Views

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

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

The activation energy for a reaction at the temperature T K was found to be `2.303RT J mol^(-1)`. The ratio of the rate constant to Arrhenius factor i

2 Answers 9 Views

The rate of a reaction doubles when the tempeature changes from 300 K to 310 K. Activation energy for the reaction is: `(R = 8.314 JK^(-1) mol^(-1), l

2 Answers 5 Views

The activation energy of a reaction is `94.14 kJ mol^(-1)` and the value of rate constant at 313 K is `1.8 xx 10^(-5)s^(-1)`. Calculate the frequency

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

The energy of activation for a reaction is `100 KJ mol^(-1)`. The peresence of a catalyst lowers the energy of activation by `75%`. What will be the e

2 Answers 8 Views

The activation energy of exothermic reaction `ArarrB` is 80 kJ `"mol"^(-1)` . The heat of reaction is `200 kJ "mol"^(-1)`. The activation energy for t

2 Answers 5 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

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

According to the available data:
`k_(1)=3.0 mol L^(-1)s^(-1), k_(2)=30 mol L^(-1)s^(-1), T_(1)=700 K, T_(2)=800K`.
R=`8.314 J K^(-1) mol^(-1)`.
`logk_(2)/k_(1)=E_(a)/(2.303R) [1/T_(1)-1/T_(2)]`
`log (30 mol L^(-1)s^(-1))/(3.0 mol L^(-1)s^(-1)) = (E_(a))/(2.303 xx 8.314 J K^(-1)mol^(-1)) [1/700K - 1/800K]`
`log 10 = (E_(a))/(2.303 xx 8.314 J mol^(-1))[(800-700)/(700 xx 800)]`
`1=E_(a)/(2.303 xx (8.314 J K^(-1)mol^(-1))[1/700K -1/800K]`
`log10 = (E_(a))/(2.303 xx 8.314 J mol^(-1))[(800-700)/(700xx800)]`
`1=(E_(a))/(2.303 xx 8.314 J mol^(-1)) xx (100)/(700 xx 800)`
`E_(a)= (2.303 xx 8.314 xx 700 xx 800)/(100)J mol^(-1)`
`=107223.99 J mol^(-1)=107.224 kJ mol^(-1)`