Home
Class 12
CHEMISTRY
The decomposition of A into product has ...

The decomposition of A into product has value of k as `4.5xx10^(3)s^(-1)` at 10 `""^(@)C` and energy of activation 60 kJ `"mol"^(-1)`. At what temperature would k be `1.5xx10^(4)s^(-1)`?

Text Solution

Verified by Experts

`k_(1)=4.5xx10^(3)s^(-1), T_(1)=10+273K=283K, k_(2)=1.5xx10^(4)s^(-1), T_(2)=?, E_(a)=60"kJ mol"^(-1)`
Applying Arrhenius equation and substituting the values, we get
`"log"(k_(2))/(k_(1))=(E_(a))/(2.303R)((T_(2)-T_(1))/(T_(1)T_(2)))`
or `"log"(1.5xx10^(4))/(4.5xx10^(3))=(E_(a))/(RT)= -(60000"J mol"^(-1))/(2.303xx8.314JK^(-1)"mol"^(-1))((T_(2)-283)/(283T_(2)))`
or `log 3.333=3133.63 ((T_(2)-283)/(283T_(2))) or (0.5228)/(3133.63)=(T_(2)-283)/(283T_(2))`
or `0.0472T_(2)=T_(2)-283 or 0.9528T_(2)=283`
or `T_(2)=(283)/(0.9528)=297K=(297-273)""^(@)C=24""^(@)C`.
Doubtnut Promotions Banner Mobile Dark
|

Topper's Solved these Questions

  • CHEMICAL KINETICS

    U-LIKE SERIES|Exercise CASE BASED/SOURCE-BASED INTEGRATED QUESTIONS|15 Videos

  • CHEMICAL KINETICS

    U-LIKE SERIES|Exercise MULTIPLE CHOICE QUESTIONS |19 Videos

  • CHEMICAL KINETICS

    U-LIKE SERIES|Exercise SELF ASSESSMENT TEST (SECTION D)|1 Videos

  • CBSE EXAMINATION PAPER 2020 (CODE No.56/4/1)

    U-LIKE SERIES|Exercise SECTION -D|3 Videos

  • CHEMISTRY IN EVERYDAY LIFE

    U-LIKE SERIES|Exercise LONG ANSWER QUESTIONS - I|21 Videos

Similar Questions

Explore conceptually related problems

The decomposition of A into products has value of k as 4.5xx10^(3)s^(-1) at 10""^(@)C and energy of activation 60" kJ mol"^(-1) . At what ttemperature would k be 1.5xx10^(4)s^(-1) ?

(a) The decomposition of A into products has a value of K as 4.5 xx 10^(3)s^(-1) " at "10^(@)C and energy of activation 60kJ "mol"^(-1) . At what temperature would K be 1.5 xx 10^(4) s^(-1) ? (b) (i) If half life period of a first order reaction is x and 3//4^(th) life period of the same reaction is y, how are x and y related to each other ? (ii) In some cases it is found that a large number of colliding molecules have energy more than threshold energy, yet the reaction is slow. Why?

Knowledge Check

  • A reaction has a rate constant k=8.54 xx 10^(-4)M s^(-1) at 45^(@) C and an activation energy, E_(a)=90.8KJ . What is the value of K at 25^(@) C?

    A
    `4.46 xx 10^(-5)M^(-1)s^(-1)`
    B
    `8.54 xx 10^(-5)M^(-1)s^(-1)`
    C
    `4.46 xx 10^(-4)M^(-1)s^(-1)`
    D
    `8.54 xx 10^(-4)M^(-1)s^(-1)`

  • The rate constant of a reaction is 1.5xx10^7 s^(-1) at 50^@C and 4.5xx10^(7)s^(-1) at 100^@C . What is the value of activation energy ?

    A
    `2.2xx10^3" J mol"^(-1)`
    B
    `2300 "J mol"^(-1)`
    C
    `2.2xx10^4" J mol"^(-1)`
    D
    `220 " J mol"^(-1)`

    • Similar Questions

    Explore conceptually related problems

    The decomposition of A into product has value of k as 4.5xx10^(3)s^(-1) at 10^(@)C and energy of activation of 60kJmol^(-1) . At what temperature would k be 1.5xx10^(4)s^(-1)?

    The decomposition of A into product has value of K as 4.5 xx 10^(3) sec^(-1) at 10^@ C and energy of activation 60 kJ/mol. At what temperature would K be 1.5 xx 10^4 "sec"^(-1) .

    (a) Illustrate graphically the effect of catalyst on activation energy. (b) Catalysts have no effect on the equilibrium constant. Why? (c ) The decomposition of A into product has value of k as 4.5xx10^(3)s^(-1) at 10^(@)C and activation energy is 60 kJ mol^(-1) . Calculate the temperature at which the value of kW(J) be 1.5xx10^(4)s^(-1) .

    (a) Illustrate graphically the effect of catalyst on activation energy. (b) Catalysts have no effect on the equilibrium constant. Why ? ( c ) The decomposition of A into product has value of k as 4.5xx10^(3)s^(-1)" at "10^(@)C and activation energy is 60" kJ mol"^(-1) . Calculate the temperature at which the value of k will be 1.5xx10^(4)s^(-1).

    The first order gaseous decomposition of N_(2)O_(4) in to NO_(2) has a rate constant value k=4.5xx10^(3)s^(-1) at 1^(@)C and energy of acitivation 58kJmol^(-1) . What should be the rise in temperature in ("^(@)C) to get the value of K=1.00xx10^(4)s^(-1) . (log2.2=0.3424)

    The rate constant for the decomposition of hydrocarbons is 2.418xx10^(-5)s^(-1) at 546K . If the energy of activation is 179.9kJ mol^(-1) , what will be the value of pre - exponential factor?

    Home
    Profile