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The activation energy for the reaction ...

The activation energy for the reaction `:` ltbr. `2Hl(g) rarr H_(2)(g)+I_(2)(g)`
is `209.5 kJ mol^(-1)` at `581K`. Calculate the fraction of molecules of reactants having energy equal to or greater than activation energy ?

A

`1.82xx10^(-18)`

B

`1.47xx10^(19)`

C

`2.67xx10^(16)`

D

`3.89xx10^(19)`

Text Solution

Verified by Experts

The correct Answer is:
B
Fraction of molecules (x) having energy equal to or greater than activation energy may be calculated as follows:
`x=n//N=e^(-Ea)"//RT"`
In `x=-E_(a)/(RT)or logx=-E_(a)/(2.303RT)`
or `logx=-(209xx10^(3)J" mol"^(-1))/(2.303xx(8.314Jk^(-1)" mol"^(-1))xx581k)`
`=-18.8323`
`x=Antilog(-18.8323)`
`=1.471xx10^(-19)`
Fraction of molecules `=1.47xx10^(-19`
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The activation energy for the reaction 2HI (g) to H_(2)(g) +I_(2)(g) , is 209.5 kJ "mol"^(-1) at 581 K. Calculate the fraction of molecules of reactants hving energy equal to or greater than activation energy.

The activation energy for the reaction, 2Hi(g) to H_(2)(g) + I_(2)(g) is 209.5 kJ moli^(-1) at 581 K. Calculate the fraction of molecules of reactants having energy equal to or greater than activation energy.

Knowledge Check

  • The activation energy for the reaction - H_(2)O_(2)toH_(2)O+(1)/(2)O_(2) is 18 K cal//mol at 300 K. calculate the fraction of molecules of reactonts having energy equal to or greater than activation energy ? Anti log (-13.02)=9.36xx10^(-14)

    A
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    B
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    C
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    D
    `5.2xx10^(-15)`

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