The activation energy for the reaction `:` . `2Hl(g) rarr H_(2)(g)+I_(2)(g)` is `209.5 kJ mol^(-1)` at `581K`. Calculate the fraction of molecules of

Question

The activation energy for the reaction `:` . `2Hl(g) rarr H_(2)(g)+I_(2)(g)` is `209.5 kJ mol^(-1)` at `581K`. Calculate the fraction of molecules of

The activation energy for the reaction `:`
. `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 ?

Monjur Alahi

4 views

2025-01-04 04:42

1 Answers

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

Salim Ahmed Kawsar

Correct Answer - `1.471xx10^(-19)`

4 views Answered 2023-02-05 15:29