Question #d0666

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Answer 1 · 2016-06-04T10:51:10

$- 130 k J m o l^{-1}$ $-130kJmol^"-1"$

$DeltaH_"reaction"=H_"product"-H_"reactant"....(1)$

$DeltaE_"activation"=E_"TS"-H_"reactant"....(2)$

Energy released during bond formation

$DeltaH_"bond formation"=H_"product"-E_"TS"$

Subtracting (2) rom(1) We get

$DeltaH_"reaction"-DeltaE_"activation"=H_"product"- E_"TS"=DeltaH_"bond formation"$

Given

$DeltaH_"reaction"=53kJmol^"-1"$

$DeltaE_"activation"=183kJmol^"-1"$

$:.DeltaH_"bond formation"=(53-183)kJmol^"-1"=-130kJmol^"-1"$