1. Tardigrade
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  3. Chemistry
  4. Calculate the standard enthalpy change (in kJ mol-1 ) for the reaction H2(g)+O2(g) xrightarrowH2O2(g) Given that bond enthalpies of H-H, text O=O, OH and OO (in kJ mol-1 ) are respectively 438, 498, 464 and 138.

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Q. Calculate the standard enthalpy change (in $kJ\, mo{{l}^{-1}} $ ) for the reaction
$ {{H}_{2}}(g)+{{O}_{2}}(g)\xrightarrow{{}}{{H}_{2}}{{O}_{2}}(g) $
Given that bond enthalpies of $ H-H,\text{ O}=O, $ $ OH $ and $ OO $ (in kJ $ mo{{l}^{-1}} $ ) are respectively $438, 498, 464$ and $138$.

KEAMKEAM 2010Thermodynamics

Solution:

$ {{H}_{2}}(g)+{{O}_{2}}(g)\xrightarrow{{}}{{H}_{2}}{{O}_{2}}(g) $
$ \Delta {{H}_{reaction}}=B{{E}_{reac\tan ts}}-B{{E}_{products}} $
$ =[BE(H-H)+BE(O=O)] $
$ -2BE(O-H)+BE(O-O)] $
$ =[438+498]-[2\times 464+138] $
$ =936-1066=-130\text{ }kJ\text{ }mo{{l}^{-1}} $