1.

heat of formation ofH_(2)O is -188 kJ // mol and H_(2)O_(2) is -286 kJ // mol . The enthaply change for thereaction2H_(2)O_(2) to 2H_(2)O+O_(2) is

Answer»

196 kJ
`-196 kJ`
984 kJ
`-984 kj`

Solution :`H_(2)+1/2 O_(2)to H_(2)O,DeltaH=-188kJ mol^(-1)`
`H_(2) + O_(2) toH_(2)O_(2) , DeltaH =286 kJ mol ^(-1)`
multiply Eqs.(i) and (ii) by 2
`2H_(2)+O_(2) to 2H_(2)O, DeltaH=-376 kJ mol^(-1)`
`2H_(2)+2O_(2) to 2H_(2)O_(2) , DeltaH =- 572 kJ mol^(-1)`
Eq .(iii) - Eq.(iv)
`2H_(2)O_(2) to 2H_(2)O + O_(2), DeltaH_(r) =+196 kJ`


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