1.

Heat of formation of `H_(2)O` is -188kJ/mol and `H_(2)O_(2)` is -286 kJ/mol. The enthalpy change for the reaction, `2H_(2)O_(2) to 2H_(2)O+O_(2)`A. 196 kJB. `-196kJ`C. `984kJ`D. `-984kJ`

Answer» Correct Answer - A
`H_(2)+(1)/(2)O_(2)toH_(2)O,DeltaH=-188" kJ "mol^(-1)` . . (i)
`H_(2)+O_(2) to H_(2)O_(2) to H_(2)O_(2), DeltaH=-286" kJ "mol^(-1)` . . . (ii)
Multiply (i) and (ii) by 2 and subtracting, we get
`2H_(2)O_(2)to 2H_(2)O +O_(2),DeltaH_(r)=+196kJ`.


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