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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` |
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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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