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)`

Question

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)`

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 kJ
B. `-196kJ`
C. `984kJ`
D. `-984kJ`

Monjur Alahi

5 views

2025-01-04 04:42

1 Answers

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Salim Ahmed Kawsar · Accepted Answer · 2023-02-05T15:29:48+06:00

Salim Ahmed Kawsar

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`.

5 views Answered 2023-02-05 15:29