The enthalpy of the reaction : `H_(2)O_(2)(l)toH_(2)O(l)+(1)/(2)O_(2)(g)` is `-23.5 kcal mol^(-1)` and the enthalpy of formation of `H_(2)O(l)` is `-6

Question

The enthalpy of the reaction : `H_(2)O_(2)(l)toH_(2)O(l)+(1)/(2)O_(2)(g)` is `-23.5 kcal mol^(-1)` and the enthalpy of formation of `H_(2)O(l)` is `-6

The enthalpy of the reaction :
`H_(2)O_(2)(l)toH_(2)O(l)+(1)/(2)O_(2)(g)` is `-23.5 kcal mol^(-1)` and the enthalpy of formation of `H_(2)O(l)` is `-68.3 kcal mol^(-1)`. The enthalpy of formation of `H_(2)O_(2)(l)` is
A. `-44.8 kcal mol^(-1)`
B. `44.8 kcal mol^(-1)`
C. `-91.8 kcal mol^(-1)`
D. `91.8 kcal mol^(-1)`

Monjur Alahi

4 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)+O_(2)toH_(2)O_(2)` is `DeltaH_(f)` of `H_(2)O_(2)`.
To get `H_(f)` for this reaction subtract equation `(i)` from `DeltaH_(f)` of `H_(2)O`.

4 views Answered 2023-02-05 15:29