The enthalpy change at `298 K` of the reaction `H_(2)O_(2)(l)toH_(2)O(l)+1//2O_(2)(g)` is `-23.5 kcal mol^(-1)` and enthalpy of formation of `H_(2)O_(

Question

The enthalpy change at `298 K` of the reaction `H_(2)O_(2)(l)toH_(2)O(l)+1//2O_(2)(g)` is `-23.5 kcal mol^(-1)` and enthalpy of formation of `H_(2)O_(

The enthalpy change at `298 K` of the reaction
`H_(2)O_(2)(l)toH_(2)O(l)+1//2O_(2)(g)` is `-23.5 kcal mol^(-1)` and enthalpy of formation of `H_(2)O_(2)(l)` is `-44.8 kcal mol^(-1)`. The enthalpy of formation of `H_(2)O(l)` is
A. `-68.3 kcal mol^(-1)`
B. `68.3 kcal mol^(-1)`
C. `-91.8 kcal mol^(-1)`
D. `91.8 kcal mol^(-1)`

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

Correct Answer - A
Given
`(i) H_(2)O_(2)(l)toH_(2)O(l)+(1)/(2)O_(2)(g)` , `DeltaH=-23.5 kcal mol^(-1)`
`(ii) H_(2)(g)+O_(2)(g)to H_(2)O_(2)(l)` : `DeltaH=-44.8 kcal mol^(-1)`
Write eqn. `(ii)` in reverse direction , we get
`(iii) H_(2)O_(2)(l)to H_(2)(g)+O_(2)(g)`, `DeltaH=+44.8 kcal mol^(-1)`
Subtract eqn. `(iii)` from `(i)`, we get
`H_(2)(g)+(1)/(2)O_(2)(g)toH_(2)O(l)`, `DeltaH=-68.4 kcal mol^(-1)`