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Calculate the standard enthalpy change (...

Calculate the standard enthalpy change (in kJ `mol^(-1)`) for the reaction
`H_(2)(g)+O_(2)(g) rarr H_(2)O_(2)(g)`
Given that bond enthalpies of `H-H, O=O`, O-H and O-O (in Kj `mol^(-1)`) are respectively 438, 498 , 464 and 138.

A

`-130`

B

`-65`

C

`+130`

D

`-334`

Text Solution

Verified by Experts

The correct Answer is:
A
`H_(2)(g) + O_(2)(g) rarr H_(2)O_(2)(g)`
`Delta H_("reaction") = BE_("reactants") - BE_("products")`
`= [BE(H-H) + BE(O=O)]`
`- 2BE(O-H) + BE(O-O)]`
`= [438 + 498] - [2 xx 464 + 138]`
`= 936 - 1066 = - 130 kJ mol^(-1)`
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