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If the end energies of H-H, Br-Br and H-...

If the end energies of H-H, Br-Br and H-Br are 433, 192 and 364 kJ `mol^(-1)` respectively, then `DeltaH^(@)` for the reaction, `H_(2)(g)+Br_(2)(g)to2HB r(g)` is

A

`+ 261` kJ

B

`- 103` kJ

C

`- 261` kJ

D

`+ 103` kJ

Text Solution

Verified by Experts

The correct Answer is:
B
`H-H+Br-Brrarr2H-Br`
`{:(433+192,2xx364),(625,728):}`
Energy absorbed = Energy released
Not energy released = 728-625=103 kJ
i.e., = `DeltaH=-103 kJ`.
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Knowledge Check

  • If the end energies of H-H, Br-Br and H-Br are 433, 1992 and 364 kJ mol^(-1) respectively, then DeltaH^(@) for the reaction, H_(2)(g)+Br_(2)(g)to2HB r(g) is

    A
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    B
    `+103kJ`
    C
    `+261kJ`
    D
    `-103kJ`

  • If the bond energies of H-H, Br- Br and H -Br are 433, 192 and 364kJ mol^(-1) respectively, DeltaH^(@) for the reaction H_(2)(g) + Br_(2)(g) rarr 2HBr(g) is

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    `- 261kJ`
    B
    `+103kJ`
    C
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  • The bond energies of H--H , Br--Br and H--Br are 433,, 192 and 364KJmol^(-1) respectively. The DeltaH^(@) for the reaction H_(2)(g)+Br_(2)(g)rarr2HBr(g) is

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    B
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    D
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