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At standard conditions, if the change in...

At standard conditions, if the change in the enthalpy for the following reaction is -109KJ`mol^(-1)`
`H_2(g) + Br_2(g) rarr 2HBr(g)`
Given that bond energy of `H_2 and Br_2` is 435kj`mol^(-1)` and 192kj`mol^(-1)`, respectively. What is the bond energy of HBr ?

A

368

B

736

C

518

D

259

Text Solution

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The correct Answer is:
A
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Calculate the enthalpy change for the reaction: H_(2) (g) + Br_(2) (g) rarr 2HBr(g) Given the bond enthalpies H_(2), Br_(2) and HBr are 435 kJ mol^(-1), 192 kJ mol^(-1) and 368 kJ mol^(-1) respectively.

Calculate the enthalpy change for the reaction . H_(2)(g) + Br_(2)(g) to 2HBr (g) Given that the bond energies of H-H, H-Br, and Br-Br are 435 kJ mol^(-1) , 364 kJ mol^(-1) , and 192 kJ mol^(-1) , respectively.

Knowledge Check

  • What will be the enthalpy change for the following reaction? H_2(g)+Br_2(g) rarr 2HBr(g) given that bond energy of H_2 ,Br_2 and HBr is 435kj/mol , 192 kj/mol , and 368 kj/mol respectively.

    A
    `(+109KJ/mol)`
    B
    (-109 Kj/mol)
    C
    (+493kj/mol)
    D
    (-493kJ/mol)

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