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  • If the bond energies of H-H, Br-Br and H-Br are 433, 192 and 364 kJ respectively, the <img alt="\Del

If the bond energies of H-H, Br-Br and H-Br are 433, 192 and 364 kJ mol^{-1} respectively, the \Delta H\degree for the reaction

H_{2\left ( g \right )}+Br_{2\left ( g \right )}\rightarrow 2HBr\left ( g \right )\, \, is

Option: 1

-261 kJ


Option: 2

+103 kJ


Option: 3

+261 kJ


Option: 4

-103 kJ


Answers (1)

best_answer

H_{2}\left ( g \right )\, +\, Br_{2}\left ( g \right )\, \rightarrow 2HBr

\Delta H^{0}\dot{}\, reaction\, =\, \left ( B.E \right )\, reactant\, -\left ( B.E \right )\, product

                                =\left [ 433+192 \right ]\, -\left [ 2\times 364 \right ]

                                 =-103\, KJ

                    Option (4) is correct answer   

Posted by

jitender.kumar

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