J.R. S. answered 04/22/22
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2 𝐻𝐵𝑟(𝑔) ==> H2(𝑔) + Br2(𝑙) ... ∆𝐻 = +72.4 kJ/mol
90.19 g Br2 x 1 mol Br2 / 159.8g x 72.4 kJ / mol = 40.9 kJ