Calculate ∆h0 for the following reaction. br2(
g. → br2(l)∆hf0 br2(
g. =30.91 kj/mol; ∆hf0 br2 (l) =0 kj/mol)

From the balance chemical reaction, it can be seen that a gaseous bromine is converted to liquid bromine. The heat of reaction is equal to the heat of the product minus the heat of the reactant. Following this, the Δhf0 is equal to,
                                       Δhf0 = (0 kJ/mol) - (30.91 kJ/mol)
                                            Δhf0 = -30.91 kJ/mol

Is there a diagram/picture you can provide? it seems this question goes along with a

Bis the answerhope it !