What is ∆h0 for this reaction?

br2 (g) → br2 (l)

∆hf ° br2 (g) = 30.91 kj/mol

∆H° of reaction is -30.91 KJ/mol.

Explanation:

Given data:

∆Hf ° of Br2 (g) = 30.91 KJ/mol

∆H° of reaction = ?

Solution:

we know that standard heat of formation for Br2 (l) is 0.0 KJ/mol.

Balance chemical equation:

Br2 (g) → Br2 (l)

Formula:

∆H° = ∆Hf ° (product) - ∆Hf °(reactant)

∆H° = 0.0 KJ/mol - 30.91 KJ/mol

∆H° = - 30.91 KJ/mol

the correct answer to this question is 2