Calculate ΔG∘rxn at 298 K for the following reaction:
I2(g)+Br2(g)⇌2IBr(g)Kp=436

ΔG°rxn = -15,1 kJ/mol

Explanation:

ΔG°rxn = -RTlnK = -(8.314 J/mol K)(298 K)×ln436

ΔG°rxn = -1.51×10^4 J/mol = -15.1 kJ/mol

an exergonic reaction, it makes sense, since K > 1

there are no statements provided but here is why:

sulfur belongs to group 16 because it can be found in nature whether it is in a free or combined state.

hope this

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