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Question

For the reaction below:

a PBr5(g) + b O2(g)  c P2O5(g) + d BrO2(g)

Substance ΔH°f (kJ/mol) S° (J/mol*K) ΔG°f (kJ/mol)

PBr5(g) -219.28 +115.19 -251.374

O2(g) 0 +202.17 0

P2O5(g). -358.27 +261.93 -436.118

BrO2(g) +16.94 +208.59 -48.817

a. What is the value of ΔG°rxn for this reaction at 25.00°C?

b. Is the reaction spontaneous (exergonic) or non-spontaneous (endergonic)?

c. How many kilojoules of energy are transferred when 21.375g of P2O5(g) is produced?

Answer 1

Balance the equation

4 PBr₅(g) + 25 O₂(g) ==> 2 P₂O₅(g) + 20 BrO₂(g) ... balanced equation

Then find the sum of ∆Gº for the products and the reactants, and take the difference.

∑products - ∑reactant

∑products = 2 x -436.118 + 20 x -48.817

∑reactants = 4 x -251.374 + 25 x 0

Do the math and get the answer to part (a)

(b). If ∆Gº is negative, the reaction is spontaneous, otherwise not

(c). Convert 21.375 g P₂O₅ to moles and then multiply that value by the answer in part (a) and divide by 2 since the answer in part (a) is for 2 mols P₂O₅