Question

Using the following thermochemical data, calculate ΔHf° of Cr2O3(s).

2CrCl3(s) + 3H2O(l) → Cr2O3(s) + 6HCl(g) ΔH° = 276.9 kJ/mol

2Cr(s) + 3Cl2(g) → 2CrCl3(s) ΔH° = -1113.0 kJ/mol

4HCl(g) + O2(g) → 2Cl2(g) + 2H2O(l) ΔH° = -202.4 kJ/mol

Answer 1

Answer - We are given the reaction with ∆H°rxn for each as follow –

2CrCl₃(s) + 3H₂O(l) ----> Cr₂O₃(s) + 6HCl(g) ΔH° = 276.9 kJ/mol ……..1

2Cr(s) + 3Cl₂(g) ----> 2CrCl₃(s) ΔH° = -1113.0 kJ/mol……………….2

4HCl(g) + O₂(g) ----> 2Cl₂(g) + 2H₂O(l) ΔH° = -202.4 kJ/mol………….3

Now we need to added all three reactions in such way that we remaining the reaction of only formation of Cr₂O₃(s).

We need the reaction of formation of Cr₂O₃(s) is

4Cr(s) + 3 O₂(g) ----> 2 Cr₂O₃(s)

So we need 4 Cr and 3 O₂ in reactant side and 2 Cr₂O₃(s) in the product side

So we need to multiply equation number 1 by 2 , for reaction 2 by 2 and for reaction 3 we need to multiply with 3. Then added all these three reaction

4CrCl₃(s) + 6H₂O(l) ----> 2 Cr₂O₃(s) + 12 HCl(g) ΔH° = 553.8 kJ/mol ……..4

4 Cr(s) + 6Cl₂(g) ----> 4 CrCl₃(s) ΔH° = -2226.0 kJ/mol……………….5

12 HCl(g) + 3 O₂(g) ----> 6Cl₂(g) + 6H₂O(l) ΔH° = -607.2 kJ/mol………….6

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4Cr(s) + 3 O₂(g) ----> 2 Cr₂O₃(s) ΔH° = -2279.4 kJ/mol

Now this is for 2 moles of Cr₂O₃(s), so for Cr₂O₃(s) ΔHf° = -2279.4 kJ/ 2 mol

                                                                                              = -1139.7 kJ/mol