Question

# Using the following thermochemical data, calculate ΔHf° of Cr2O3(s). 2CrCl3(s) + 3H2O(l) → Cr2O3(s) + 6HCl(g)...

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 - 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

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