Question

show me step by step calculations on enthalpy of formation of hydrogen peroxide

Answer 1

Enthalpy of formation of liquid hydrogen peroxide, H₂O₂(l), at 25°C

we can use the following thermochemical equations :

H2(g) + 1/2 O₂(g) ===> H₂O(g)     ; ΔΗ= -241.82 kJ/mol    ---- i

2H(g) + O(g) ===> H2O(g)     ; ΔΗ= -926.92 kJ/mol    ..... ii

2H(g) + 2O(g) ===> H₂O₂(g)     ; ΔΗ= -1070.60 kJ/mol     ..... iii

2O (g) ===> O₂(g) ; ΔΗ = -498.34 kJ/mol                    ........ iv

H₂O₂(l) ===> H₂O₂(g) ; ΔΗ = 51.46 kJ/mol               ........ v

=== from (iii) - (ii)

H₂O(g) + O(g) ===> H₂O₂(g)     ; ΔΗ= - 143.68 kJ/mol    ..... vi

(vi) - 1/2( iv)

H₂O(g) + 1/2 O₂(g) ===> H₂O₂(g)     ; ΔΗ= 105.49 kJ/mol    .... vii

(vii) - (v)

H₂O(g) + 1/2 O₂(g) ===> H₂O₂(l)     ; ΔΗ= 54.3 kJ/mol    .... viii

(viii) + (i)

H₂O(g) + 1/2 O₂(g) ===> H₂O₂(l)     ; ΔΗ= 54.3 kJ/mol

+H2(g) + 1/2 O₂(g) ===> H₂O(g)     ; ΔΗ= -241.82 kJ/mol  

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H₂(g) + O₂(g) ===>H₂O₂(l)     ; ΔΗ= - 187.79 kJ/mol