Assume that you measured ?Hrxn= -1290 kJ for the reaction: 2 HCN(g) + 5/2 O2(g) à2...

Assume that you measured ?H_rxn= -1290 kJ for the reaction:

2 HCN(g) + 5/2 O₂(g) à2 CO₂(g) + N₂(g) + H₂O(l)

Use your measured ?H_rxnwith literature values of ?H_f(CO₂(g)) = -394 kJ/mol

and ?H_f(H₂O(l)) = -286 kJ/mol to calculate the experimental ?H_ffor HCN.

Expert Solution

Given:

?Ho rxn = -1290.0 KJ/mol

Hof(O2(g)) = 0.0 KJ/mol

Hof(CO2(g)) = -394.0 KJ/mol

Hof(H2O(l)) = -286.0 KJ/mol

Balanced chemical equation is:

2 HCN(g) + 5 O2(g) ---> 2 CO2(g) + N2 (g) + H2O(l)

?Ho rxn = 2*Hof(CO2(g)) + 1*Hof(N2(g)) + 1*Hof(H2O(l)) - 2*Hof( HCN(g)) - 5*Hof(O2(g))

-1290.0 = 2*(-394.0) + 1*(0) + 1*(-286.0) - 2*Hof(HCN(g)) - 5*(0.0)

Hof(HCN(g)) = 108 KJ/mol

Answer: 108 KJ/mol

queen_honey_blossom answered 2 years ago

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