Question

Find ΔH0 , ΔS0 , and ΔG0 for the reaction at 298 K.

CH4 (g) + NH3 (g) à HCN (g) + 3 H2 (g)

Answer 1

Given reaction,

CH₄ (g) + NH₃ (g) ------------> HCN (g) + 3 H₂ (g)

Using thermodynamic table of standard enthalpies, entropies of formation of species concenrned, Standard enthapy change, entropy and then using Gibbs equation Standard Gibbs energy change is calculated,

1) Standard enthalpy of reaction (H^orxn)

H^orxn = Hf^o(Products) -  Hf^o(Ractant)

H^orxn = [Hf^o(HCN) + 3 x Hf^o(H₂)] - [Hf^o(NH3) + Hf^o(CH₄)]

H^orxn = [(+130.5) + 3 x (0)] - [(-46.19) + (-74.85)]

H^orxn = 251.54 kJ/mol

H^orxn = 251540 J/mol

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2)

Standard entropy change of reaction (S^orxn)

S^orxn = Sf^o(Products) -   Sf^o(Ractant)

S^orxn = [Sf^o(HCN) + 3 x Sf^o(H₂)] - [Sf^o(NH3) + Sf^o(CH₄)]

H^orxn = [(+113.01) + 3 x (130.6)] - [(192.3) + (186.2)]

H^orxn = 126.31 J/K.mol.

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3) Standard Gibbs Energy change (G^orxn)

Gibb's equation states,

G^orxn = H^orxn - TS^orxn

Using above calculated values,

G^orxn =  251540 - 298 x 126.31

G^orxn = 213899.62 J/mol

G^orxn = 213.89962 kJ/mol

G^orxn = 213.9 kJ/mol ................(rounded)