25℃ to oxidize sulfide in natural waters with nitrate

25℃ to oxidize sulfide in natural waters with nitrate

 

 

Is it possible at 25℃ to oxidize sulfide in natural waters with nitrate?

Typical concentrations that exist are 10^(-4) M of reacting species and pH 8.

Assume that ionic strength effects are negligible. The reaction is

H^+ + NO3^- + HS^- + H2O ⇌ SO4^2- + NH4^+

 

------------------------

 

H^+(aq) + NO3^-(aq) + HS^-(aq) + H2O(l)

⇌ SO4^2-(aq) + NH4^+(aq)

 

 

 

ΔG°_rxn = [생성물들의 ΔG°_f 합] – [반응물들의 ΔG°_f 합]

( 참고 https://ywpop.tistory.com/7393 )

 

= [ΔG°_f(SO4^2-) + ΔG°_f(NH4^+)]

– [ΔG°_f(H^+) + ΔG°_f(NO3^-) + ΔG°_f(HS^-) + ΔG°_f(H2O)]

= [(–744.6) + (–79.4)] – [(0) + (–111.3) + (12.1) + (–237.2)]

= –487.6 kJ/mol

( ΔG°_f 값은 Mcmurry 일반화학 6판 부록에서 참고함. )

 

 

 

Q = [SO4^2-] [NH4^+] / [H^+] [NO3^-] [HS^-]

= [(10^(-4)) (10^(-4))] / [(10^(-8)) (10^(-4)) (10^(-4))]

= 10^8

 

 

 

ΔG = ΔG° + RT lnQ

( 참고 https://ywpop.tistory.com/10336 )

 

= (–487600) + (8.314) (273.15 + 25) ln(10^8)

= –441938.46 J/mol

= –441.9 kJ/mol

 

 

 

ΔG < 0 이므로,

자발적 반응.

the oxidation of sulfide by nitrate is

a thermodynamically spontaneous

and thus possible process.

 

 

 

또는

K = e^[–ΔG° / RT]

= e^[–(–487600) / ((8.314) (273.15 + 25))]

= 2.68×10^85

 

 

 

K >>> Q 이므로,

정반응 쪽으로 반응이 일어난다.

( 참고 https://ywpop.tistory.com/10539 )

 

 

 

답: 가능하다. It is possible.

 

 

 

 

[키워드] 자유 에너지 기준, ΔG 계산 기준, standard Gibbs free energy of formation