Question

please help find the equilibrium concentrations of H2 and S2. Kc=1.67*10^-7

11 /19 consider the von for the decomposition of H₂S: 2H₂Scq, 2H, ags + Suces Kr. 1.67x10 3 a 0.5L ron vessel initially contains 0.0125 mol of H₂S @ 800°C. Find the equilibrium concentrations of H, & S.. 0.025 2 H, SO 0.025 M -2x 2 Hoy + Scg ом OM [HS] = mol = 0.0175 mol 0.5L

Answer 1

decomposition neachon Of H2S Kc = 1.67X10 2H sig) - 2+2 (g) +518) = 0.0125 0.025M of ths = no. of moles of H₂S 0.5 Initial concentration Volume = 24 (g) + 5₂ (g). 2H₂5 (g) 0.025 M nitial 2x x final 0.025-2 beheren 2 is in more for ke Writing the equation ke= [52] [tb]² [H₂S]² 1.67X107 = (x) (2x) ² (0.025-2x32 it, Assuming, since 2x is very small with respect to 0.025, neglecting 4x3 1.67X107. (0.025R = 230 1:67X10*x (0.025) 4

x= 0.00029 M at Equilibrium, 240.00029 2 0.00058 M concentration Of Hb = 2x = concentration of S₂ = x= 0.00029 m.
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