For the reactions system $2H_2(g) + S_2(g) rightleftharpoons 2H_2S(g)$ , a 1.00 liter vessel is found to contain 0.50 moles of $H_2$ 0.020 moles of $S_2$ , and 68,5 moles of $H_2S$ . How do you calculate the numerical value of the Keq of this system?

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Answer 1 · 2016-09-20T04:35:42

You have the numerical values to calculate the equilibrium.

$2H_2(g) + S_2(g) rightleftharpoons 2H_2S(g)$

$K_(eq)$ $=$ $([H_2S]^2)/([H_2]^2[S_2])$ $=$ $((68.5*mol*L^-1)^2)/((0.50*mol*L^-1)^2(0.020*mol*L^-1))$

$=$ $"A large number"$

I don't believe this. Disulfur is an exceptionally high energy molecule, and I do not believe that these figures are authentic.

For the reactions system 2H_2(g) + S_2(g) rightleftharpoons 2H_2S(g)2H_2(g) + S_2(g) rightleftharpoons 2H_2S(g), a 1.00 liter vessel is found to contain 0.50 moles of H_2H_2 0.020 moles of S_2S_2, and 68,5 moles of H_2SH_2S. How do you calculate the numerical value of the Keq of this system?