For the equilibrium: 2 H₂S(g) → 2H2(g) + S₂(g); Kc = 4.5 x 10³, at a certain instant, what will happen to the reaction when 0.010 mol of H₂S, 1.0 mol of H₂ and 1.5 mols of S2 were all added to a 1-liter reaction vessel? Nothing, the system is at equilibrium More hydrogen gas will be formed More sulfur gas will be formed The amount of hydrogen formed will be half the amount of sulfur formed. More hydrogen sulfide will be formed.

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For the equilibrium: 2 H₂S(g) → 2H2(g) + S₂(g); Kc = 4.5 x 10³, at a certain instant,
what will happen to the reaction when 0.010 mol of H₂S, 1.0 mol of H₂ and 1.5 mols
of S₂ were all added to a 1-liter reaction vessel?
Nothing, the system is at equilibrium
More hydrogen gas will be formed
More sulfur gas will be formed
The amount of hydrogen formed will be half the amount of sulfur formed.
More hydrogen sulfide will be formed.
Transcribed Image Text:For the equilibrium: 2 H₂S(g) → 2H2(g) + S₂(g); Kc = 4.5 x 10³, at a certain instant, what will happen to the reaction when 0.010 mol of H₂S, 1.0 mol of H₂ and 1.5 mols of S₂ were all added to a 1-liter reaction vessel? Nothing, the system is at equilibrium More hydrogen gas will be formed More sulfur gas will be formed The amount of hydrogen formed will be half the amount of sulfur formed. More hydrogen sulfide will be formed.
Expert Solution
Step 1

•Here Kc is given.

•To find out direction of equilibrium we have to calculate reaction quotient (Q) at certain instant.

If =>  Q>Kc:- Than reaction goes backwards

          Q<Kc :- Than reaction goes forward 

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