Question

For the reactions system 2H2(g) + S2(g) 2H2S(g), a 1.00-liter vessel is found to contain 0.5 mole of H2 , 0.02 mole of S2, and 68.5 moles of H2S. What is the equilibrium constant expression? What are the chemical formulas, not numbers. (Enter subscripts after the letters: for example, H2O = Hs2O. Also, don't forget to use the proper chemical shorthand for chemical symbols. For instance, chlorine = Cl but not cl or cL.) Please and thank you!! :)

Answer 1

The equilibrium constant expression for the reaction 2H₂(g) + S₂(g) → 2H₂S(g) is Keq = [H₂S]2 / ([H₂]2 × [S₂]), where the brackets indicate equilibrium concentrations of the gases in molarity.

Step-by-step explanation:

The equilibrium constant expression for a reaction system describes how the concentrations of reactants and products relate to each other when the reaction is at equilibrium. For the reaction 2H₂(g) + S₂(g) → 2H₂S(g), the equilibrium constant expression, Keq, can be written as:

Keq = [H₂S]2 / ([H₂]2 × [S₂])

In this expression, [H₂], [S₂], and [H₂S] represent the equilibrium concentrations of hydrogen, sulfur, and hydrogen sulfide gases, respectively. Given that the reaction takes place in a 1.00-liter vessel, the equilibrium constant can be directly related to the molarity of the gas species. In the scenario provided, you use the molarities 0.5 M for H₂, 0.02 M for S₂, and 68.5 M for H₂S.

Answer 2

First you calculate the molarity of all components in equilibrium :

Volume = 1.00 liter

For H₂ :

M = n / V

M = 0.5 / 1.00

M = 0,5 mol/L^-1

For S₂ :

M = 0.02 / 1.00

M = 0.02 mol/L^-1

For H₂S :

M = 68.5 / 1.00

M = 68.5 mol/L^-1

the equilibrium constant expression :

Kc = products / reagents

Reagents products
↓ ↓
2 H₂(g) + S₂(g) = 2 H₂S(g)

Kc = [ H₂S ]^2 / [ H₂ ]^2 * [ S2 ]

Kc = [ 68.5 ]^2 / [ 0.5 ]^2 * [ 0.02 ]

Kc = [ 4692.25] / [ 0.25 ] * [ 0.02 ]

kc = [ 4692.25] / [ 0.005 ]

kc = 988450

hope this helps!