Please explain A mixture of 0.50 moles of H_2S, 0.75 moles of H_2, and 0.85 mole

Question

Please explain

A mixture of 0.50 moles of H_2S, 0.75 moles of H_2, and 0.85 moles of S_2 are placed in a 10.0 L container. Which of the following statements is true if the following reaction has a K_c value of K_c = 4.0 times 10^-8 at a temperature of 750 degree C? 2 H_2S (g) doubleheadarrow 2 H_2 (g) S_2 (g) The reaction shifts to the right to establish equilibrium The concentrations of the products will decrease, while the concentration of the reactants will increase to establish equilibrium The reaction is already at equilibrium The concentrations of H_2S and H_2 will decrease since they are larger than the concentration of S_2 The value of K will change as the reaction proceeds toward equilibrium

Explanation / Answer

[H2S] = number of mol / volume

= 0.50 mol / 10.0 L

= 0.050 M

[H2] = number of mol / volume

= 0.75 mol / 10.0 L

= 0.075 M

[S2] = number of mol / volume

= 0.85 mol / 10.0 L

= 0.085 M

Qc = [H2]^2 [S2] / [H2S]^2

   = (0.075)^2 * (0.085) / (0.050)^2

   = 0.191

Since Kc< QC, the reaction will move to reactant side to reestablish equilibrium

A) is wrong since reaction move to left

B) True since reaction move to left

C) false

D) False. Concentration of H2 and S2 will decrease

E) False. Value of K is constant

Answer: B

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