Part B The following reaction was performed in a sealed vessel at 742 °C:
H2(g) +12(g) =
=2HI(g)
Initially, only H, and I were present at concentrations of (H2) = 3.90M and [12] = 2.85 M. The equilibrium
concentration of I, is 0.0100 M. What is the equilibrium constant, Kc, for the reaction at this temperature?
Express your answer numerically.
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It decreases by 4 is your answer mark brainliest 

Answer is: the equilibrium  shifts toward the reactants, to the left side of chemical reaction.it is because lead(ii) nitrate dissociates in water: pb(no₃)₂ → pb²⁺(aq) + 2no₃⁻(aq).according to lechatelier's principle, the position of equilibrium moves to counteract thechange (more lead cations, common ion), the position of equilibrium will move so that the concentration ofproducts decrease and concentration of reactants of chemical reaction increase.read more on -