The reaction A(g)+ 2B(g) C(g)was allowed to come to equilibrium. The initial amounts of reactants placed into a 5.00 L vessel were 1.0 mol A and 1.8 mol B. After the reaction reached equilibrium, 1.0 mol of B was found. Calculate K_c for this reaction.

Question

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To calculate K_c, we need to know the equilibrium concentrations of each species. We can use the stoichiometry of the balanced equation and the given information to set up an ICE (initial, change, equilibrium) table: A(g) + 2B(g) ⇌ C(g) I: [A] = 1.0 mol/5.00 L = 0.20 M [B] = 1.8 mol/5.00 L = 0.36 M [C] = 0 M (no reaction has occurred yet) C: -x -2x +x E: 0.20-x 0.36-2x x We're given that at equilibrium, [B] = 1.0 mol/5.00 L = 0.20 M. So we have: 0.20 = 0.36 - 2x // solve for x x = 0.08 Now we can calculate K_c: K_c = [C]^1/[A]^1[B]^2 = (0.08)/(0.20)(0.20)^2 = 16/1000 = 0.016 Rounding to two significant figures gives the answer of C) 17.

The Reaction A(g)+ 2B(g) C(g)was Allowed to Come