At a certain temperature the reaction CO₂(g) + H₂(g) ⇄ CO(g) + H₂O(g) ₂
Has K_c = 2.50. If 2.00 mol of carbon dioxide and 1.5 mol of hydrogen are placed in a 5.00 L vessel and equilibrium is established, what will be the concentration of carbon monoxide?

Question

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The balanced equation for the reaction is: CO₂(g) + H₂(g) ⇄ CO(g) + H₂O(g)₂ The equilibrium constant expression is: K_c = [CO][H₂O] / [CO₂][H₂] We are given the value of K_c and the initial concentrations of the reactants, which are: [CO₂] = 2.00 mol / 5.00 L = 0.400 M [H₂] = 1.5 mol / 5.00 L = 0.300 M Since the reaction stoichiometry is 1:1 for CO and CO₂, at equilibrium, the concentration of CO will be equal to the concentration of CO₂ that reacted, which is (0.400 - x) M, where x is the equilibrium concentration of H₂O. We can use an ICE table to determine the equilibrium concentration of H₂O: | | CO₂ | H₂ | CO | H₂O | |--------|-----------------|-----------------|---------|----------------| | Initial | 0.400 M | 0.300 M | 0 | 0 | | Change | -x | -x | +x | +x | | Equil. | 0.400 - x M | 0.300 - x M | x M | x M | Substituting the equilibrium concentrations into the equilibrium constant expression, we get: 2.50 = x^2 / ((0.400 - x)(0.300 - x)) Solving for x gives: x = 0.209 M Therefore, the equilibrium concentration of CO is: [CO] = 0.400 - x = 0.191 M The correct answer is (C).

At a Certain Temperature the Reaction CO2(g) + H2(g) ⇄

At a Certain Temperature the Reaction CO₂(g) + H₂(g) ⇄