At a certain temperature the reaction CO₂(g) + H₂(g) CO(g) + H₂O(g) 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?

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First, we need to write out the balanced chemical equation and the expression for the equilibrium constant, K: CO₂(g) + H₂(g) 11eb16b2_ffc5_451a_984d_ff4375913d9e_TB7799_11 CO(g) + H₂O(g) K_c = [CO][H₂O] / [CO₂][H₂] The initial concentrations are: [CO₂] = 0 mol / 5.00 L = 0 M [H₂] = 1.5 mol / 5.00 L = 0.30 M [CO] = 2.00 mol / 5.00 L = 0.40 M [H₂O] = 0 M (none initially) Let x be the amount of CO and H₂O that reacts to form CO₂ and H₂ at equilibrium. Then the equilibrium concentrations can be written as: [CO₂] = 0 + x [H₂] = 0.30 - x [CO] = 0.40 + x [H₂O] = x Substituting these into the equilibrium constant expression gives: 2.50 = (0.40 + x)(x) / ((0 + x)(0.30 - x)) Simplifying and solving for x gives: x = 0.135 M Therefore, the equilibrium concentrations are: [CO₂] = 0.135 M [H₂] = 0.165 M [CO] = 0.535 M [H₂O] = 0.135 M The concentration of carbon monoxide is 0.535 M, which corresponds to choice C.

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

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