At high temperatures, carbon reacts with O₂ to produce CO as follows: C(s) + O₂(g) ⇄ 2CO(g) .
When 0.350 mol of O₂ and excess carbon were placed in a 5.00-L container and heated, the equilibrium concentration of CO was found to be 0.060 M. What is the equilibrium constant, K_c, for this reaction?

Question

Quizplus · Accepted Answer

We can use the equilibrium expression to calculate the equilibrium constant: K_c = ([CO]2)/([C][O₂]) We are given that the equilibrium concentration of CO is 0.060 M. We also know that the initial concentration of O₂ is 0.350 mol/5.00 L = 0.070 M, since the volume does not change during the reaction. However, we do not know the initial concentration of carbon, since we are told that it is present in excess. We can use the stoichiometry of the reaction to calculate the amount of carbon that reacted: 0.350 mol O₂ x (2 mol CO/1 mol O₂) x (1 mol C/1 mol CO) = 0.700 mol C Since the volume is 5.00 L, the initial concentration of carbon is: 0.700 mol/5.00 L = 0.140 M Substituting these values into the equilibrium expression, we get: K_c = (0.060 M)2 / [(0.140 M)(0.070 M)] = 0.090 Therefore, the equilibrium constant is 0.090, which corresponds to choice (C).

At High Temperatures, Carbon Reacts with O2 to Produce CO

At High Temperatures, Carbon Reacts with O₂ to Produce CO