At a certain temperature,K_c equals 1.4 × 10² for the reaction: 2 CO(g) + O₂(g) ⇌ 2 CO₂(g) .
If a 3.50-L flask contains 0.400 mol of CO₂ and 0.100 mol of O₂ at equilibrium,how many moles of CO are also present in the flask?

Question

Quizplus · Accepted Answer

From the given equilibrium constant, we can write the expression for the equilibrium constant in terms of the concentrations of the reactants and products: K = [CO₂]2 /[CO][O₂] At equilibrium, let the concentration of CO be x, then we have: [CO₂] = 0.400 mol / 3.50 L = 0.114 mol/L [O₂] = 0.100 mol / 3.50 L = 0.0286 mol/L Substituting these concentrations and the given equilibrium constant into the equilibrium expression, we get: 1.4 × 10² = (0.114 mol/L)2 / (x)(0.0286 mol/L) Solving for x, we get: x = 0.200 mol/L Therefore, the number of moles of CO present in the 3.50-L flask is: n(CO) = (0.200 mol/L)(3.50 L) = 0.700 mol Thus, the answer is B (0.200 mol)

At a Certain Temperature,Kc Equals 1

At a Certain Temperature,K_c Equals 1