At a certain temperature,K_c equals 1.4 × 10² for the reaction: 2 CO(g) + O₂(g) ⇌ 2 CO₂(g) .
If a 5.00-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

Given the equilibrium constant (K) is 1.4 × 10^12 for the reaction 2 CO(g) + O_2(g) ⇌ 2 CO_2(g), and the amounts of CO_2 and O_2 at equilibrium are 0.400 mol and 0.100 mol respectively in a 5.00 L flask, we can calculate the concentration of these substances ([CO_2] = 0.400 mol / 5.00 L = 0.080 M, [O_2] = 0.100 mol / 5.00 L = 0.020 M). Using the equilibrium expression K = [CO_2]^2 / ([CO]^2[O_2]), we can solve for [CO] knowing K and the concentrations of CO_2 and O_2. Rearranging and solving for [CO] gives a concentration that, when multiplied by the volume of the flask (5.00 L), gives the moles of CO present, which is approximately 0.239 mol.

At a Certain Temperature,Kc Equals 1

At a Certain Temperature,K_c Equals 1