The reaction: 2 co(l) + o2(g) ⇄2 co2(g) with a h = −25.0 kj/mol, is at equilibrium. which of the following lists three ways this reaction can be shifted to the right? a. remove co2, lower pressure and remove cob. remove co2, raise pressure and add coc. raise temperature, lower pressure and remove o2d. add o2, raise pressure and lower temperaturee. remove co2, increase volume and lower temperature

Option B and D both have 3 ways to shift the reaction to the right

Explanation:

The reaction  2 CO(l) + O2(g) ⇄2 CO2(g) is exothermic because ΔH = -25 kJ < 0 This means heat will be released

Therefore, increasing the temperature will shift the equilibrium to the left, while decreasing the temperature will shift the equilibrium to the right.

A. remove CO2, lower pressure and remove CO

⇒ Lowering the pressure will shift the equilibrium to the side with most moles of gas : On the left side there are 3 moles, on the right side 2 moles. By lowering the pressure, the equilibrium will shift to the left.

B  remove CO2, raise pressure and add CO

⇒ By raising the pressure, the equilibrium will shift to the side with the lesser amount of moles of gas. This is the right side.

⇒ Remove CO2: the equilibrium will shift to the side of CO2 ,so the reaction will shift toward products to replace the product removed. (The right side).

⇒ Add CO: the equilibrium will shift to the side of CO2, so the reaction will shift toward the products side to reduce the added CO. (The right side).

C raise temperature, lower pressure and remove O2

⇒ Increasing the temperature will shift the equilibrium to the left

D add O2, raise pressure and lower temperature

⇒ decreasing the temperature will shift the equilibrium to the right

⇒ By raising the pressure, the equilibrium will shift to the side with the lesser amount of moles of gas. This is the right side.

⇒ Add O2: the equilibrium will shift to the side of CO2, so the reaction will shift toward the products side to reduce the added O2. (The right side).

E  remove CO2, increase volume and lower temperature

⇒ decreasing the temperature will shift the equilibrium to the right

⇒ Remove CO2: the equilibrium will shift to the side of CO2 ,so the reaction will shift toward products to replace the product removed. (The right side).

⇒ Increase volume : with a pressure decrease due to an increase in volume, the side with more moles is more favorable. The equilibrium will shift to the left side.

a 20-cm-long, cylindrical graphite (pure carbon) rod is inserted into an oxidizing atmosphere at 1145 k and 1.013 x 105 pa pressure. the oxidizing process is limited by the diffu- sion of oxygen counterflow to the carbon monoxide that is formed on the cylindrical surface.

a