# Consider the equilibrium system CO(g) + Fe_3O_4(s) ⇌ CO_2(g) + 3FeO(s) How does the equilibrium position shift as a result of each of the following disturbances?

## Left Right Not CO2 is removed by adding solid NaOH. Left Right Not Additional Fe3O4(s) is added to the system. Left Right Not Dry ice is addded at a constant T. 4 Left Right Not The volume is reduced by one-half at a constant T. Left Right Not CO is added.

Jun 6, 2017

1. Right
2. Not
3. Left
4. Not
5. Right

#### Explanation:

1. The equilibrium shifts rightwards (to the products) as the concentration of one of the (gaseous) product decreases.

2. Change in the amount of $F {e}_{3} {O}_{4}$ will not affect the equilibrium as it is in the solid state. (Only change in concentration of gaseous or aqueous reactants affects the equilibrium.)

3. Dry ice (solid carbon dioxide) sublimate readily to produce gaseous $C {O}_{2}$, increasing the concentration of products. (Also, the reaction is endothermic. http://www.chemteam.info/Thermochem/Hess-3eq-prob1-10.html)

4. There's an equal number of gas molecules on both sides of the reaction. Therefore changes in volume have no influence on the equilibrium. (It does speed up reactions of both directions, anyway)

5. Adding $C O \left(g\right)$ will increase the concentration of the reactants.