Did I answer this equilibrium question correctly?
2CO(g) + O2(g) <-> 2CO2(g)
1) How will increasing the concentration of CO shift the equilibrium?
a] to the right [i chose this]
b] to the left
c] no effect
2] how will increasing the concentration of CO2 shift the equilibrium?
a] to the right
b] to the left [i chose this]
c] no effect
3] How will decreasing the volume of the container shift the equilibrium?
a] to the right
b] to the left
c] no effect [i chose this]
Please let me know! thank you!
1 and 2 ok.
3 is no.
Decreasing the volume means increasing pressure. When pressure is increased the reaction shifts to the side with smaller number of gaseous moles. You have 3 moles on left and 2 on the right; therefore, it shifts to the right.
ohhh it was no effect and no effect for 1 and 3 i got it
Ohh i understand Dr. Bob. So in terms of adding more.. would that be the same as increasing concentration? For example:
S(s) + O2(g) <-> SO2
1] how will adding more S(s) shift the equilibrium?
a] to the right [this?]
b] to the left
c] no effect
2] How will removing some SO2(g) shift the equilibrium?
a] to the right [this?]
b] to the left
c] no effect
3] How will decrasing the volume of the container shift the equilibrium?
a] to the right [thissss!]
b] to the left
c] no effect
Remember solids, the sulfur in this case, doesn't enter into the K expression. Adding another bit of S or taking a little out won't change the equilibrium. And for the pressure change, you have 1 mole gas on the left and 1 mole on the right so there is no effect.
For question 1, you correctly chose option "a" - to the right. Increasing the concentration of CO will shift the equilibrium to the right, favoring the formation of more CO2. According to Le Chatelier's principle, if you increase the concentration of one reactant (CO in this case), the system will try to counteract the change by favoring the forward reaction to consume more of that reactant.
For question 2, you chose option "b" - to the left. However, increasing the concentration of CO2 will actually have no effect on the equilibrium position. This is because CO2 is a product of the reaction, not a reactant. Changing the concentration of a product does not affect the equilibrium position according to Le Chatelier's principle.
For question 3, you correctly chose option "c" - no effect. Changing the volume of the container has no effect on the equilibrium position in this case. The given reaction does not involve any gases other than CO, O2, and CO2. Changing the volume does not affect the concentrations of these gases, so the equilibrium position remains unchanged.
Overall, you answered the first and third questions correctly.