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.
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!
4 answers
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
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
ohhh it was no effect and no effect for 1 and 3 i got it
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.