Asked by emma
What happens to the following reaction at equilibrium if the pressure is decreased?
2H"subtext"2(g) + O"subtext"2(g) 2H"subtext"2O(g)
2H"subtext"2(g) + O"subtext"2(g) 2H"subtext"2O(g)
Answers
Answered by
DrBob222
You could save a lot of time by using this:
2H2 + O2 ==> 2H2O. Everyone knows that the number two following H and O are subscripts. We CAN write them as subscripts but no one wants to take the time to do it.
2H<sub>2</sub> + O<sub>2</sub> ===> 2H<sub>2</sub>O
Here is the rule to follow:
An increase in pressure for a gaseous system in equilibrium causes the equilibrium to shift in the direction of FEWER moles. Decrease in P does the opposite.
2H2 + O2 ==> 2H2O. Everyone knows that the number two following H and O are subscripts. We CAN write them as subscripts but no one wants to take the time to do it.
2H<sub>2</sub> + O<sub>2</sub> ===> 2H<sub>2</sub>O
Here is the rule to follow:
An increase in pressure for a gaseous system in equilibrium causes the equilibrium to shift in the direction of FEWER moles. Decrease in P does the opposite.
Answered by
emma
so it shifts to the right because Q < K correct?
Answered by
DrBob222
I don't think so.
H2 + O2 ==> 2H2O
If P increases, it shifts to side with fewer moles of gases. There are 3 mols on the left; two moles on the right so for increase P it shifts to the right. A decrease in P must shift to the left.
H2 + O2 ==> 2H2O
If P increases, it shifts to side with fewer moles of gases. There are 3 mols on the left; two moles on the right so for increase P it shifts to the right. A decrease in P must shift to the left.
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