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How do I find the equilibrium concentrations for the following equation?X2(g)+3Y2(g)...
How do I find the equilibrium concentrations for the following equation?
X2(g)+3Y2(g) reacts to form 2XY3(g). X2 and Y2 = 2moles. They are added to a 1.0 L container. The Y2 equilibrium concentration was .8 mol. I understnd what my initial concentrations are and that as the concentration of the reactants goes down the conctration () of the product must go up. What I don't understand is how to use the ration of say X2:Y2 (1/3) to find the change in  for X2.
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X2 + 3Y2 ------>2XY3
The reaction is in gas phase.
The initial amount of X2 and Y2 are 2 mols. They are put into a 1.0 L container. Therefore their initial concentrations are 2.0 mol/L.
The equillibrium concentration of Y2 is 0.8 mol/L. Therefore the reacted amount of Y2 at the equlibrium is,
= 2.0 - 0.8
= 1.2 mol/L
According to stoichiometry of the equation,
1 mol of X2 reacts with 3 mols of Y2.
X2:Y2 = 1:3
Therefore, 1 mol of Y2 reacts with 1/3 mols of X2.
X2:Y2 = 1/3:1
Therefore amount of X2 reacted = 1/3 x (amount of Y2 reacted)
= 1/3 x 1.2 mol/L
= 0.4 mol/L
Therefore X2 concentration left in container at the equilibrium is,
= 2.0 - 0.4
= 1.6 mol/L
According to stochiometry, for 1mol of X2 reacted, it will produce 2 mols of XY3.
X2:XY3 = 1:2
Therefore, amount of XY3 produced from 0.4 mol of X2,
= 0.4 x 2
= 0.8 mol/L
Therefore, the equilibrium concentraions are,
X2 = 1.6 mol/L
Y2 = 0.8 mol/L
XY3 = 0.8 mol/L
Posted by thilina-g on May 19, 2012 at 5:49 AM (Answer #1)
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