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How do I find the equilibrium concentrations for the following equation?X2(g)+3Y2(g)...

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quinnellow | Student, College Freshman | eNoter

Posted May 18, 2012 at 11:45 PM via web

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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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thilina-g | College Teacher | (Level 1) Educator

Posted May 19, 2012 at 5:49 AM (Answer #1)

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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


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