Question

Item 30 < 30 of 37 Constants Periodic Table Learning Goal: To determine equilibrium concentrations from initial conditions. The reversible reaction Calculating equilibrium concentrations when the net reaction proceeds forward Consider mixture B, which will cause the net reaction to proceed forward. net + XY(aq) = X(aq) + Y(aq) has a reaction quotient Qc defined as Qc = Concentration (M) initial: change: equilibrium: XY] 0.500 X 0.100 Y] 0.100 + 0.100+ + 0.500 – 2 0.100+2 The change in concentration, 2, is negative for the reactants because they are consumed and positive for the products because they are produced. Because the reaction is reversible, both the forward and reverse reactions will occur simultaneously. The reaction will eventually reach equilibrium, at which point the Part B Figure 1 of 1 > Based on a K: value of 0.250 and the given data table, what are the equilibrium concentrations of XY, X, and Y. respectively? Express the molar concentrations numerically. View Available Hint(s) IVO ALOM O 2 ? (XY). (X). [Y] = Equilibrium Reaction forms products Reaction forms reactants Submit Previous Answers Request Answer X Incorrect; Try Again

Answer 1

Part.B :-

ICE table is :

....................XY <------------------> X .................+..................Y

Initial.............0.500 ...................0.100................................0.100

Change...........-x...........................+x..................................+x

Equilibrium.....(0.500-x) .............(0.100+x)...........................(0.100+x)

x = Degree of dissociation

Expression of Equilibrium constant i.e. K (which is equal to the product of the molar concentration of products divided by product of the molar concentration of reactants raise to power their stoichiometric coefficient when reaction is at equilibrium stage).

k = [X].[Y]/[XY]

0.250 = (0.100+x)²/(0.500-x)

0.250(0.500-x) = (0.100+x)²

0.125 - 0.250 x = x² + 0.01 + 0.2 x

x² + 0.2 x + 0.250 x + 0.01 - 0.125 = 0

x² + 0.45 x - 0.115 = 0

On solving

x = 0.18197

Therefore, Equilibrium concentration of XY = 0.500 - 0.18197 = 0.318 M Equilibrium concentration of X = 0.100 + 0.18197 = 0.282 M Equilibrium concentration of Y = 0.100 + 0.18197 = 0.282 M

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Part.C :-

ICE table is :

....................XY <------------------> X .................+..................Y

Initial.............0.200 ...................0.300................................0.300

Change...........+x...........................-x..................................-x

Equilibrium.....(0.200+x) .............(0.300-x).........................(0.300-x)

x = Degree of dissociation

Expression of Equilibrium constant i.e. K (which is equal to the product of the molar concentration of products divided by product of the molar concentration of reactants raise to power their stoichiometric coefficient when reaction is at equilibrium stage).

k = [X].[Y]/[XY]

0.250 = (0.300-x)²/(0.200+x)

0.250(0.200+x) = (0.300-x)²

0.05 + 0.250 x = x² - 0.600 x + 0.09

x² - 0.600 x - 0.250 x + 0.09 - 0.05 = 0

x² - 0.850 x + 0.04 = 0

On solving

x = 0.05

Therefore, Equilibrium concentration of XY = 0.200 + 0.05 = 0.250 M Equilibrium concentration of X = 0.300 - 0.05 = 0.250 M Equilibrium concentration of Y = 0.300 - 0.05 = 0.250 M

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