Equilibrium Fundamentals of Equilibrium Concentration Calculations 7 of 41> Part
ID: 1090158 • Letter: E
Question
Equilibrium Fundamentals of Equilibrium Concentration Calculations 7 of 41> Part B Constants Periodic Table Loarning Goal: To determine equilibrium concentrations from initial conditions. The reversible reaction Based on a Ke value of 0.220 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) has a reaction quotient c defined as C[XY Because the reaction is reversible, both the forward and reverse reactions will occur simultaneously. The reaction wil eventualy reach equilibrium, at which point the concentraions do not change, and Q is equal to a constant known as K,c Submit Calculating equilibrium concentrations when the net reaction proceeds in reverse Consider mixture C, which will cause the net reaction to proceed in reverse. net Concentration (M) XYX initial change equilibrium 0.200 0.300 0.300 0.200+zx 0.300- 0.300 2 Figure 1 of 1 The change in concentration, z, is positive for the reactants because they are produced and negative for the products because they are consumed Part C of XY, X, and Y, respectively? Based on a Ke value of 0.220 and the data table given, what are the equilibrium Express the molar concentrations numerically. View Available Hint(s) Reaction forms forms reactaTs SubmitExplanation / Answer
The equilirbium realtion fitting the reaction equation
XY X + Y
is
Kc = [X][Y]/[XY]
When you apply the expresiions for the equilibrium concentrations from the last row of you table and the value for Kc you get:
0.220 = (0.300- x)(0.300 -x) / (0.200 +x)
<=>
0.220(0.200 + x) = (0.300 - x)(0.300 - x)
<=>
0.044 +0.220x = 0.09 - 0.6x - x²
<=>
x² -0.82x +0.046 = 0
The solutions to this quadratic equation are:
x = -0.41 ±( (0.41)² -0.046 )
X=0.41±0.349428104
<=>
x = 0.759
x = 0.06057
use second x value which is minimum,so
Hence equilibrium concentrations are:
[XY] = 0.200 + 0.06057 = 0.26057
[X] = 0.300 – 0.06057 = 0.23943
[Y] = 0.300 - 0.06057 = 0.23943
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