± Fundamentals of Equilibrium Concentration Calculations The reversible reaction
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Question
± Fundamentals of Equilibrium Concentration Calculations
The reversible reaction
XY(aq)X(aq)+Y(aq)
has a reaction quotient Qc defined as
Qc=[X][Y][XY]
Because the reaction is reversible, both the forward and reverse reactions will occur simultaneously. The reaction will eventually reach equilibrium, at which point the concentrations do not change, and Qc is equal to a constant known as Kc.
Figure
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Part B
Based on a Kc value of 0.190 and the given data table, what are the equilibrium concentrations of XY, X, and Y, respectively?
Express the molar concentrations numerically.
Calculating equilibrium concentrations when the net reaction proceeds in reverse
Consider mixture C, which will cause the net reaction to proceed in reverse.
Concentration (M)initial:change:equilibrium:[XY]0.200+x0.200+xnet[X]0.300x0.300x+[Y]0.300x0.300x
The change in concentration, x, is positive for the reactants because they are produced and negative for the products because they are consumed.
Part C
Based on a Kc value of 0.190 and the data table given, what are the equilibrium concentrations of XY, X, and Y, respectively?
Express the molar concentrations numerically.
Initial concentrations (M) Mixture [XY] [X] [Y] A 0.100 0 0 B 0.500 0.100 0.100 C 0.200 0.300 0.300 Reaction forms products Reaction forms reactants EquilibriumExplanation / Answer
The reaction happening will be
XY(aq) ------ X(aq) + Y(aq)
K = [X][Y]/[XY]
Part A
For mixture A
Q = [0][0]/0.100 = 0
Since K>Q, reaction will form products and move to right
For mixture B
Q = [0.100][0.100]/0.500 = 0.02
Since K>Q, reaction will form products and move to right
For mixture C
Q = [0.30][0.30]/0.20 = 0.45
Since K<Q, reaction will form reactants and move to left
Part B
For mixture C
XY ----- X + Y
Initial 0.200 0.300 0.300
Final (0.200+x) (0.300-x) (0.300-x)
K = (0.300-x)^2/(0.200+x) = 0.190
x = 0.0724
[X] = [Y] = 0.3 - 0.0724 = 0.2276
[XY] = 0.2 + 0.0724 = 0.2724M
Part C
For mixture A
XY ----- X + Y
Initial 0.100 0 0
Final (0.100-x) x x
K = x^2/(0.100-x) = 0.190
x = 0.0724
[X] = [Y] = 0.0724
[XY] = 0.1 - 0.0724= 0.0276M
For mixture B
XY ----- X + Y
Initial 0.500 0.100 0.100
Final (0.500-x) (0.1+x) (0.1+x)
K = (0.1+x)^2/(0.500-x) = 0.190
x = 0.1557
[X] = [Y] = 0.2557
[XY] = 0.5 - 0.1557 = 0.3443M
For mixture C
XY ----- X + Y
Initial 0.200 0.300 0.300
Final (0.200+x) (0.300-x) (0.300-x)
K = (0.300-x)^2/(0.200+x) = 0.190
x = 0.0724
[X] = [Y] = 0.3 - 0.0724 = 0.2276
[XY] = 0.2 + 0.0724 = 0.2724M
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