Answer:
Kc = 3.1x10²
Explanation:
At equilibrium, the velocity of product formation is equal to the velocity of reactants formation. For a generic reaction, the equilibrium constant (Kc) is:
aA + bB ⇄ cC + dD
[tex]Kc = \frac{[C]^c*[D]^d}{[A]^a*[B]^b}[/tex]
Where [X] is the molar concentration of X, and the solid substances are not considered (because it's activity is 1, for the other substances, the activity is substituted for the molar concentration, which forms the equation above).
For the reaction given, let's make an equilibrium chart:
Fe³⁺(aq) + SCN⁻(aq) ⇄ FeSCN²⁺(aq)
1.1*10⁻³ 8.2*10⁻⁴ 0 Initial
-x -x +x Reacts (stoichiometry is 1:1:1)
1.1*10⁻³ -x 8.2*10⁻⁴ -x x Equilibrium
x = 1.8*10⁻⁴ M, so the molar concentrations at equilibrium are:
[Fe⁺³] = 1.1*10⁻³ - 1.8*10⁻⁴ = 9.2*10⁻⁴ M
[SCN⁻] = 8.2*10⁻⁴ - 1.8*10⁻⁴ = 6.4*10⁻⁴ M
[FeSCN⁺²] = 1.8*10⁻⁴ M
Kc = [FeSCN⁺²]/([Fe⁺³]*[SCN⁻])
Kc = (1.8*10⁻⁴)/(9.2*10⁻⁴*6.4*10⁻⁴)
Kc = 306 = 3.1x10²
The equilibrium constant is 310.
Equilibrium constant of a reaction can be obtained from the equilibrium concentrations of the species in the reaction
We have to set up the ICE table as shown below;
Fe³⁺(aq) + SCN⁻(aq) ⇄ FeSCN²⁺(aq)
Initial: 1.1×10−3 M 8.2×10−4 M 1.8×10−4 M
Reacting: -x -x +x
Equilibrium: 1.1×10−3 M - x 8.2×10−4 M - x 1.8×10−4 M + x
Since x = 1.8×10−4 M
Equilibrium concentration of Fe³⁺ = 9.2*10⁻⁴ M
Equilibrium concentration of SCN⁻ = 6.4*10⁻⁴ M
Equilibrium concentration of FeSCN²⁺ = 1.8*10⁻⁴ M
The equilibrium constant is obtained from the expression;
Kc = [FeSCN²⁺] /[Fe³⁺] [SCN⁻]
Substituting values into the given equation, we have;
Kc = [1.8×10−4 M]/[ 9.2*10⁻⁴ M] [6.4*10⁻⁴ M]
Kc = 310 (to 2sf)
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