Consider the following reaction:
fe3+(aq)+scn−(aq)⇌fescn2+(aq)
a solution is made containing an initial [fe3+] of 1.1×10−3 m and an initial [scn−] of 8.2×10−4 m. at equilibrium, [fescn2+]= 1.8×10−4 m .
calculate the value of the equilibrium constant (kc).
express your answer using two significant figures.

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

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 answer to your question is c.

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