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Relation Between Chemical Kinetics and Chemical Equilibrium

There is a direct relationship between the rate constant for the forward and reverse directions of a chemical reaction and the equilibrium constant for that reaction. That stands to reason because a large forward rate constant will lead to a larger K. A large backward rate constant will lead to a small K. for a first order chemical reaction,

A ↔ B, rate_f  = k_f [A] and K_f  = [B]/[A]

For the reverse reaction, B ⇔ A, rate_r = k_r[B] K_r  = [A]/[B]

By definition, at equilibrium the two rates, but not the k_f and the k_r, are the same. (Remember that the rates depend on the k values and on the A and B concentrations.)

If rate_f  = rate_r, then k_f [A] = k_r[B]

k_f /k_r = [A]/[B] which is equal to the equilibrium constant, K_f. A large forward rate constant leads to a high rate of formation of B and a large value of K_f. Similarly, k_r / k_f = [B]/[A] = K_r.

Predicting The Direction of a Reaction

The reaction quotient, Q_c can help keep track of reaction direction on changing some concentrations.

K_c is the constant at equilibrium; Q _c is the value obtained with the nonequilibrium initial or forced values given by the problem. If K_c and Q_c are equal, then the system is already at equilibrium. If Q_c > K_c, then the reaction must shift to the left before equilibrium is reached (products must back react to become reactants). If Q_c < K_c, then the system must shift to the right for the system to attain equilibrium (reactants must turn into products).

Ex: The Ka for the dissociation of acetic acid is 1.8 x 10^-5. HA(aq) ⇔ A^-1(aq) + H⁺¹(aq)

If a system contains [A^-1 ] = 2.0 x 10^-4M, [H⁺¹] = 1.5 x 10^-5 M and [HA] = 0.55 M, determine whether the system is in equilibrium or not and in what direction it must shift to put it into equilibrium. Q_c = 5.5 x 10^-9 (compared to K_a = 1.8 x 10^-5) so the system must shift to the right (more product) to attain equilibrium.

Calculating Equilibrium Concentrations

Use the Law of Mass action equation

                         K = [C]^c [D]^d/ [A]^a [B]^b

If all the reactant and product concentrations are known and the balanced equation is known, then a value for K can be easily calculated.

When K is known, the equilibrium concentration values can be calculated knowing one or more initial concentration values. Use the balanced equation to relate initial and equilibrium values. The resulting equation may involve the quadratic formula.

Ex: For the dissociation of any weak acid,  HA(aq) ⇔ H⁺¹(aq) + A^-1(aq)

• = [H⁺¹] · [A^-1]/ [HA] Suppose K = 4.2 x 10^-7and imagine a system where the initial HA concentration is 0.100M. What are the equilibrium concentrations of HA, H⁺¹ and A^-1?

At equilibrium, let [A^-1] = [HA] = X. Let [HA] = 0.100-X

(Sometimes, X is so small that the [HA] change can be ignored and equilibrium [HA] = 0.001M also. See if this is true in this case.)

4.4 x 10^-7 = X²/(0.100) X² = 4.4 x 10^-8 X = 2.1 x 10^-4M = [A^-1]

        = [HA] [HA] = (0.100 -X) = 0.100 – 0.00021 ≈ 0.001M (The assumption holds.)

To check the answer, plug the numbers into the K expression to see if the original K value is obtained.

Other calculations may be more difficult and require use of the quadratic formula.

       Example : Ca⁺²  + EDTA^-2 ⇔  CaEDTA K = 4.80

Suppose initial [Ca⁺²] = 0.1000 M and initial [EDTA^-2] = 0.0500M, what are the equilibrium concentrations of Ca⁺², EDTA^-2 and CaEDTA?

Let X = [CaEDTA], 0.0100 – X = [Ca⁺²] and 0.0500 -X = [EDTA²].

= [CaEDTA] / [Ca⁺²] · [EDTA^-2] = 9.80

 = X/(0.0100 – X)(0.0500 – X)

  X = 9.80(5.00 x 10^-4 – 0.0600X + X²)

X = 4.90 x 10^-3 – 0.588X + 9.80X² 9.80X² – 1.588X + 4.90 x 10^-3

Using the quadratic equation, X =- 0.159 (physically impossible) and

X = 0.0032 (a plausible value) [CaEDTA] = 0.0032M, [Ca⁺²] = 0.0068M and [EDTA^-2] = 0.0468M