Rate Law Equation for the Second Order Reaction

Rate Law Equation for the Second Order Reaction:

Let us consider the general second order reaction (in which both reactants are of the same type)-

A + A ———-> Products

According to the law of mass action, Rate (dx/dt) ∝ [A]2

Let the initial concentration of ‘A’ be ‘a’ moles per litre. Suppose after ‘t’ seconds ‘x’ moles of ‘A’ disintegrate into the product.

∴ Concentration of ‘A’ after time ‘t’ seconds = (a – x) moles/litre
dx/dt ∝ (a – x)2 or dx/dt = k2 (a – x)2 ……….(i) [where k2 is the second order rate constant]

Equation (i) is a differential form of the rate law equation. To get its integrated form, separate variables.

dx/(a – x)2 = k2 dt

Integrating the above equation both sides, we get-
∫dx/(a – x)2 = k2 ∫dt
⇒ [(a – x)-2+1/(-2 + 1)] d(a – x)/dt = k2t + I [where ‘I’ is constant of integration]
⇒ [-1/(a – x)] (-1) = k2t + I
⇒ 1/(a – x) = k2t + I ……….(ii)

When t = 0 sec, x = 0,
∴ 1/a = k2 x 0 + I
⇒ I = 1/a ……….(iii)

Put the value of ‘I’ from equation (iii) in equation (ii), and we get-
1/(a – x) = k2t + 1/a
⇒ [1/(a – x)] – [1/a] = k2t
⇒ (a – a + x)/a(a – x) = k2t
⇒ x/a(a – x) = k2t
⇒ k2 = (1/t) [x/a(a – x)]
⇒ k2 = x/at(a – x) ……….(iv)

This equation (iv) is called the integrated rate law equation for second order reaction when both the reactants are the same.

Units of second order rate constant (k2) = (1/sec) X (moles / litre / moles / litre / moles / litre) = litre mol-1 sec-1.

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