## 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 = k_{2} (a – x)^{2} ……….(i) [where k_{2} 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} = k_{2} dtIntegrating the above equation both sides, we get- ∫dx/(a – x) ^{2} = k_{2} ∫dt⇒ [(a – x) ^{-2+1}/(-2 + 1)] d(a – x)/dt = k_{2}t + I [where ‘I’ is constant of integration]⇒ [-1/(a – x)] (-1) = k _{2}t + I⇒ 1/(a – x) = k _{2}t + I ……….(ii)When t = 0 sec, x = 0, ∴ 1/a = k _{2} x 0 + I⇒ I = 1/a ……….(iii) Put the value of ‘I’ from equation (iii) in equation (ii), and we get- 1/(a – x) = k _{2}t + 1/a⇒ [1/(a – x)] – [1/a] = k _{2}t⇒ (a – a + x)/a(a – x) = k _{2}t⇒ x/a(a – x) = k _{2}t⇒ k _{2} = (1/t) [x/a(a – x)]⇒ k _{2} = 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 (k _{2}) = (1/sec) X (moles / litre / moles / litre / moles / litre) = litre mol^{-1} sec^{-1}. |