Rate Law Equation for the First Order Reaction

Rate Law Equation for the First Order Reaction:

Consider the general reaction-

A ———-> Products

Let the initial concentration of ‘A’ is ‘a’ moles per litre, i.e. when t = 0 sec. Suppose after time ‘t’ second, ‘x’ moles of products are formed, then the concentration of reactant ‘A’ after ‘t’ second is (a – x) moles per litre. Applying rate law of equation, we get-

dx/dt ∝ (a – x)
or dx/dt = k1 (a – x) ……….(i) [where k1 is the first order rate constant]

Equation (i) is the differential form of the rate law equation. Thus, to get its integrated equation, rearrange the variables in equation (i), and we get-
or dx/(a – x) = k1dt ……….(ii)

Integrating equation (ii), we get-
∫dx/(a – x) = k1 ∫dt ⇒ -ln (a – x) = k1t + I ……….(iii)
Where ‘I’ is constant of integration and needs to be evaluated.

When t = 0 sec ; x = 0
∴ -ln (a – 0) = 0 + I ⇒ -ln (a) = I ……….(iv)

Put equation (iv) in equation (iii), we get-
-ln (a – x) = k1t – ln (a)
⇒ ln (a) – ln (a – x) = k1t
⇒ ln a/(a – x) = k1t
⇒ k1 = (1/t) ln a/(a – x)
⇒ k1 = (1/t) (2.303) log a/(a – x)
k1 = (2.303/t) log a/(a – x) ……….(v)
It is an integrated rate law equation for first order reaction. It is used to determine the unit of rate constant.

Hence, units of rate constant (k) = (2.303 log/sec) X (mole / litre / moles / litre)
Units of rate constant (k1) = sec-1

Hence, first order reaction has units which are independent of the concentration of reactants.

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