## 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 = k _{1} (a – x) ……….(i) [where k_{1} 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) = k _{1}dt ……….(ii)Integrating equation (ii), we get- ∫dx/(a – x) = k _{1} ∫dt ⇒ -ln (a – x) = k_{1}t + 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) = k _{1}t – ln (a) ⇒ ln (a) – ln (a – x) = k _{1}t⇒ ln a/(a – x) = k _{1}t⇒ k _{1} = (1/t) ln a/(a – x)⇒ k _{1} = (1/t) (2.303) log a/(a – x)⇒ k ……….(v)_{1} = (2.303/t) log a/(a – x)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 (k _{1}) = sec^{-1} |

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