Question

Derive the relation C_t = C₀ e^−kt for a first order reaction.

Answer 1

Consider the first-order reaction

\[\ce{A -> products}\]

The differential rate law is

\[\ce{- \frac{dC}{dt} = kC}\]    ...(1)

where:

C = concentration of A at time t

C₀ = concentration of A at t = 0

k = first-order rate constant

Seperate variables

\[\ce{\frac{dC}{C} = -k dt}\]    ...(2)

Integrating Eq. (2) between limits C = C₀ at t = 0 and C = C_t at t = t:

\[\int\limits_{C_0}^{C_t}\frac {dC}{C} = -k\int\limits_{0}^{t}dt\]

ln C_t​ − ln C₀​ = −kt

\[\ce{ln \frac{C_t}{C_0} = -kt}\]

Taking antilogarithm

\[\ce{\frac{C_t}{C_0} = e^{-kt}}\]

C_t​ = C₀​ e^−kt

This is the integrated rate equation for a first-order reaction, showing that the concentration decays exponentially with time.