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Chemistry · Ch 6 — Chemical Kinetics

Integrated rate law for the first order reactions in solution

6.5.1

Integrated rate law for the first order reactions in solution

Consider the first order reaction

A⟶product...(6.3)\mathrm{A \longrightarrow product} \qquad \text{...(6.3)}

The differential rate law is given by

rate=−d[A]dt=k [A]...(6.4)\text{rate} = -\frac{\mathrm{d[A]}}{\mathrm{d}t} = k\,\mathrm{[A]} \qquad \text{...(6.4)}

where [A] is the concentration of the reactant at time tt. Rearranging Eq. (6.4),

d[A][A]=−k dt...(6.5)\frac{\mathrm{d[A]}}{\mathrm{[A]}} = -k\,\mathrm{d}t \qquad \text{...(6.5)}

Let [A]0\mathrm{[A]_0} be the initial concentration of the reactant A at time t=0t = 0. Suppose [A]t\mathrm{[A]_t} is the concentration of A at time tt. Equation (6.5) is integrated between the limits [A]=[A]0\mathrm{[A]} = \mathrm{[A]_0} at t=0t = 0 and [A]=[A]t\mathrm{[A]} = \mathrm{[A]_t} at t=tt = t:

∫[A]0[A]td[A][A]=−k∫0tdt\int_{\mathrm{[A]_0}}^{\mathrm{[A]_t}} \frac{\mathrm{d[A]}}{\mathrm{[A]}} = -k \int_{0}^{t} \mathrm{d}t

On integration,

[ln⁡[A]][A]0[A]t=−k (t)0 t\Big[\ln \mathrm{[A]}\Big]_{\mathrm{[A]_0}}^{\mathrm{[A]_t}} = -k\,(t)_0^{\,t}

Substitution of the limits gives ln⁡[A]t−ln⁡[A]0=−k t\ln\mathrm{[A]_t} - \ln\mathrm{[A]_0} = -k\,t,

orln⁡[A]t[A]0=−kt...(6.6)\text{or}\quad \ln\frac{\mathrm{[A]_t}}{\mathrm{[A]_0}} = -kt \qquad \text{...(6.6)}

ork=1t ln⁡[A]0[A]t\text{or}\quad k = \frac{1}{t}\,\ln\frac{\mathrm{[A]_0}}{\mathrm{[A]_t}}

Converting ln⁡\ln to log⁡10\log_{10}, we write

k=2.303t log⁡10[A]0[A]t...(6.7)k = \frac{2.303}{t}\,\log_{10}\frac{\mathrm{[A]_0}}{\mathrm{[A]_t}} \qquad \text{...(6.7)}

Eq. (6.7) gives the integrated rate law for first order reactions. The rate law can be written in the following forms:

i. Eq. (6.6) is ln⁡[A]t[A]0=−kt\ln\dfrac{\mathrm{[A]_t}}{\mathrm{[A]_0}} = -kt. By taking the antilog of both sides we get

[A]t[A]0=e−kt     or     [A]t=[A]0 e−kt...(6.8)\frac{\mathrm{[A]_t}}{\mathrm{[A]_0}} = e^{-kt} \;\;\text{ or }\;\; \mathrm{[A]_t} = \mathrm{[A]_0}\,e^{-kt} \qquad \text{...(6.8)} …

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