Answer to Question #110619 in Physical Chemistry for DEEPAK SINGH

Question

derive integrated rate law for the first order reaction.

Answer 1

Answer:

Let's write a general equation for the first order reaction:

A $\rightarrow$ products

If at time zero (t=0) the concentration of A was equal to $a$ and after certain time it decreased by $x$, the rate of the reaction $v$ is:

$v = -\frac{\text{d}(a-x)}{\text{d}t} = \frac{\text{d}x}{\text{d}t}$

According to the law of mass action, the rate of the reaction $v$ is proportional to the product of the concentrations of the reagents, raised to the powers of stoichiometric coefficients:

$v = k(a-x)$ .

Combine the right sides of the equations:

$\frac{\text{d}x}{\text{d}t} = k(a-x)$ .

Separate the variables:

$\frac{\text{d}x}{a-x} = k\text{d}t$ .

Integrate the both sides and get a form of the integrated rate law for the first order reaction:

$\int_{0}^{x}\frac{dx}{a-x} = \int_{0}^{t}k\text{d}t$

$\text{ln}\frac{a}{a-x} = kt$

Rearrange and get other forms of the integrated rate law for the first order reaction:

$a-x = ae^{-kt}$

$x = a(1-e^{-kt})$

Rearrange to get an expression for the rate constant of the first-order reaction:

$k = \frac{1}{t}\text{ln}\frac{a}{a-x}$

Using $a-x = a/2$ at $t = \tau_{1/2}$ , we can also determine the half life of the reaction:

$\tau_{1/2} = \frac{\text{ln}2}{k}$