Question

drive the rate equation K = 2.303/t *log a/a-x for first order

Answer 1

Answer:

The rate equation of first-order reaction is $k=\frac{2.303}{t}log\frac{[a]}{[a-x]}$.

Explanation:

In this class of reactions, the rate of the reaction is proportional to the first power of the concentration of the reactant R. For example,

R⇒P

Where,

R=reactant

P=product

$Rate=\frac{d[R]}{dt}=k[R]$           (1)

or $\frac{d[R]}{[R]}=-Kdt$                 (2)

By integrating this equation we get,

$ln[R]=-kt+I$                 (3)

"I" is the constant of integration and its value can be determined easily.

When t=0, R=[R]₀, where [R]₀ is the initial concentration of the reactant.

Therefore equation (3) can be written as

$ln[R]_0=-k\times 0+I$

$ln[R]_0=I$            (4)

By substituting equation (4) in (3) we get;

$ln[R]=-kt+ln[R]_0$      (5)

Rearranging equation (5) we get;

$ln\frac{[R]}{[R]_0}=-kt$           (6)

For any equation, we know that,

$lnx=log\times x \times 2.303=2.303logx$  (7)

Using equation (7) in equation (6) we get;

$2.303log\frac{[R]}{[R_0]}=-kt$      (8)

Equation (8) can also be written as;

$kt=2.303log\frac{[R_0]}{[R]}$

$k=\frac{2.303}{t}log\frac{[R_0]}{[R]}$            (9)

As per the question,

R₀=a

R=a-x

Therefore equation (9) can be written as;

$k=\frac{2.303}{t}log\frac{[a]}{[a-x]}$           (10)

Hence, the rate equation of first-order reaction is $k=\frac{2.303}{t}log\frac{[a]}{[a-x]}$