Question

The first order integrated rate equation is

• A. $k=\frac{x}{t}$
• B. $k=\frac{2.303}{t}log\frac{a}{a-x}$
• C. $k=\frac{1}{t}ln\frac{a}{a-x}$
• D. $k=\frac{1}{t}\frac{a}{a\left(a-x\right)}$

Answer 1

Answer: (C)

For first order,

rate $=\frac{d[R]}{dt}=k[R]$

or $\frac{d[R]}{[R]}=kdt...(i)$

On integrating Eq. (i)

$\int \frac{d[R]}{[R]}=k\int dt$

$ln[R]=-kt+C...(ii)$

At t = O,[R] = $[R_0]$

[where, R = final concentration, i.e., a - x and $R_0$ is the initial

concentration, i.e., a.]

$ln[R_0]=C$

On putting the value of C in Eq. (ii), we get

$ln[R]=-kt+ln[R_0]$

$-kt=ln[R]-ln[R_0]$

$kt=ln[R_0]-ln[R]$

or $k=\frac{1}{t}ln\frac{[R_0]}{[R]}$

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