Question

The first order integrated rate equation is:

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

Answer 1

Answer: (C)

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

For first order,

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

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

On integrating Eq. (i)

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

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

At t = 0, [R] = [R${}_0$]

[where, R = final concentration, ie, $\alpha -x$ and $R_0$ is the initial concentration, ir, $\alpha$].

ln $\left[R_0\right]$ = C

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

ln[R] = - kt + ln$\left[R_0\right]$

- kt = ln [R] - ln$\left[R_0\right]$

kt = ln$\left[R_0\right]$ - ln [R]

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

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