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]} = k \, dt ... (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} $