Question

Derive integrated rate equation for rate constant K for a first order reaction how will you Express it in exponential form

Answer 1

Answer:

Go through the image attached below.

Explanation:

The integrated forms of the rate law can be used to find the population of reactant at any time  after the start of the reaction. Plotting in[A] with respect to time for a first-order reaction gives a straight line with the slope of the line equal to –k.

This general relationship, wherein an amount changes at a rate that relies upon its quick esteem, is said to pursue an exponential law.

Answer 2

we have to derive rate of equation for a first order reaction, where rate constant K is given.

$A\rightarrow\textbf{product}$

rate of first order reaction, r = -$\frac{d[A]}{dt}$ = K[A]

⇒ $\int\limits^{[A]_t}_{[A]_0}{\frac{1}{[A]}}\,d[A]=-K\int\limits^t_0\,dt$

⇒$\left[ln[A]]^{[A]_t}_{[A]_0}\right]=-Kt$

⇒$ln[A]_t-ln[A]_0=-Kt$

⇒$ln\left[\frac{[A]_t}{[A]_0}\right]=-Kt$

⇒$ln\left[\frac{[A]_0}{[A]_t}\right]=Kt$

hence, rate of reaction for first order reaction is given as $ln\left[\frac{[A]_0}{[A]_t}\right]=Kt$ or, $K=\frac{2.303}{t}log\left[\frac{[A]_0}{[A]_t}\right]$

now, $ln\left[\frac{[A]_t}{[A]_0}\right]=-Kt$

⇒$\frac{[A]_t}{[A]_0}=e^{-Kt}$

⇒$[A]_t=[A]_0e^{-Kt}$ This is the expression for first order reaction in exponential form.