Question

Which of the following is a correct expression for the integrated rate law of first order reaction : $A\to \text{Products}$ ?
Here,
$a=[A{]}_0=\text{Initial concentration of the reactant A}$
$(a-x)=\left[A\right]=\text{Concentration of the reactant A at time t}$

• A. $\left[A\right]=[A{]}_0-kt$
• B. $\left[A\right]=[A{]}_0{e}^{-kt}$
• C. $[A{]}_0=[A]e^{-kt}$
• D. $\left[A\right]=[A{]}_0+kt$

Answer 1

The correct option is B $\left[A\right]=[A{]}_0{e}^{-kt}$

For a first order reaction:
$A\to \text{Products}t=0at=ta-x$
a is intial concentration i.e. $[A{]}_0$ and (a-x) is the concentration at time t i.e. $\left[A\right]$
$-\frac{d\left[A\right]}{dt}=k\left[A\right]$
$-\frac{d(a-x)}{dt}=k(a-x)$
a is a constant value
$\frac{dx}{dt}=k(a-x)$
$\frac{dx}{a-x}=k.dt$
$-\underset{0}{\overset{x}{\int }}\frac{dx}{a-x}=k\underset{0}{\overset{t}{\int }}dt$
$ln\frac{a}{a-x}=kt$
$2.303log\frac{a}{a-x}=kt$
This is the integrated rate law of first order reaction
$ln\frac{[A{]}_0}{\left[A\right]}=kt$
$\Rightarrow \left[A\right]=[A{]}_0{e}^{-kt}$
So, the correct option is (b)