Question

For a second order reaction plots are made for $\frac{1}{\left[A\right]}$vs time for the reaction, $2A\to$Product. Pick up the correct sentences.

• A. the graph will show straight line with slop K
• B. the graph will show straight line wiith intercept$[A{]}_0$
• C. the graph will show straight line with slop $[A{]}_0$
• D. the graph will show straight line with Intercept ${\frac{1}{\left[A\right]}}_0$

Answer 1

The correct option is A the graph will show straight line with slop K

For the second-order reaction of the type

$2A⟶$ products

We have,

Rate=$r=\frac{dx}{dt}=K(a-x{)}^2$ $-\left(i\right)$

where, $a$= initial concentration of $A$

$x$= concentration of product formed after time $t$

$(a-x)$= concentration remaining at time $t$

Now, separating variable of $e{q}^n\left(i\right)$ & integrating :-

$\int \frac{dx}{(a-x{)}^2}=\int Kdt$

${\left[\frac{-1}{a-x}\right]}^{(-1)}=Kt+C$ ; $C$ is integration constant

$\Rightarrow$ $\frac{1}{a-x}=Kt+C$

Now, $(a-x)=\left[A\right]=$ concentration at any time $t$

So,

$\frac{1}{\left[A\right]}=Kt+C$

$\Downarrow$ $\Downarrow$

$y$ $=Kx+C$

considering $\frac{1}{\left[A\right]}$ as $y$ and $t$ as $x$; we have

graph between $\frac{1}{\left[A\right]}$ and time as a straight line with slope $K$