Question

For the reaction $A\to$ products, the following data is given for a particular run.
time(min): 0 5 15 35
$\frac{1}{\left[A\right]}\left(M^{-1}\right)$: 1 2 4 8
Determine the order of the reaction.

Answer 1

The given reaction is: $A⟶$ Products

Now, we have for second order reaction,

$\frac{1}{\left[A\right]}=Kt+\frac{1}{[A{]}_0}$

where, $[A{]}_0$ and $\left[A\right]$ are concentrated at $t=0$ and $t=t$ respectively.

Now, from given data,

at $t=0$, $\frac{1}{[A{]}_0}=1m^{-1}$

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

At, $t=5$ min, $\frac{1}{\left[A\right]}=2M^{-1}$

$\therefore 2=5K+1\Rightarrow K=+\frac{1}{5}$

$\Rightarrow \frac{1}{\left[A\right]}=\frac{1}{5}t+1$ $-\left(A\right)$

Putting rest of the values given in the data we see that it stisfies $e{q}^n\left(A\right)$ so the given reaction is of second order.