Question

For the hypothetical reaction, $2A+B⟶$ Products, the following data are obtained:

Expt. No.	Initial conc. of $\left(A\right)$ $\left(mol{L}^{-1}\right)$	Initial conc. of $\left(B\right)$ $\left(mol{L}^{-1}\right)$	Initial rate $mol$ $L^{-1}$ $s^{-1}$
$1.$	$0.10$	$0.20$	$3\times {10}^2$
$2.$	$0.30$	$0.40$	$3.6\times {10}^3$
$3.$	$0.30$	$0.80$	$1.44\times {10}^4$
$4.$	$0.10$	$0.40$	______
$5.$	$0.20$	$0.60$	______
$6.$	$0.30$	$1.20$	______

Find out how the rate of the reaction depends upon the concentration of $A$ and $B$ and fill in the blanks.

Answer 1

From experiments (2) and (3), it is clear than when concentration of $A$ is kept constant and that of $B$ is doubled, the rate increases four times. This shows that the reaction is of second order with respect to $B$.

Similarly, from experiments (1) and (2), it is observed that when the concentration of $A$ is increased three times and that of $B$ two times, the rate becomes twelve times. Hence, the reaction is first order with respect to $A$.

Thus, the rate law for the reaction is:
Rate $=k\left[A\right]{\left[B\right]}^2$

Substituting the values of expt. (1) in the rate equation,

$3\times {10}^2=k\left[0.10\right]{\left[0.20\right]}^2$

or $k=\frac{3\times {10}^2}{\left[0.10\right]{\left[0.20\right]}^2}$

$=7.5\times {10}^4$ $L^2$ ${mol}^{-2}$ $s^{-1}$

Expt. (4): Rate $=k\left[0.10\right]{\left[0.40\right]}^2$

$=7.5\times {10}^4\times 0.10\times 0.40\times 0.40$

$=1.2\times {10}^3$ $mol$ $L^{-1}$ $s^{-1}$

Expt. (5): Rate $=k\left[0.20\right]{\left[0.60\right]}^2$

$=7.5\times {10}^4\times 0.20\times 0.60\times 0.60$

$=5.4\times {10}^3$ $mol$ $L^{-1}$ $s^{-1}$

Expt. (6): Rate $=k\left[0.30\right]{\left[1.20\right]}^2$

$=7.5\times {10}^4\times 0.30\times 1.20\times 1.20$

$=3.24\times {10}^4$ $mol$ $L^{-1}$ $s^{-1}$