Question

The experimental data for the reaction, $2A+B_2\to 2AB$ are as follows. Write probable rate expression;

$\left[A\right]$ $mollitr{e}^{-1}$	$\left[B_2\right]$ $mollitr{e}^{-1}$	$Rate\times {10}^4$ $mollitr{e}^{-1}{\sec }^{-1}$
$0.50$	$0.50$	$1.6$
$0.50$	$1.00$	$3.2$
$1.00$	$1.00$	$3.2$

Answer 1

Rate of reaction$=K{\left[A\right]}^m{\left[B_2\right]}^n$

Now, according to given data,

Thus $1.6\times {10}^4=K\left(0.5{)}^m\right(0.5{)}^n$-------(i)

$3.2\times {10}^4=K\left(0.5{)}^m\right(1.0{)}^n$-----------(ii)

$3.2\times {10}^4=K\left(1.0{)}^m\right(1.0{)}^n$-----------(iii)

By (i) and (ii) $n=1$

By (ii) and (iii) $m=0$

$\therefore rate=\frac{dx}{dt}=K{\left[B_2\right]}^1$