Question

For the reaction, $2NO+H_2\to N_{2}O+H_{2}O$ the value of -dp/dt is found to be $1.50Torr{s}^{-1}$ for a pressure of 359 Torr of $NO$ and $0.25Torr{s}^{-1}$ for a pressure of 152 Torr, the pressure of $H_2$ being constant. On the other hand, when the pressure of $NO$ is kept constant, \dp/dt is $1.60Torr{s}^{-1}$ for a hydrogen pressure of 289 Torr and $0.79Torr{s}^{-1}$ for a pressure of 147 Torr. The over all order of the reaction is :

Answer 1

Rate law would be represented as $\frac{-dp}{dt}=rate=k\left[P_{NO}{]}^a\right[P_{H_2}{]}^b$
Using the date given in the problem, keeping pressure o $H_2$ constant, we get
$1.50=k[359{]}^a$ .....(i)
$0.25=k[152{]}^a$ .....(ii)
Dividing (i) by (ii), $\frac{1.50}{0.25}={\left(\frac{359}{152}\right)}^a$
Taking log of both sides,
$log6=alog\frac{359}{152}$ $\therefore a\approx 2$
Similarly, by keeping the pressure of $NO$ constant, rate law is given by
$1.60=k\left[P_{NO}{]}^a\right[289{]}^b$ .....(iii)
$0.79=k\left[P_{NO}{]}^a\right[147{]}^b$ .....(iv)
Dividing (iii) by (iv) $\frac{1.60}{0.79}={\left(\frac{289}{147}\right)}^b$
Taking log of both sides
$log\left(\frac{1.60}{0.79}\right)=blog\left(\frac{289}{147}\right)$ $\therefore b\approx 1$
$\therefore \text{overall order of the reaction}=(a+b)=(2+1)=3$