(3)

Rate = $k{\left[A\right]}^x$ ${\left[B\right]}^y$

$2\times {10}^{-3}=k{[0.1]}^x$ ${[0.1]}^y$                  ...(i)

$4\times {10}^{-3}=k{[0.2]}^x$ ${[0.1]}^y$                     ...(ii)

$1.6\times {10}^{-2}=k{[0.2]}^x$ ${[0.2]}^{{}^y}$               ...(iii)

Dividing eq. (i) by (ii), we get

$\frac{1}{2}={\left(\frac{{\displaystyle 1}}{{\displaystyle 2}}\right)}^x \Rightarrow {\left(\frac{1}{2}\right)}^1={\left(\frac{1}{2}\right)}^{\mathrm{x}}\Rightarrow x=1$

Similarly, dividing eq. (ii) by (iii), we get

$\frac{1}{4}={\left(\frac{{\displaystyle 1}}{{\displaystyle 2}}\right)}^y \Rightarrow {\left(\frac{1}{2}\right)}^2={\left(\frac{1}{2}\right)}^{\mathrm{y}} \Rightarrow y=2$

Thus, order of the reaction with respect to A and B, respectively are 1 and 2.

(4)

$r=k{\left[A\right]}^{-1/2}{\left[B\right]}^{3/2}$

Order $=-\frac{1}{2}+\frac{3}{2}=\frac{2}{2}=1$

(1)

$r_1=k{\left[A\right]}^2\left[B\right]$

Keeping conc. of B constant and tripling conc. of A, new rate would be

$r_2=k{\left[3A\right]}^{{}^2}\left[B\right]=9k{\left[A\right]}^{{}^2}\left[B\right]$

$\therefore r_2=9\times r_1$

(3)

Note: Correct the reactions given in question as

$X_2\rightleftharpoons X+X$                                      (fast)

$X+Y_2\to XY+Y$                           (slow)      

Slow step is the rate determining step.

$\text{Rate}=k\left[X\right]\left[Y_2\right]$                        ...(i)

Equilibrium constant for fast step, $K=\frac{{\left[X\right]}^2}{\left[X_2\right]}$

$\left[X\right]=\sqrt{K\left[X_2\right]}$

By substituting [X] in equation (i), we get $\text{Rate}=k\sqrt{K\left[X_2\right]}$ $\left[Y_2\right]=k\text{'}{\left[X_2\right]}^{1/2}\left[Y_2\right]$

$\therefore \text{Order of reaction}=\frac{1}{2}+1=\frac{3}{2}=1.5$

(1)

At low pressure, rate is proportional to the surface coverage and is of first order, while at high pressure, it follows zero order kinetics due to complete coverage of surface area.

(3)

Reaction is of zero order as the unit of rate constant is mol ${\mathrm{L}}^{-1}$ ${\mathrm{s}}^{-1}$.

Concentration of $B=k\times t=0.6\times {10}^{-3}\times 20\times 60=0.72$ $\mathrm{M}$