For the reaction, 2A → B, rates= k[A]². If the concentration of reactant is doubled, then the:

Identify the set of correct statements & choose the correct answer from the options given below:

At high pressure, the following reaction is zero order:

The correct statements among the following is:

A graph of volume of hydrogen released vs time for the reaction between zinc and dil. HCl is given in the graph below.
                         
The correct statement among the following based on the graph given above is:

$$\begin{gathered} 1.<mspace\; width="1em"></mspace>\mathrm{Average}<mspace\; width="1em"></mspace>\mathrm{rate}<mspace\; width="1em"></mspace>\mathrm{upto}<mspace\; width="1em"></mspace>40s\text{\hspace{0.17em}is\hspace{0.17em}}\frac{V_3-V_2}{40} \\ 2.<mspace\; width="1em"></mspace><mspace\; width="1em"></mspace>\mathrm{Average}<mspace\; width="1em"></mspace>\mathrm{rate}<mspace\; width="1em"></mspace>\mathrm{upto}<mspace\; width="1em"></mspace>40s\text{\hspace{0.17em}is\hspace{0.17em}}\frac{V_3-V_2}{40-30} \\ 3.<mspace\; width="1em"></mspace>\mathrm{Average}<mspace\; width="1em"></mspace>\mathrm{rate}<mspace\; width="1em"></mspace>\mathrm{upto}<mspace\; width="1em"></mspace>40s\text{\hspace{0.17em}is\hspace{0.17em}}\frac{V_3}{40} \\ 4.<mspace\; width="1em"></mspace>\mathrm{Average}<mspace\; width="1em"></mspace>\mathrm{rate}<mspace\; width="1em"></mspace>\mathrm{upto}<mspace\; width="1em"></mspace>40s\text{\hspace{0.17em}is\hspace{0.17em}}\frac{V_3-V_1}{40-20} \end{gathered}$$

Consider the reaction A$\to$B. The concentration of both the reactant and the product varies exponentially with time.

The graph that accurately depicts how reactant and product concentrations change with time is:

1.		2.
3.		4.

Match the items in Column I with Column II:

Codes:

Consider the following graph:
      

The instantaneous rate of reaction at t = 600 sec will be:

$1.-\hspace{0.17em}4.75$ $\times {10}^{-4}$ $\mathrm{mol}$ ${\mathrm{L}}^{-1}{\mathrm{s}}^{-1}$
$2.$ $5.75\times {10}^{-5}$ $\mathrm{mol}$ ${\mathrm{L}}^{-1}{\mathrm{s}}^{-1}$
$3.$ $$ $6.75\times {10}^{-6}$ $\mathrm{mol}$ ${\mathrm{L}}^{-1}{\mathrm{s}}^{-1}$
$4.$ $-6.75\times {10}^{-6}$ $\mathrm{mol}$ ${\mathrm{L}}^{-1}{\mathrm{s}}^{-1}$

Match the graphical variations with their descriptions given above and identify the correct codes below: