Exercise-4

For the reaction ${\mathrm{N}}_2+3{\mathrm{H}}_2\to$ the rate expression is $-\mathrm{d}\left[{\mathrm{NH}}_3\right]/\mathrm{dt}=\mathrm{k}\left[{\mathrm{H}}_2\right]\left[{\mathrm{N}}_2\right]$

The correct statement is.

$\mathrm{I}$. The reaction is not elementary.

$\mathrm{II}$.  The reaction is of second order.

$\mathrm{III}$. $-\mathrm{d}\left[{\mathrm{H}}_2\right]/\mathrm{dt}=-\mathrm{d}\left[{\mathrm{NH}}_3\right]/\mathrm{dt}$.

Which of the energy values marked as $\mathrm{I}, \mathrm{II}$ and $\mathrm{III}$ in the following diagram, will change by the addition of a suitable catalyst?

Urea, $\mathrm{CO}{\left({\mathrm{NH}}_2\right)}_2,$ decomposes at $90°\mathrm{C}$ as ${\left[\mathrm{CO}\left({\mathrm{NH}}_2\right)\right]}_2\left(\mathrm{aq}\right)\to {\mathrm{NH}}_4^{+}\left(\mathrm{aq}\right)+{\mathrm{OCN}}^{-}\left(\mathrm{aq}\right).$ Experimental data obtained for the reaction is given in the following plot;

From the graph it can be inferred that-

The decomposition of ${\mathrm{H}}_2{\mathrm{O}}_2$ can be followed by titration with ${\mathrm{KMnO}}_4$ and is found to be a first order reaction. The rate constant is $4.5\times {10}^{-2}.$ In an experiment, the initial titre value was $25 \mathrm{mL}.$ The titre value will be $5 \mathrm{mL}$ after a lapse of:

In a second order reaction, $20\%$ of a substance is dissociated in $40 \min .$ The time taken by $80\%$ of its dissociation is:

A simple mechanism for enzyme-catalyzed reaction is given by the following set of equations:

$\underset{\left(\mathrm{enzyme}\right)}{\mathrm{E}}+\underset{\left(\mathrm{reactant}\right)}{\mathrm{S}}\rightleftharpoons \underset{(\mathrm{intermediate}-1)}{\mathrm{ES}}$

$\underset{(\mathrm{intermediate}-1)}{\mathrm{ES}}\rightleftharpoons \underset{(\mathrm{intermediate}-2)}{\mathrm{EP}}$

$\underset{(\mathrm{intermediate}-2)}{\mathrm{EP}}\rightleftharpoons \underset{\left(\mathrm{enzyme}\right)}{\mathrm{E}}+\underset{\left(\mathrm{product}\right)}{\mathrm{P}}$

This is known as the Michaelis-Menten mechanism. The potential energy diagram is shown in the figure. Which of the following sets of identifications is correct? (Assume that the temperature and pressure are constant).

A reaction, the rate constant is expressed as $\mathrm{k}={\mathrm{Ae}}^{-40000/\mathrm{T}}.$ The energy of the activation is:

Consider the following reactions at $300 \mathrm{K}.$

$\mathrm{A}\to \mathrm{B}$ (uncatalysed reaction)

$\mathrm{A}\stackrel{\mathrm{catalyst}}{\to }\mathrm{B}$ (catalyst reaction)

The activation energy is lowered by $8.314 \mathrm{KJ} {\mathrm{mol}}^{-1}$ the catalysed reaction. How many times is the rate of this catalysed reaction greater than that of uncatalysed reaction? (Given ${\mathrm{e}}^{3.33}=28$)