1

### AIPMT 2002

2A $\to$ B + C

It would be a zero order reaction when
A
the rate of reaction is proportional to square of concentration of A
B
The rate of reaction remains same at any concentration of A
C
the rate remains unchanged at any concentration of B and C
D
the rate of reaction doubles if concentrations of B is increased to double.

## Explanation

2A $\to$ B + C

If it is zero order reaction r = k [A]o.

i.e the rate remains same at any concentration of 'A'. i.e independent upon concentration of A.
2

### AIPMT 2001

For the reaction;

2N2O5 $\to$ 4NO2 + O2 rate and rate constant are 1.02 $\times$ 10$-$4 and 3.4 $\times$ 10$-$5 sec$-$1 respectively, then concentration of N2O5 at that time will be
A
1.732
B
3
C
1.02 $\times$ 10$-$4
D
3.4 $\times$ 105

## Explanation

For the reaction;

2N2O5 $\to$ 4NO2 + O2

This is a first order reaction.

$\therefore$ rate = k [N2O5] ;

[N2O5] = ${{rate} \over k}$

= ${{1.02 \times {{10}^{ - 4}}} \over {3.4 \times {{10}^{ - 5}}}}$

= 3
3

### AIPMT 2001

When a bio-chemical reaction is carried out in laboratory, outside the human body in absence of enzyme, then rate of reaction obtained is 10$-$6 times, the activation energy of reaction in the presence of enzyme is
A
6/RT
B
P is required
C
different from E$a$ obtained in laboratory
D
can't say anything.

## Explanation

From Arrhenius equation, k = Ae –Ea/RT

The activation energy of the reaction in the presence of enzyme is different from Ea obtained in laboratory.
4

### AIPMT 2000

For the reaction H+ + BrO$_3^ -$ + 3Br$-$ $\to$ 5Br2 + H2O
which of the following relation correctly represents the consumption and formation of products.
A
${{d\left[ {B{r^ - }} \right]} \over {dt}} = - {3 \over 5}{{d\left[ {B{r_2}} \right]} \over {dt}}$
B
${{d\left[ {B{r^ - }} \right]} \over {dt}} = {3 \over 5}{{d\left[ {B{r_2}} \right]} \over {dt}}$
C
${{d\left[ {B{r^ - }} \right]} \over {dt}} = - {5 \over 3}{{d\left[ {B{r_2}} \right]} \over {dt}}$
D
${{d\left[ {B{r^ - }} \right]} \over {dt}} = {5 \over 3}{{d\left[ {B{r_2}} \right]} \over {dt}}$

## Explanation

$- {1 \over 3}$${{d\left[ {B{r^ - }} \right]} \over {dt}} = + {1 \over 5}{{d\left[ {B{r_2}} \right]} \over {dt}}$

$\Rightarrow$ ${{d\left[ {B{r^ - }} \right]} \over {dt}} = - {3 \over 5}{{d\left[ {B{r_2}} \right]} \over {dt}}$