1
MCQ (Single Correct Answer)

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
MCQ (Single Correct Answer)

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
MCQ (Single Correct Answer)

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
MCQ (Single Correct Answer)

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

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

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

EXAM MAP

Joint Entrance Examination

JEE Advanced JEE Main

Medical

NEET

Graduate Aptitude Test in Engineering

GATE CE GATE ECE GATE ME GATE IN GATE EE GATE CSE GATE PI