1

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.
2

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}}$

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

AIPMT 2000

How enzymes increases the rate of reactions
A
by lowering activation energy
B
by increaing activation energy
C
by changing equilibrium constant
D
by forming enzyme substrate complex.

Explanation

Enzymes act like catalyst in biochemical reactions. Presence of an enzyme increases the rate of reaction by lowering the activation energy of the reactant.