1
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.
2
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

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

$$ \Rightarrow $$ $${{d\left[ {B{r^ - }} \right]} \over {dt}} = - {3 \over 5}{{d\left[ {B{r_2}} \right]} \over {dt}}$$
3
MCQ (Single Correct Answer)

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.

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