AIPMT 2011 Mains

The rate of the reaction : 2N₂O₅ $$ \to $$ 4NO₂ + O₂
can be written in three ways.

$${{ - d\left[ {{N_2}{O_5}} \right]} \over {dt}} = k\left[ {{N_2}{O_5}} \right]$$

$${{d\left[ {N{O_2}} \right]} \over {dt}} = k'\left[ {{N_2}{O_5}} \right];\,\,$$ $${{d\left[ {{O_2}} \right]} \over {dt}} = k''\left[ {{N_2}{O_5}} \right]$$

The relationship between k and k' and between k and k'' are

$$k' = 2k ; k'' = k$$

$$k' = 2k ; k'' = k/2$$

$$k' = 2k ; k'' = 2k$$

$$k' = k ; k'' = k$$

Explanation

Rate of disappearance of reactants = Rate of appearance of products

$$ - {1 \over 2}$$$${{ - d\left[ {{N_2}{O_5}} \right]} \over {dt}}$$ = $${1 \over 4}$$$${{d\left[ {N{O_2}} \right]} \over {dt}}$$ = $${{d\left[ {{O_2}} \right]} \over {dt}}$$

$$ \Rightarrow $$ $${1 \over 2}$$$$k\left[ {{N_2}{O_5}} \right]$$ = $${1 \over 4}$$$$k'\left[ {{N_2}{O_5}} \right]$$ = $$k''\left[ {{N_2}{O_5}} \right]$$

$$ \Rightarrow $$ $${k \over 2} = {{k'} \over 4} = k''$$

$$ \Rightarrow $$ $$k' = 2k ; k'' = k/2$$

MCQ (Single Correct Answer)

AIPMT 2011 Prelims

Which one of the following statements for the order of a reaction is incorrect?

Order can be determined only experimentally.

Order is not influenced by stoichiometric coefficient of the reactants.

Order of a reaction is sum of power to the concentration terms of reactants to express the rate of reaction.

Order of reaction is always whole number.

Explanation

Order of a reaction is not always whole number. It can be zero, or fractional also.

MCQ (Single Correct Answer)

AIPMT 2010 Prelims

During the kinetic study of the reaction, 2A + B $$ \to $$ C + D, following results were obtained

Run	[A]/mol L^$$-$$1	[B]/mol L^$$-$$1	Initial rate of formation of D/mol L^$$-$$1 min^$$-$$1
I.	0.1	0.1	6.0$$ \times $$10^$$-$$3
II.	0.3	0.2	7.2$$ \times $$10^$$-$$2
III.	0.3	0.4	2.88$$ \times $$10^$$-$$1
IV.	0.4	0.1	2.40$$ \times $$10^$$-$$2

Based on the above data which one of the following is correct?

Rate = k[A]²[B]

Rate = k[A][B]

Rate = k[A]²[B]²

Rate = k[A][B]²

Explanation

Rate = k[A]^x [B]^y

For the given situations

(I) rate = k(0.1)^x (0.1)^y = 6.0$$ \times $$10^$$-$$3

(II) rate = k(0.2)^x (0.3)^y = 7.2$$ \times $$10^$$-$$2

(III) rate = k(0.3)^x (0.4)^y = 2.88$$ \times $$10^$$-$$1

(IV) rate = k(0.4)^x (0.1)^y = 2.40$$ \times $$10^$$-$$2

Dividing eq. (I) by eq. (IV) we get

$${\left( {{{0.1} \over {0.4}}} \right)^x}{\left( {{{0.1} \over {0.1}}} \right)^y} = {{6.0 \times {{10}^{ - 3}}} \over {2.4 \times {{10}^{ - 2}}}}$$

$$ \Rightarrow $$ $${\left( {{1 \over 4}} \right)^x} = {\left( {{1 \over 4}} \right)^1}$$

$$ \Rightarrow $$ x = 1

On dividing eq. (II) by eq. (III) we get

$${\left( {{{0.3} \over {0.3}}} \right)^x}{\left( {{{0.2} \over {0.4}}} \right)^y} = {{7.2 \times {{10}^{ - 2}}} \over {2.88 \times {{10}^{ - 1}}}}$$

$$ \Rightarrow $$ $${\left( {{1 \over 2}} \right)^y} = {1 \over 4}$$

$$ \Rightarrow $$ y = 2

$$ \therefore $$ Rate = k[A][B]²

MCQ (Single Correct Answer)

AIPMT 2010 Prelims

For the reaction N₂O_5(g) $$ \to $$ 2NO_2(g) + 1/2O_2(g)
the value of rate of disappearance of N₂O₅ is given as 6.25 $$ \times $$ 10^$$-$$3 mol L^$$-$$1 s^$$-$$1. The rate of formation of NO₂ and O₂ is given respectively as

6.25 $$ \times $$ 10^$$-$$3 mol L^$$-$$1 s^$$-$$1 and
6.25 $$ \times $$ 10^$$-$$3 mol L^$$-$$1 s^$$-$$1

1.25 $$ \times $$ 10^$$-$$2 mol L^$$-$$1 s^$$-$$1 and
3.125 $$ \times $$ 10^$$-$$3 mol L^$$-$$1 s^$$-$$1

6.25 $$ \times $$ 10^$$-$$3 mol L^$$-$$1 s^$$-$$1 and
3.125 $$ \times $$ 10^$$-$$3 mol L^$$-$$1 s^$$-$$1

1.25 $$ \times $$ 10^$$-$$2 mol L^$$-$$1 s^$$-$$1 and
6.25 $$ \times $$ 10^$$-$$3 mol L^$$-$$1 s^$$-$$1

Explanation

N₂O_5(g) $$ \to $$ 2NO_2(g) + 1/2O_2(g)

$$ - {{d\left[ {{N_2}{O_5}} \right]} \over {dt}} = {1 \over 2}{{d\left[ {N{O_2}} \right]} \over {dt}} = 2{{d\left[ {{O_2}} \right]} \over {dt}}$$

$$ \Rightarrow $$ $${{d\left[ {N{O_2}} \right]} \over {dt}} = - 2{{d\left[ {{N_2}{O_5}} \right]} \over {dt}}$$

= 2 $$ \times $$ 6.25 $$ \times $$ 10 mol l^-1 sec^-1

= 1.25 $$ \times $$ 10^$$-$$2 mol L^$$-$$1 s^$$-$$1

$${{d\left[ {{O_2}} \right]} \over {dt}} = - {1 \over 2}{{d\left[ {{N_2}{O_5}} \right]} \over {dt}}$$

= $${{6.25 \times {{10}^{ - 3}}} \over 2}$$

= 3.125 $$ \times $$ 10^$$-$$3 mol L^$$-$$1 s^$$-$$1