1
MCQ (Single Correct Answer)

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
A
$$k' = 2k ;  k'' = k$$
B
$$k' = 2k ;  k'' = k/2$$
C
$$k' = 2k ;  k'' = 2k$$
D
$$k' = k ;  k'' = k$$

Explanation

Rate of disappearance of reactants = Rate of appearance of products

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

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

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

$$ \Rightarrow $$ $$k' = 2k ;  k'' = k/2$$
2
MCQ (Single Correct Answer)

AIPMT 2011 Prelims

Which one of the following statements for the order of a reaction is incorrect?
A
Order can be determined only experimentally.
B
Order is not influenced by stoichiometric coefficient of the reactants.
C
Order of a reaction is sum of power to the concentration terms of reactants to express the rate of reaction.
D
Order of reaction is always whole number.

Explanation

Order of a reaction is not always whole number. It can be zero, or fractional also.
3
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?
A
Rate = k[A]2[B]
B
Rate = k[A][B]
C
Rate = k[A]2[B]2
D
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
4
MCQ (Single Correct Answer)

AIPMT 2010 Prelims

For the reaction N2O5(g) $$ \to $$  2NO2(g) + 1/2O2(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
A
6.25 $$ \times $$ 10$$-$$3 mol L$$-$$1 s$$-$$1 and
6.25 $$ \times $$ 10$$-$$3 mol L$$-$$1 s$$-$$1
B
1.25 $$ \times $$ 10$$-$$2 mol L$$-$$1 s$$-$$1 and
3.125 $$ \times $$ 10$$-$$3 mol L$$-$$1 s$$-$$1
C
6.25 $$ \times $$ 10$$-$$3 mol L$$-$$1 s$$-$$1 and
3.125 $$ \times $$ 10$$-$$3 mol L$$-$$1 s$$-$$1
D
1.25 $$ \times $$ 10$$-$$2 mol L$$-$$1 s$$-$$1 and
6.25 $$ \times $$ 10$$-$$3 mol L$$-$$1 s$$-$$1

Explanation

N2O5(g) $$ \to $$  2NO2(g) + 1/2O2(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

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