NEW
New Website Launch
Experience the best way to solve previous year questions with mock tests (very detailed analysis), bookmark your favourite questions, practice etc...
1

### AIPMT 2012 Prelims

In a reaction, A + B $$\to$$ product, rate is doubled when the concentration of B is doubled, and rate increases by a factor of 8 when the concentration of both the reactants (A and B) are doubled, rate law for the reaction can be written as
A
rate = k[A][B]2
B
rate = k[A]2[B]2
C
rate = k[A][B]
D
rate = k[A]2[B]

## Explanation

Rate of reaction for A + B $$\to$$ Product

Rate = k[A]x[B]y …(1)

where, x and y are order w.r.t. A and B respectively. When the concentration of only B is doubled, the rate is doubled, so

R’ = k [A]x [2B]y = 2R …(2)

If concentration of both the reactants A and B are doubled then the rate increases by a factor of 8 so

R’’ = k[2A]x[2B]y = 8R ...(3)

= k2x 2y [A]x [B]y = 8R …(4)

From equation (1) and (2), we get

$${{2R} \over R} = {{{{\left[ A \right]}^x}{{\left[ {2B} \right]}^y}} \over {{{\left[ A \right]}^x}{{\left[ B \right]}^y}}}$$

$$\Rightarrow$$ 2 = 2y

$$\Rightarrow$$ y = 1

From equation (1) and (4), we get

$${{8R} \over R} = {{{2^x}{2^y}{{\left[ A \right]}^x}{{\left[ B \right]}^y}} \over {{{\left[ A \right]}^x}{{\left[ B \right]}^y}}}$$

$$\Rightarrow$$ 8 = $${{2^x}{2^y}}$$

Substituting the value of y gives

$${{2^x}{2^1}}$$ = 8

$$\Rightarrow$$ $${{2^x}}$$ = 4

$$\Rightarrow$$ x = 2

By replacing the values of x and y in

rate law; rate = k[A]2[B]
2

### AIPMT 2011 Mains

The unit of rate constant for a zero order reaction is
A
mol L$$-$$1 s$$-$$1
B
L mol$$-$$1 s$$-$$1
C
L2 mol$$-$$2 s$$-$$1
D
s$$-$$1

## Explanation

Rate = K[A]0

Unit of k = mol L–1 sec–1
3

### AIPMT 2011 Mains

The half-life of a substance in a certain enzyme-catalysed reaction is 138 s. The time required for the concentration of the substance to fall from 1.28 mg L$$-$$1 to 0.04 mg L$$-$$1 is
A
414 s
B
552 s
C
690 s
D
276 s

## Explanation

We know, $$N = {N_0}{\left( {{1 \over 2}} \right)^n}$$

Give N0(original amount) = 1.28 mg L–1

N (amount of substance left after time T) = 0.04 mg L–1

$$\therefore$$ $${{0.04} \over {1.28}} = {\left( {{1 \over 2}} \right)^n}$$

$$\Rightarrow$$ $${1 \over {32}} = {\left( {{1 \over 2}} \right)^n}$$

$$\Rightarrow$$ n = 5

$$\therefore$$ Time required = 5 × t1/2 = 5 × 138 = 690 s
4

### 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

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

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

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

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

### Joint Entrance Examination

JEE Main JEE Advanced WB JEE

### Graduate Aptitude Test in Engineering

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

NEET

Class 12