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1

AIPMT 2012 Prelims

MCQ (Single Correct Answer)
In a zero-order reaction, for every 10oC rise of temperature, the rate is doubled. If the temperature is increased from 10oC to 100oC, the rate of the reaction will become
A
256 times
B
512 times
C
64 times
D
128 times

Explanation

For energy 10° rise in temperature the rate of reaction doubles. So, rate = 2n
when, n = 1 rate = 21 = 2

when, temperature is increased from 10°C to 100°C change in temperature = 100 – 10 = 90°C

i.e., n = 9

So, rate = 29 = 512 times
2

AIPMT 2012 Prelims

MCQ (Single Correct Answer)
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]
3

AIPMT 2011 Mains

MCQ (Single Correct Answer)
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
4

AIPMT 2011 Mains

MCQ (Single Correct Answer)
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

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