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1

NEET 2016 Phase 1

MCQ (Single Correct Answer)
The rate of first-order reaction is 0.04 mol L$$-$$1 s$$-$$1 at 10 seconds and 0.03 mol L$$-$$1 s$$-$$1 at 20 seconds after initiation of the reaction. The half-life period of the reaction is
A
44.1 s
B
54.1 s
C
24.1 s
D
34.1 s

Explanation

For the first order reaction

A $$ \to $$ Product

Rate $$ \propto $$ [A]

k = $${{2.303} \over {{t_2} - {t_1}}}\log {{{{\left( {rate} \right)}_1}} \over {{{\left( {rate} \right)}_2}}}$$

= $${{2.303} \over {20 - 10}}\log {{0.04} \over {0.03}}$$

= 0.0287 sec-1

$${t_{{1 \over 2}}} = {{0.693} \over k} = {{0.693} \over {0.0287}}$$ = 24.14 sec
2

AIPMT 2015 Cancelled Paper

MCQ (Single Correct Answer)
When initial concentration of a reactant is doubled in a reaction, its half-life period is not affected. The order of the reaction is
A
second
B
more than zero but less than first
C
zero
D
first.

Explanation

Half-life period of a first order reaction is independent of initial concentration,

$${t_{{1 \over 2}}} = {{0.693} \over k}$$
3

AIPMT 2015 Cancelled Paper

MCQ (Single Correct Answer)
The activation energy of a reaction can be determined from the slope of which of the following graphs?
A
ln k vs. $${1 \over T}$$
B
$${T \over {\ln \,k}}$$ vs. $${1 \over T}$$
C
ln k vs. $$T$$
D
$${{\ln k} \over T}$$ vs. $$T$$

Explanation

According to Arrhenius equation,

$$k = A{e^{ - {{{E_a}} \over {ET}}}}$$

Taking natural log on both the sides we get,

ln k = ln A $${ - {{{E_a}} \over {ET}}}$$ ...........(1)

Comparing (1) with standard form of equation of line

y = mx + C

We get Slope, m = $${ - {{{E_a}} \over R}}$$

Hence, if ln k is plotted against 1/T, slope of the line will be $${ - {{{E_a}} \over R}}$$.
4

AIPMT 2015

MCQ (Single Correct Answer)
The rate constant of the reaction A $$ \to $$ B is 0.6 $$ \times $$ 10$$-$$3 mol L$$-$$1 s$$-$$1. If the concentration of A is 5 M, then concentration of B after 20 minutes is
A
3.60 M
B
0.36 M
C
0.72 M
D
1.08 M

Explanation

For zero order reaction unit of rate constant is mole per second.

$$ \therefore $$ For zero order

x = kt

$$ \Rightarrow $$ x = 0.6 × 10–3 × 20 × 60 = 0.72 M

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