1
MCQ (Single Correct Answer)

AIPMT 2015 Cancelled Paper

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 $${ - {{{E_a}} \over R}}$$.
2
MCQ (Single Correct Answer)

AIPMT 2015

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
3
MCQ (Single Correct Answer)

NEET 2013 (Karnataka)

A reaction is 50% complete in 2 hours and 75% complete in 4 hours. The order of reaction is
A
1
B
2
C
3
D
0

Explanation

For a first order reaction,

t75% = 2 $$ \times $$ t50%
4
MCQ (Single Correct Answer)

NEET 2013

What is the activation energy for a reaction if its rate doubles when the temperature is raised from 20oC to 35oC?
(R = 8.314 J mol$$-$$1 K$$-$$1)
A
34.7 kJ mol$$-$$1
B
15.1 kJ mol$$-$$1
C
342 kJ mol$$-$$1
D
269 kJ mol$$-$$1

Explanation

$$\log {{{k_2}} \over {{k_1}}} = {{{E_a}} \over {2.303R}}\left( {{1 \over {{T_1}}} - {1 \over {{T_2}}}} \right)$$

Initial temperature, T1 = 20 + 273 = 293 K

Final temperature, T2 = 35 + 273 = 308 K

R = 8.314 JK–1 mol–1

As rate becomes double on raising temperature

$$ \therefore $$ r2 = 2r1

As rate constant, k $$\infty $$ r

k2 = 2k1

$$ \therefore $$ $$\log 2 = {{{E_a}} \over {2.303 \times 8.314}}\left( {{1 \over {293}} - {1 \over {308}}} \right)$$

$$ \Rightarrow $$ $$0.301 = {{{E_a}} \over {19.147}} \times {{15} \over {293 \times 308}}$$

$$ \Rightarrow $$ E$$a$$ = 34673 J mol–1 = 34.7 kJ mol–1

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