1
MCQ (Single Correct Answer)

NEET 2016 Phase 1

The addition of a catalyst during a chemical reaction alters which of the following quantities?
A
Enthalpy
B
Activation energy
C
Entropy
D
Internal energy

Explanation

A catalyst provides an alternate path to the reaction which has lower activation energy.
2
MCQ (Single Correct Answer)

NEET 2016 Phase 2

The decomposition of phosphine (PH3) on tungsten at low pressure is a first-order reaction. It is because the
A
rate is proportional to the surface coverage
B
rate is inversely proportional to the surface coverage
C
rate is independent of the surface coverage
D
rate of decomposition is very slow.

Explanation

At low pressure, rate is proportional to the surface coverage and is of first order while at high pressure it follows zero order kinetic due to complete coverage of surface area.
3
MCQ (Single Correct Answer)

NEET 2016 Phase 1

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
4
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 $${ - {{{E_a}} \over R}}$$.

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