Chemical Kinetics — Page 3 | NEET Chemistry

Q21

The decomposition of phosphine (PH 3 ) 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.

Correct Answer

Option A

Solution

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.

Q22

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

Correct Answer

Option C

Solution

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

Q23

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

Correct Answer

Option B

Solution

A catalyst provides an alternate path to the reaction which has lower activation energy.

Q24

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

Correct Answer

Option C

Solution

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

Q25

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

Correct Answer

Option A

Solution

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}}

.

Q26

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.

Correct Answer

Option D

Solution

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

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

Q27

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

Correct Answer

Option A

Solution

For a first order reaction, t 75% = 2 $\times$ t 50%

Q28

For a reaction between A and B the order with respect to A is 2 and the other with respect to B is 3. The concentrations of both A and B are doubled, the rate will increase by a factor of

A 12

B 16

C 32

D 10

Correct Answer

Option C

Solution

Rate 1 = k[A] 2 [B] 3 when concentrations of both A and B are doubled then Rate 2 = k[2A] 2 [2B] 3 = 32 k[A] 2 [B] 3 $\therefore$ Rate will increase by a factor of 32.

Q29

What is the activation energy for a reaction if its rate doubles when the temperature is raised from 20 o C to 35 o C? (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

Correct Answer

Option A

Solution

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

Initial temperature, T 1 = 20 + 273 = 293 K Final temperature, T 2 = 35 + 273 = 308 K R = 8.314 JK –1 mol –1 As rate becomes double on raising temperature $\therefore$ r 2 = 2r 1 As rate constant, k $\infty$ r k 2 = 2k 1 $\therefore$

\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

Q30

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]

Correct Answer

Option D

Solution

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) = k2 x 2 y [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 = 2 y $\Rightarrow$ 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]