Chemical Kinetics — Page 4 | JEE Chemistry

Q31

The rate of a reaction A doubles on increasing the temperature from 300 to 310 K. By how much, the temperature of reaction B should be Increased from 300 K so that rate doubles if activation energy of the reaction B is twice to that of reaction A.

A 9.84 K

B 4.92 K

C 2.45 K

D 19.67 K

Correct Answer

Option B

Solution

For reaction A, T1 = 300 K, T2 = 310 K, k2 = 2 k1

\log {{{k_2}} \over {{k_1}}} = {{{E_{{a_1}}}} \over {2.303R}}\left[ {{1 \over {{T_1}}} - {1 \over {{T_2}}}} \right]

$\therefore$

\log {{2{k_1}} \over {{k_1}}} = \log 2 = {{{E_{{a_1}}}} \over {2.303R}}\left[ {{1 \over {300}} - {1 \over {310}}} \right]

...(1) For reaction B, T1 = 300 K, T2 = ?, k2 = 2k1, Ea2 = 2Ea1

\log {{2{k_1}} \over {{k_1}}} = \log 2 = {{2{E_{{a_1}}}} \over {2.303R}}\left[ {{1 \over {300}} - {1 \over {{T_2}}}} \right]

From eq. (i) and (ii), we get

{{2{E_{{a_1}}}} \over {2.303R}}\left[ {{1 \over {300}} - {1 \over {{T_2}}}} \right] = {{{E_{{a_1}}}} \over {2.303R}}\left[ {{1 \over {300}} - {1 \over {310}}} \right]

$\Rightarrow$

2\left[ {{1 \over {300}} - {1 \over {{T_2}}}} \right] = \left[ {{1 \over {300}} - {1 \over {310}}} \right]

$\Rightarrow$ T2 =

{{{300 \times 310} \over {610}}}

$\times$ 2 = 304.92 K $\therefore$ Increased temperature = (304.92 – 300) = 4.92 K

Q32

The rate of a reaction quadruples when the temperature changes from 300 to 310 K. The activation energy of this reaction is : (Assume activation energy and preexponential factor are independent of temperature; ln 2 = 0.693; R = 8.314 J mol−1 K−1)

A 107.2 kJ mol

-

1

B 53.6 kJ mol

-

1

C 26.8 kJ mol

-

1

D 214.4 kJ mol

-

1

Correct Answer

Option A

Solution

According to Arrhenius equation,

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

Substituting the given values in the equation, we get

\log 4 = {{{E_a}} \over {2.303 \times 8.314\,J\,{K^{ - 1}}mo{l^{ - 1}}}}\left( {{{310 - 300} \over {300 \times 310}}} \right)

{E_a} = {{0.602 \times 2.303 \times 8.314 \times 300 \times 310} \over {10}}

= 107197.12 J/mol $-$ 1 or 107.2 kJ/mol $-$ 1

Q33

Two reactions R1 and R2 have identical pre-exponential factors. Activation energy of R1 exceeds that of R2 by 10 kJ mol–1. If k1 and k2 are rate constants for reactions R1 and R2 respectively at 300 K, then ln(k2/k1) is equal to : (R = 8.314 J mol–1 K–1)

A 12

B 6

C 4

D 8

Correct Answer

Option C

Solution

We know, from arrhenius equation, k = A.

{e^{{{ - {E_a}} \over {RT}}}}

$\therefore$ k1 = A.

{e^{{{ - {E_{{a_1}}}} \over {RT}}}}

......(1) k2 = A.

{e^{{{ - {E_{{a_2}}}} \over {RT}}}}

......(2) On dividing equation (2) by (1), we get

{{{k_2}} \over {{k_1}}} = {e^{{{\left( {{E_{{a_1}}} - {E_{{a_2}}}} \right)} \over {RT}}}}

$\Rightarrow$

\ln \left( {{{{k_2}} \over {{k_1}}}} \right) = {{\left( {{E_{{a_1}}} - {E_{{a_2}}}} \right)} \over {RT}}

=

{{10,000} \over {8.314 \times 300}}

= 4

Q34

If 50% of a reaction occurs in 100 second and 75% of the reaction occurs in 200 secod, the order of this reaction is :

A Zero

B 1

C 2

D 3

Correct Answer

Option B

Solution

Assume initial concentration of the reactant = 1 M After 100 second, concentration becomes of the reactant = 1 $\times$

{{50} \over {100}}

= 0.5 M After 200 second, concentration of the reactant becomes = 1 $\times$

{{25} \over {100}}

= 0.25 M.

So, in first 100 second reactant concentration becomes half of initial and in the second 100 second concentration becomes 0.5 M to 0.25 M, which is also half of 0.5 M.

So, after each 100 second period, concentration of reactant becomes half of initial concentration.

So, 100 second is the half life period and it is independent of concentration of reactant.

This is characteristic of first order reaction.

Q35

For a first order reaction, A

\to

P, t1/2 (half-life) is 10 days The time required for

{1 \over 4}

th conversion of A (in days) is : (ln 2 = 0.693, ln 3 = 1.1)

A 5

B 3.2

C 4.1

D 2.5

Correct Answer

Option C

Solution

For first order reaction, The half life, t

{1 \over 2}

=

{{0.693} \over k}

Here given t

{1 \over 2}

= 10 days

\therefore\,\,\,

k =

{{0.693} \over {10}}

= 0.0693 days $-$ 1 Now, the time required for

{1 \over 4}

th conversion of A is , t =

{{2.303} \over k}

log10

\left( {{a \over {a - x}}} \right)

=

{{2.303} \over {0.0693}}

log

\left( {{1 \over {1 - {1 \over 4}}}} \right)

=

{{2.303} \over {0.0693}}

log

\left( {{4 \over 3}} \right)

= 4.1 days.

Q36

If a reaction follows the Arrhenius equation, the plot ln k vs

{1 \over {\left( {RT} \right)}}

gives straight line with a gradient (

-

y) unit. The energy required to active the reactant is :

A y unit

B y/R unit

C yR unit

D

-

y unit

Correct Answer

Option A

Solution

According to Arrhenius equation, k = A

{e^{ - {{{E_a}} \over {RT}}}}

or ln k = ln A -

{{{{E_a}} \over {RT}}}

Comparing the above equation with straight line equation, y = mx + c, we get, slope or gradient (m) = –Ea and Intercept (c) = ln A Also given that slope or gradient (m) = -y $\therefore$ -y = –Ea $\Rightarrow$ Ea = y So the activation energy of the reactant, Ea = y unit

Q37

NO2 required for a reaction is produced by the decomposition of N2O5 in CCl4 as per the equation, 2N2O5(g)

\to

4NO2(g) + O2(g). The initial concentration of N2O5 is 3.00 mol L–1 and it is 2.75 mol L–1 after 30 minutes. The rate of formation of NO2 is :

A 2.083 × 10–3 mol L–1 min–1

B 8.333 × 10–3 mol L–1 min–1

C 4.167 × 10–3 mol L–1 min–1

D 1.667 × 10–2 mol L–1 min–1

Correct Answer

Option D

Solution

Rate of disappearance of N2O5 =

- {{\Delta \left[ {{N_2}{O_5}} \right]} \over {\Delta \left[ t \right]}}

=

- {{\left[ {2.75 - 3} \right]} \over {30}}

=

+ {1 \over {120}}

mol L-1 min-1 Rate of formation of NO2 =

+ {{\Delta \left[ {N{O_2}} \right]} \over {\Delta \left[ t \right]}}

As

- {1 \over 2}{{\Delta \left[ {{N_2}{O_5}} \right]} \over {\Delta \left[ t \right]}} = + {1 \over 4}{{\Delta \left[ {N{O_2}} \right]} \over {\Delta \left[ t \right]}}

$\Rightarrow$

{{\Delta \left[ {N{O_2}} \right]} \over {\Delta \left[ t \right]}}

=

+ {1 \over {120}} \times {4 \over 2}

= 0.01667 = 1.667 × 10–2 mol L–1 min–1

Q38

For the reaction of H2 with I2, the rate constant is 2.5 × 10–4 dm3 mol–1s–1 at 327°C and 1.0 dm3 mol–1 at 527°C. The activation energy for the reaction, in kJ mole–1 is : (R = 8.314 JK–1 mol–1 )

A 59

B 166

C 72

D 150

Correct Answer

Option B

Solution

From Arrhenius equation, we get

\log {{{K_2}} \over {{K_1}}} = {{{E_a}} \over {2.303R}}\left( {{1 \over {{T_1}}} - {1 \over {{T_2}}}} \right)

$\Rightarrow$

\log {1 \over {2.5 \times {{10}^{ - 4}}}} = {{{E_a}} \over {2.303 \times 8.314}}\left( {{1 \over {600}} - {1 \over {800}}} \right)

$\Rightarrow$ 3.6 =

{{{E_a}} \over {2.303 \times 8.314}} \times {{200} \over {600 \times 800}}

$\Rightarrow$ E

a

= 166 kJ/mol

Q39

For a reaction scheme

A\overset{{{k_1}}}\longrightarrow B\overset{{{k_2}}}\longrightarrow C

, if the rate of formation of B is set to be zero then the concentration of B is given by :

A

{k_1}{k_2}[A]

B

\left( {{{{k_1}} \over {{k_2}}}} \right)[A]

C

({k_1} + {k_2})[A]

D

({k_1} - {k_2})[A]

Correct Answer

Option B

Solution

Given, the rate of formation of B is set to be zero. $\therefore$

{{d\left[ B \right]} \over {dt}} = 0

For this reaction,

A\mathrel{\mathop{\kern0pt\longrightarrow} \limits_{{R_1}}^{{k_1}}} B

Rate of reaction for this reaction (R1) =

{k_1}\left[ A \right]

For this reaction,

B\mathrel{\mathop{\kern0pt\longrightarrow} \limits_{{R_2}}^{{k_2}}} C

Rate of reaction for this reaction (R2) =

{k_2}\left[ B \right]

Net rate of formation of B =

{{d\left[ B \right]} \over {dt}}

= R1 - R2 As

{{d\left[ B \right]} \over {dt}} = 0

$\therefore$ R1 - R2 = 0 $\Rightarrow$

{k_1}\left[ A \right]

-

{k_2}\left[ B \right]

= 0 $\Rightarrow$

\left[ B \right] = {{{k_1}} \over {{k_2}}}\left[ A \right]

Q40

Decomposition of X exhibits a rate constant of 0.05

\mu

g/year. How many year are required for the decomposition of 5

\mu

g of X into 2.5

\mu

g?

A 50

B 20

C 25

D 40

Correct Answer

Option A

Solution

According to unit of rate constant it is a zero order reaction, then half life of zero order reaction. t1/2 =

{{{a_0}} \over {2k}} = {5 \over {2 \times 0.05}}

= 50 years