Chemical Kinetics — Page 6 | NEET Chemistry

Q51

The rate of a first order reaction is 1.5

\times

10

-

2 mol L

-

1 min

-

1 at 0.5 M concentration of the reactant. The half-life of the reaction is

A 0.383 min

B 23.1 min

C 8.73 min

D 7.53 min

Correct Answer

Option B

Solution

For a first order reaction, A $\to$ products Rate(r) = k[A] $\Rightarrow$ k =

{r \over {\left[ A \right]}}

$\Rightarrow$ k =

{{1.5 \times {{10}^{ - 2}}} \over {0.5}} = 3 \times {10^{ - 2}}

So,

{t_{1/2}} = {{0.693} \over {3 \times {{10}^{ - 2}}}}

= 23.1 min

Q52

The temperature dependence of rate constant (k) of a chemical reaction is written in terms of Arrhenius equation,

k = A \cdot {e^{ - E{}^ * /RT}}

. Activation energy (E

*

) of the reaction can be calculated by plotting

A

k\,\,vs\,\,T

B

k\,\,vs\,\,{1 \over {\log T}}

C

\log \,k\,\,vs\,\,{1 \over T}

D

\log \,k\,\,vs\,{1 \over {\log T}}

Correct Answer

Option C

Solution

Arrhenius equation k =

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

$\Rightarrow$ log k = log A -

{{{{E_a}} \over {2.303RT}}}

Comparing it with equation of straight line i.e., y = mx + C On plotting log k vs

{1 \over T}

, we get a straight line, the slope indicates the value of activation energy.

Q53

If the rate of the reaction is equal to the rate constant, the order of the reaction is

A 0

B 1

C 2

D 3

Correct Answer

Option A

Solution

As r = k[A] n if n = 0 r = k[A] 0 or r = k thus for zero order reactions rate is equal to the rate constant.

Q54

The reaction A

\to

B follows first order kinetics. The time taken for 0.8 mole of A to produce 0.6 mole of B is 1 hour. What is the time taken for conversion of 0.9 mole of A to produce 0.675 mole of B?

A 1 hour

B 0.5 hour

C 0.25 hour

D 2 hours

Correct Answer

Option A

Solution

For first order reaction k =

{{2.303} \over t}\log {{{{\left[ A \right]}_0}} \over {{{\left[ A \right]}_t}}}

$\Rightarrow$ k =

{{2.303} \over 1}\log {{0.8} \over {0.2}}

= 2.303 log 4 ....(

1) Let t 1 hour is required for changing the concentration of A from 0.9 mole to 0.675 mole of B.

Remaining mole of A = 0.9 – 0.675 = 0.225 $\therefore$ k =

{{2.303} \over {{t_1}}}\log {{0.9} \over {0.225}}

.....(2) From equation (1) and (2)

{{2.303} \over {{t_1}}}\log {{0.9} \over {0.225}}

= 2.303 log 4 $\Rightarrow$

{{t_1}}

= 1 hr

Q55

The activation energy for a simple chemical reaction A

\rightleftharpoons

B is E

a

in forward direction. The activation energy for reverse reaction

A is negative of E

a

B is always less than E

a

C can be less than or more than E

a

D is always double of E

a

Correct Answer

Option C

Solution

The energy of activation of reverse reaction is less than or more than energy of activation E

a

of forward reaction

\Delta H = {\left( {{E_a}} \right)_F} - {\left( {{E_a}} \right)_R}

Because it depends upon the nature of reaction. If

{\left( {{E_a}} \right)_F} > {\left( {{E_a}} \right)_R}

, reaction is endothermic. or

{\left( {{E_a}} \right)_F} < {\left( {{E_a}} \right)_R}

, reaction is exothermic.

Q56

2A

\to

B + C It would be a zero order reaction when

A the rate of reaction is proportional to square of concentration of A

B The rate of reaction remains same at any concentration of A

C the rate remains unchanged at any concentration of B and C

D the rate of reaction doubles if concentrations of B is increased to double.

Correct Answer

Option B

Solution

2A $\to$ B + C If it is zero order reaction r = k [A] o . i.e the rate remains same at any concentration of 'A'. i.e independent upon concentration of A.

Q57

When a bio-chemical reaction is carried out in laboratory, outside the human body in absence of enzyme, then rate of reaction obtained is 10

-

6 times, the activation energy of reaction in the presence of enzyme is

A 6/RT

B P is required

C different from E

a

obtained in laboratory

D can't say anything.

Correct Answer

Option C

Solution

From Arrhenius equation, k = Ae –E a /RT The activation energy of the reaction in the presence of enzyme is different from E a obtained in laboratory.

Q58

For the reaction; 2N 2 O 5

\to

4NO 2 + O 2 rate and rate constant are 1.02

\times

10

-

4 and 3.4

\times

10

-

5 sec

-

1 respectively, then concentration of N 2 O 5 at that time will be

A 1.732

B 3

C 1.02

\times

10

-

4

D 3.4

\times

10 5

Correct Answer

Option B

Solution

For the reaction; 2N 2 O 5 $\to$ 4NO 2 + O 2 This is a first order reaction. $\therefore$ rate = k [N 2 O 5 ] ; [N 2 O 5 ] =

{{rate} \over k}

=

{{1.02 \times {{10}^{ - 4}}} \over {3.4 \times {{10}^{ - 5}}}}

= 3

Q59

How enzymes increases the rate of reactions

A by lowering activation energy

B by increaing activation energy

C by changing equilibrium constant

D by forming enzyme substrate complex.

Correct Answer

Option A

Solution

Enzymes act like catalyst in biochemical reactions.

Presence of an enzyme increases the rate of reaction by lowering the activation energy of the reactant.

Q60

For the reaction H + + BrO

_3^ -

+ 3Br

-

\to

5Br 2 + H 2 O which of the following relation correctly represents the consumption and formation of products.

A

{{d\left[ {B{r^ - }} \right]} \over {dt}} = - {3 \over 5}{{d\left[ {B{r_2}} \right]} \over {dt}}

B

{{d\left[ {B{r^ - }} \right]} \over {dt}} = {3 \over 5}{{d\left[ {B{r_2}} \right]} \over {dt}}

C

{{d\left[ {B{r^ - }} \right]} \over {dt}} = - {5 \over 3}{{d\left[ {B{r_2}} \right]} \over {dt}}

D

{{d\left[ {B{r^ - }} \right]} \over {dt}} = {5 \over 3}{{d\left[ {B{r_2}} \right]} \over {dt}}

Correct Answer

Option A

Solution

- {1 \over 3}

{{d\left[ {B{r^ - }} \right]} \over {dt}} = + {1 \over 5}{{d\left[ {B{r_2}} \right]} \over {dt}}

$\Rightarrow$

{{d\left[ {B{r^ - }} \right]} \over {dt}} = - {3 \over 5}{{d\left[ {B{r_2}} \right]} \over {dt}}