Chemical Kinetics — Page 8 | JEE Chemistry

Q71

Consider the reaction, 2A + B

\to

products. When concentration of B alone was doubled, the half-life did not change. When the concentration of A alone was doubled, the rate increased by two times. The unit of rate constant for this reaction is

A L mol-1 s-1

B no unit

C mol L-1 s-1

D s-1

Correct Answer

Option A

Solution

Rate = k [A]x[B]y When [B] is doubled, keeping [A] constant half-life of the reaction does not change.

For a first order reaction

{t_{1/2}} = {{0.693} \over K}\,\,

i.e. for a first order reaction

{t_{1/2}}

does not depend up on the concentration.

Hence the reaction is first order with respect to B.

Now when [A] is doubled, keeping [B] constant, the rate also doubles.

Hence the reaction is first order with respect to A.

$\therefore$

\,\,\,\,\,

Rate = k [A]1[B]1 Order of reaction

=1+1=2

Now for a nth order reaction, unit of rate constant is (L)n–1 (mol)1–n s–1 when n = 2, unit of rate constant is

L\,\,mo{l^{ - 1}}\,{\sec ^{ - 1}}.

Q72

Isotope(s) of hydrogen which emits low energy

\beta

-

particles with t1/2 value > 12 years is/are

A Protium

B Tritium

C Deuterium

D Deuterium and Tritium

Correct Answer

Option B

Solution

_1^1

H and

_1^2

H are stable while

_1^3

H is radioactive.

Q73

A student has studied the decomposition of a gas AB

_3

at 25

^\circ

C. He obtained the following data. .tg .tg p (mm Hg) 50 100 200 400 relative t

_{1/2}

(s) 4 2 1 0.5 The order of the reaction is

A 2

B 0.5

C 1

D 0 (zero)

Correct Answer

Option A

Solution

\begin{aligned} & \mathrm{t}_{1 / 2} \propto\left(\mathrm{P}_{\mathrm{o}}\right)^{1-\mathrm{n}} \\\\ & \frac{\left(\mathrm{t}_{1 / 2}\right)_1}{\left(\mathrm{t}_{1 / 2}\right)_2}=\frac{\left(\mathrm{P}_o\right)_1^{1-\mathrm{n}}}{\left(\mathrm{P}_{o_2}\right)_2^{1-\mathrm{n}}} \\\\ & \Rightarrow\left(\frac{4}{2}\right)=\left(\frac{50}{100}\right)^{1-\mathrm{n}} \\\\ & \Rightarrow 2=\left(\frac{1}{2}\right)^{1-\mathrm{n}} \\\\ & \Rightarrow 2=(2)^{\mathrm{n}-1} \\\\ & \Rightarrow \mathrm{n}-1=1 \\\\ & \Rightarrow \mathrm{n}=2 \end{aligned}

Q74

Drug

X

becomes ineffective after

50 \%

decomposition. The original concentration of drug in a bottle was

16 \mathrm{mg} / \mathrm{mL}

which becomes

4 \mathrm{mg} / \mathrm{mL}

in 12 months. The expiry time of the drug in months is _________. Assume that the decomposition of the drug follows first order kinetics.

A 12

B 3

C 6

D 2

Correct Answer

Option C

Solution

Original concentration,

a = 16

mg/mL Concentration at time t = 12 months,

a - x = 4

mg/mL For first order kinetics, the equation for late constant

k = {{2.303} \over t}\log {a \over {a - x}}

a $\to$ Initial concentration $a-x\to$ Concentration at time, t

= {{2.303} \over {12\,months}}\log {{16\,mg/mL} \over {4\,mg/mL}}

= {{2.303} \over {12}}\log 4

= 0.1155\,mon}

Drug is ineffective after 50% decomposition.

So, the expiry time is after 50% decomposition.

The time at which 50% decomposition occurs in a reaction is known as half life

({t_{1/2}})

. So,

{t_{1/2}}

is the expiry time of the drug.

{t_{1/2}}

formula is

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

Substitute k = 0.1155 months $^{-1}$ ,

{t_{1/2}} = {{0.613} \over {0.1155}}

months $^{-1}$ = 6 months Expiry time can also be calculate as,

t = {{2.303} \over k}\log {a \over {a - x}}

Initial is taken as 100% The percent at expiry time is 50%

= {{2.303} \over {0.1155\,mon}} \times \log {{100\% } \over {50\% }}

= {{2.303} \over {0.1155}} \times \log {{100} \over {50}}

= {{2.303} \over {0.1155}} \times 0.301

= 6.002

= 6 months Correct answer is option (3) 6

Q75

Rate of a reaction can be expressed by Arrhenius equation as:

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

$ In this equation, E represents

A the energy above which all the colliding molecules will react

B the energy below which colliding molecules will not react

C the total energy of the reacting molecules at a temperature, T

D the fraction of molecules with energy greater than the activation energy of the reaction

Correct Answer

Option A

Solution

In Arrhenius equation

K = A\,{e^{ - E/RT}},\,\,E

is the energy of activation, which is required by the colliding molecules to react resulting in the formation of products.

Q76

At 518oC the rate of decomposition of a sample of gaseous acetaldehyde initially at a pressure of 363 Torr, was 1.00 Torr s–1 when 5% had reacted and 0.5 Torr s–1 when 33% had reacted. The order of the reaction is

A 0

B 2

C 3

D 1

Correct Answer

Option B

Solution

For a nth order reaction, the rate of reaction at time t , Rate = K [Pt] n Here Pt = pressure at time t, k = constant.

Note : Here instead of concentration of product, pressure of product is given.

When 5% is reacted at a rate 1 Toss S $-$ 1 Then un-reacted is 95%..

As initial pressure is 363 Torr then after 5% reaction completed the pressure will be = 363 $\times$

{{95} \over {100}}

Torr.

\therefore\,\,\,

1 = K

{\left[ {363 \times {{95} \over {100}}} \right]^n}

. . . . . . . .(

1) When 33% is reacted at a rate 0.5 , Torr S $-$ 1 then un-reacted is 67% So, after 33% reaction completion, the pressure is =

363 \times {{67} \over {100}}

Torr.

\therefore\,\,\,

0.5 = K

{\left[ {363 \times {{67} \over {100}}} \right]^n}.......

(2) Dividing (1) by (2), we get

{1 \over {0.5}} = {{{{\left[ {363 \times {{95} \over {100}}} \right]}^n}} \over {{{\left[ {363 \times {{67} \over {100}}} \right]}^n}}}

\Rightarrow \,\,2 = {\left[ {{{95} \over {67}}} \right]^n}

\Rightarrow \,\,

2 =

{\left[ {1.41} \right]^n}

\Rightarrow \,\,

2 =

{\left[ {\sqrt 2 } \right]^n}

\Rightarrow \,\,\,\,2 = {2^{{n \over 2}}}

\therefore\,\,\,

{n \over 2}

= 1

\Rightarrow \,\,\,\,

n = 2 This is a 2nd order reaction.

Q77

Rate law for a reaction between

A

and

B

is given by

\mathrm{r}=\mathrm{k}[\mathrm{~A}]^{\mathrm{n}}[\mathrm{~B}]^{\mathrm{m}}

If concentration of

A

is doubled and concentration of

B

is halved from their initial value, the ratio of new rate of reaction to the initial rate of reaction

\left(\dfrac{r_2}{r_1}\right)

is

A

(\mathrm{n}-\mathrm{m})

B

2^{(\mathrm{n}-m)}

C

\dfrac{1}{2^{m+n}}

D

(\mathrm{m}+\mathrm{n})

Correct Answer

Option B

Solution

\mathrm{r}_1=\mathrm{k}[\mathrm{~A}]^{\mathrm{n}}[\mathrm{~B}]^{\mathrm{m}}

Now $A$ is doubled \& B is halved in concentration

\Rightarrow \mathrm{r}_2=\mathrm{k} 2^{\mathrm{n}}[\mathrm{~A}]^{\mathrm{n}} \cdot \frac{[\mathrm{~B}]^{\mathrm{m}}}{2^{\mathrm{m}}}

Now $\dfrac{r_2}{r_1}=2^{(n-m)}$

Q78

In a reaction

A+B \rightarrow C

, initial concentrations of

A

and

B

are related as

[A]_0=8[B]_0

. The half lives of

A

and

B

are 10 min and 40 min , respectively. If they start to disappear at the same time, both following first order kinetics, after how much time will the concentration of both the reactants be same?

A 20 min

B 40 min

C 80 min

D 60 min

Correct Answer

Option B

Solution

The initial conditions are: $[A]_0 = 8[B]_0$ Half-life of A, $\left[\text{t}_{1/2}\right]_{A} = 10 \text{ min}$ Half-life of B, $\left[\text{t}_{1/2}\right]_{B} = 40 \text{ min}$ Both reactants follow first-order kinetics, and we want to find the time, $t$ , when $[A]_t = [B]_t$ .

Step-by-step Derivation: Write the First-order Rate Equation: For first-order reactions, the concentration $[X]_t$ at time $t$ is given by: $[X]_t = [X]_0 e^{-k_X t}$ where $k_X$ is the rate constant for substance $X$ .

Express Half-life through Rate Constant: For first-order reactions, the rate constant $k$ is related to the half-life $\text{t}_{1/2}$ by: $k = \dfrac{\ln 2}{\text{t}_{1/2}}$ Substituting the known half-lives: $k_A = \dfrac{\ln 2}{10}$ $k_B = \dfrac{\ln 2}{40}$ Set Up the Equality from the Condition $[A]_t = [B]_t$ : $[A]_0 e^{-k_A t} = [B]_0 e^{-k_B t}$ By inserting the initial condition $[A]_0 = 8[B]_0$ , it becomes: $8[B]_0 e^{-k_A t} = [B]_0 e^{-k_B t}$ Simplify and Solve for $t$ : Divide both sides by $[B]_0$ : $8 e^{-k_A t} = e^{-k_B t}$ Taking the natural logarithm of both sides: $\ln 8 = -k_A t + k_B t$ Replace $k_A$ and $k_B$ with their expressions: $\ln 8 = \left(\dfrac{\ln 2}{10} - \dfrac{\ln 2}{40}\right) t$ Simplify Further: $\ln 8 = \ln 2 \left(\dfrac{1}{10} - \dfrac{1}{40}\right) t$ Simplify $\left(\dfrac{1}{10} - \dfrac{1}{40}\right)$ to get: $\dfrac{1}{10} - \dfrac{1}{40} = \dfrac{4 - 1}{40} = \dfrac{3}{40}$ So, $\ln 8 = \ln 2 \times \dfrac{3}{40} \times t$ Final Expression for $t$ : $t = \dfrac{\ln 8}{\ln 2 \times \dfrac{3}{40}}$ Calculate the value of $t$ using $\ln 8 = 3 \ln 2$ : $t = \dfrac{3 \ln 2}{\ln 2 \times \dfrac{3}{40}} = \dfrac{40}{1} = 40 \text{ min}$ Therefore, after 40 minutes, the concentration of both reactants A and B will be the same.

Q79

For the following reactions

A\overset{{700K}}\longrightarrow {\mathop{\rm Product}\nolimits}

A\mathrel{\mathop{\kern0pt\longrightarrow} \limits_{catalyst}^{500K}} {\mathop{\rm Product}\nolimits}

it was found that Ea is decreased by 30 kJ/mol in the presence of catalyst. If the rate remains unchanged, the activation energy for catalysed reaction is (Assume pre exponential factor is same):

A 198 kJ/mol

B 135 kJ/mol

C 105 kJ/mol

D 75 kJ/mol

Correct Answer

Option D

Solution

K1 = A

{e^{ - {{{E_a}} \over {R \times 700}}}}

K2 = A

{e^{ - {{\left( {{E_a} - 30} \right)} \over {R \times 500}}}}

$\because$ Rate is same $\therefore$ Rate constant will also be same K1 = K2 A

{e^{ - {{{E_a}} \over {R \times 700}}}}

= A

{e^{ - {{\left( {{E_a} - 30} \right)} \over {R \times 500}}}}

$\Rightarrow$

{{{\left( {{E_a} - 30} \right)} \over {R \times 500}}}

=

{{{{E_a}} \over {R \times 700}}}

$\Rightarrow$ 5Ea = 7Ea – 210 $\Rightarrow$ 210 = 2Ea $\Rightarrow$ Ea = 105 kJ/mole $\therefore$ Activation energy in the presence of catalyst = 105 – 30 = 75 kJ/mol

Q80

If half-life of a substance is 5 yrs, then the total amount of substance left after 15 years, when initial amount is 64 grams is

A 16 grams

B 2 grams

C 32 grams

D 8 grams

Correct Answer

Option D

Solution

{t_{1/2}} = 5\,\,

years,

T=15

years hence - total number of half life periods

= {{15} \over 3} = 3.

$\therefore$ Amount left

= {{64} \over {{{\left( 2 \right)}^3}}} = 8g