Chemical Kinetics — Page 9 | NEET Chemistry

Q81

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

Q82

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.

Q83

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.

Q84

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

Q85

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.

Q86

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

Q87

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

Q88

The integrated rate equation is Rt = log C0 - log Ct . The straight line graph is obtained by plotting

A time vs log Ct

B

{1 \over {time}}

vs Ct

C time vs Ct

D

{1 \over {time}}

vs

{1 \over {{C_t}}}

Correct Answer

Option A

Solution

Rt = \log {C_o} - {{\mathop{\rm logC}\nolimits} _t}

It is clear from the equation that if we plot a graph between

\log \,{C_t}

and time, a straight line with a slope equal to

- {k \over {2.303}}

and intercept equal to

\log \,\left[ {{A_o}} \right]

will be obtained.

Q89

N2O5 decomposes to NO2 and O2 and follows first order kinetics. After 50 minutes, the pressure inside the vessel increases from 50 mmHg to 87.5 mmHg. The pressure of the gaseous mixture after 100 minute at constant temperature will be :

A 175.0 mmHg

B 116.25 mmHg

C 136.25 mmHg

D 106.25 mmHg

Correct Answer

Option D

Solution

N2O5 $\rightleftharpoons$ 2NO2 +

{1 \over 2}

O2 At t = 0 50 0 0 At t = 50 min 50 $-$ P1 2P1

{{{P_1}} \over 2}

Total pressure after 50 min, = 50 $-$ P1 + 2P1 +

{{{P_1}} \over 2}

= 87.5

\therefore\,\,\,

50 +

{{3{P_1}} \over 2}

= 87.5 $\Rightarrow$

\,\,\,

P1 = 25 So, 50 min is the Half $-$ life period of the reaction.

Then 100 min is two half life .tg .tg N2O5 $\rightleftharpoons$ 2NO2 +

{1 \over 2}

O2 At t = 100 min 50 $-$ P2 2P2

{{{P_2}} \over 2}

\therefore\,\,\,

50 $-$ P2 =

{{25} \over 2}

P2 = 37.5

\therefore\,\,\,

Total pressure at t = 100 min = 50 $-$ P2 + 2P2 +

{{{P_2}} \over 2}

= 50 +

{3 \over 2}

$\times$ 37.5 = 106.25 mm of Hg

Q90

For a first order reaction, the time required for completion of 90% reaction is 'x' times the half life of the reaction. The value of 'x' is (Given : ln 10 = 2.303 and log 2 = 0.3010)

A 1.12

B 2.43

C 3.32

D 33.31

Correct Answer

Option C

Solution

\mathrm{A} \rightarrow

Products For a first order reaction,

\mathrm{t}_{1 / 2}=\frac{\ln 2}{\mathrm{k}}=\frac{0.693}{\mathrm{k}}

Time for

90 \%

conversion,

t_{90 \%}=\frac{1}{k} \ln \frac{100}{10}=\frac{\ln 10}{k}=\frac{2.303}{k}

t_{90\%}=\frac{2.303}{0.693} t_{1 / 2}=3.32 t_{1 / 2}