MCQ Questions for Class 12 Engineering Chemistry Chemical Kinetics Quiz 15

For the following first-order competing reaction:

$A+Reagent\rightarrow product$

$B+Reagent\rightarrow Product$
the ratio of

${ K }_{ 1 }/{ K }_{ 2 }$ , if only

$50$ % of

$B$ will have been reacted when

$94$ % of

$A$ has been reacted is

$\left( \log { 2 } =0.3,\log { 3 } 0.48 \right)$

0%
$4.06$
0%
$0.246$
0%
$8.33$
0%
$0.12$

In the given first-order sequential reactions:

$A\xrightarrow [ ]{ { K }_{ 1 } } B\xrightarrow [ ]{ { K }_{ 2 } } C\xrightarrow [ ]{ { K }_{ 3 } } D$ , what is the ratio of number of atoms of

$A$ to the number of atoms of

$B$ after long time interval starting with pure

$A$ ?

$\left( { K }_{ 1 }\cfrac { \ln { 2 } }{ 1200 } ;{ K }_{ 2 }=\cfrac { \ln { 2 } }{ 30 } \right)$

0%
$0.67$
0%
$10$
0%
$20$
0%
$40$

Two reactions: (I)

$A\rightarrow products$ and (II)

$B\rightarrow products$ , follow first-order kinetics. The rate of reacion-I is doubled when temperature is raised from

$300$ to

$310K$ . The half-life for this reaction at

$310K$ is

$30min$ . At the same temperature,

$B$ decomposes twice as fast as

$A$ . If the energy of activation for the reaction II is half that of reaction I, the rate constant of reaction II at

$300K$ is

0%
$0.0233{ min }^{ -1 }$
0%
$0.0327{ min }^{ -1 }$
0%
$0.0164{ min }^{ -1 }$
0%
$0.0654{ min }^{ -1 }$

Explanation

For reaction (1)

$A \rightarrow Products$

$T_{1}$ = 300K ,

$T_{2}$ = 310K

Given :

$(\frac{k_{2}(310)}{k_{1}(300)})_{A} = 2$

Using Arrhenius equation for reaction (1):

$log (\frac{k_{2}(310)}{k_{1}(300)})_{A} = \frac{E_{a}(A)}{2.303R} (\frac{T_{2}-T_{1}}{T_{2}T_{1}})$

$log 2 = frac{E_{a}(A)}{2.303R} (\frac{T_{2}-T_{1}}{T_{2}T_{1}})$

(For reaction 2)

$\rightarrow$ Products

Since at 310 K, B decomposes twice as fast as A.

$\because [K_{2}(310K)]_{A} = 0.0231 min^{-1}$

$\because [K_{2}(310K)]_{B} = 2 \times 0.0231 min^{-1}$

Using Arrhenius equation for reaction (2)

$log (\frac{k_{2}(310)}{k_{1}(300)})_{B} = \frac{E_{a}(B)}{2.303R} (\frac{T_{2}-T_{1}}{T_{2}T_{1}})$

Given :

$(E_{a})_{B} = \frac{1}{2}(E_{a})_{A}$

Substittute the value of above equation and Comparing equations , we get

$(\frac{k_{2}(310)}{k_{1}(300)})_{B} = \frac{1}{2} \times log 2$

$(\frac{0.0462}{k_{1}(300)})_{B} = 1.414$

$\therefore {k_{1}(300)} = 0.0327 min^{-1}$

Surface-catalysed reactions that are incorporated by the product, obey the differential rate expression,

$\cfrac { dy }{ dt } =\cfrac { k\left[ { C }_{ 0 }-y \right] }{ 1+by }$ , where

${ C }_{ 0 }=$ initial concentration and

$k$ and

$b$ are constants. The half-life of reaction is

0%
$\left( 1+{ C }_{ 0 }b \right) \ln { 2 } -\cfrac { { C }_{ 0 }b }{ 2 }$
0%
$\left( 1-{ C }_{ 0 }b \right) \ln { 2 } +\cfrac { { C }_{ 0 }b }{ 2 }$
0%
$\left( 1-{ C }_{ 0 }b \right) \ln { 2 } -\cfrac { { C }_{ 0 }b }{ 2 }$
0%
$\left( 1+{ C }_{ 0 }b \right) \ln { 2 } +\cfrac { { C }_{ 0 }b }{ 2 } \quad$

A substance undergoes first-order decomposition. The decomposition follows two parallel first-order reaction with

${ K }_{ 1 }=1.26\times { 10 }^{ -4 }{ s }^{ -1 }$ for the formation of

$B$ and

${ K }_{ 2 }=3.15\times { 10 }^{ -4 }{ s }^{ -1 }$ for the formation of

$C$ . The percentage distribution of

$B$ and

$C$ are

0%
$29$ % $B$ , $71$ % $C$
0%
$75$ % $B$ , $25$ % $C$
0%
$90$ % $B$ , $10$ % $C$
0%
$60$ % $B$ , $40$ % $C$

Rate of an uncatalyzed first-order reaction at

$T$

$K$ is half of the rate of a catalyzed reaction at

$0.5T$

$T$ . If the catalyst lowers the threshold energy by

$20\ kcal$ , what is the activation energy of the uncatalyzed reaction?

$\left( T=300K,\ln { 2 } =0.7 \right)$

0%
$39.58\ kcal/mol$
0%
$19.58\ kcal/mol$
0%
$40.42\ kcal/mol$
0%
$20.42\ kcal/mol$

Explanation

Given ,

$\frac{k_{2}}{k_{1}} = 2$

$T_1 = 300\ K$

$T_{2} = \frac{T_{1}}{2}$

As we know,

$log(\frac{k_{2}}{k_{1}}) = \frac{E_{2}-E_{1}}{R}(\frac{1}{T_{2}}-\frac{1}{T_{1}})$

Putting values in the above equation, we get;

$log(2) = \frac{20-E_{1}}{1.987}(\frac{2}{T_1}-\frac{1}{T_1})$

$0.7 = \frac{20-E_{1}}{1.987}(\frac{2}{T_1}-\frac{1}{T_1})$

After solving, we get ;

Activation energy of the uncatalyzed reaction,

$E_{1} = 39.58\ kcal/mol$

Hence , option A is correct .

A first-order reaction:

$A\rightarrow B$ , activation energy is

$4.8kcal/mol$ . When a

$20$ % solution of

$A$ was kept at

${ 27 }^{ o }C$ for

$21.6min$ ,

$75$ % decomposition took place. What will be the percent decomposition in

$8.0min$ in a

$30$ % solution maintained at

${ 47 }^{ o }C$ ? Assume that activation energy remains constant in this range of temperature.

$(e=2.7)$

0%
$25$ %
0%
$50$ %
0%
$75$ %
0%
$87.5$ %

Decomposition of a non-volatile solute

$A$ into another non-volatile solute

$B$ and

$C$ , in aqueous solution follows first-order kinetics as:

$A\rightarrow 2B+C$
When one mole of

$A$ is dissolved in

$180g$ water and left for decomposition, the vapour pressure of solution was found to be

$20mm$

$Hg$ after

$12h$ . What is the vapour pressure of solution after

$24h$ ?
Assume constant temperature of

${ 25 }^{ o }C$ , throughout. The vapour pressure of water at

${ 25 }^{ o }C$ is

$24mm$

$Hg$

0%
$18mm$ $Hg$
0%
$19.2mm$ $Hg$
0%
$10mm$ $Hg$
0%
$16mm$ $Hg$

What is overall activation energy of reaction, if steps (1) to (4) are much faster than (5)?

0%
$\cfrac { { E }_{ { a }_{ 5 } }.{ E }_{ { a }_{ 3 } }.{ E }_{ { a }_{ 1 } }^{ 1/2 } }{ { E }_{ { a }_{ 4 } }.{ E }_{ { a }_{ 2 } }^{ 1/2 } }$
0%
${ E }_{ { a }_{ 5 } }+{ E }_{ { a }_{ 3 } }+\cfrac { 1 }{ 2 } { E }_{ { a }_{ 1 } }-{ E }_{ { a }_{ 2 } }- { E }_{ { a }_{ 4 } }$
0%
${ E }_{ { a }_{ 1 } }+{ E }_{ { a }_{ 3 } }+{ E }_{ { a }_{ 5 } }-{ E }_{ { a }_{ 2 } }-{ E }_{ { a }_{ 4 } }\quad$
0%
$-{ E }_{ { a }_{ 5 } }-{ E }_{ { a }_{ 3 } }-\cfrac { 1 }{ 2 } { E }_{ { a }_{ 1 } }+{ E }_{ { a }_{ 2 } }+\cfrac { 1 }{ 2 } { E }_{ { a }_{ 4 } }$

Arrhenius equation gives the change in rate constant (and hence rate of reaction) with temperature. If the activation energy of the reaction is found to be equal to

$RT$ , then

0%
The rate of reaction does not depend upon initial concentration
0%
The rate constant becomes about $35$ % of the Arrhenius constant $A$
0%
The rate constant becomes equal to $73$ % of the Arrhenius constant $A$
0%
The rate of the reaction becomes infinite or zero

Explanation

$k = Ae^{\dfrac{-E_{a}}{RT}}$

when

$E_{a} = RT$

$k = \dfrac{A}{e}$

$\Rightarrow k=0.37 \times A$

Hence the rate constant becomes approx. 35% of A.

The plot of concentration of a reactant vs time for a chemical reaction is shown below :
The order of this reaction with respect to the reactant is:

0%
$0$
0%
$1$
0%
$2$
0%
not possible to determine from this plot

Explanation

The order of this reaction with respect to the reactant is zero-order reaction is reaction in which the rate does not vary with the increase or decrease in the concentration of the reactants.

Therefore the rate of these reactions is always equal to the rate constant (

$K$ ) of the specific reactions (since the rate of these reactions is proportional to the 0th power of reactants concentration).

The Differential form of a zero-order reaction can be written as

Rate

$=\dfrac{dA}{dt}=k[A]^{0}=k$

Hence option A is correct.

Under what condition the order of the reaction,

$2HI(g)\rightarrow H_2(g)+I_2(g)$ , is zero.

0%
At high temperature
0%
At high partial pressure of HI
0%
At low partial pressure of HI
0%
At high partial pressure of $H_2$
0%
At high partial pressure of $I_2$

Explanation

The reaction,

$2HI(g)\overset{\Delta}{\longrightarrow}H_2(g)+I_2(g)$

is of zero-order in which HI is present at high partial pressure.

Desorption of a gas from metal surface follows first-order kinetics. The rate constant of desorption can be given by Arrhenius equation. If the desorption of hydrogen on manganese is found to increase

$10$ times on increasing the temperature from

$600$ to

$1000 K$ , the activation energy of desorption is :

0%
$6.0\ kcal/mol$
0%
$6.9\ kcal/mol$
0%
$3.0\ kcal/mol$
0%
$57.4\ kcal/mol$

For a first-order reaction,

0%
The degree of dissociation is equal to $\left(1-{e}^{-kt}\right)$
0%
A plot of reciprocal concentration of the reactant vs time gives a straight line
0%
The time taken for completion of $75\%$ reaction is thrice the ${t}_{1/2}$ of the reaction
0%
The pre-exponential factor in the Arrhenius equation has the dimension of time, ${T}^{-1}$

Explanation

(a) is correct because degree of dissociation =

$1-e^{-kt}$ at any time t.

(b) is wrong because plot of

$\log{\left[A\right]}$ vs

$t$ is a straight line

$75\%$ reaction is two half life.

(d) is correct because in

$k = Ae^{{-E_{a}}/{RT}}$ ,

$E_{a/RT}$ is dimensionless hence

$A$ has the unit of

$K$ .

Option A and D are correct.

For the combustion of carbon,

$\quad \Delta H=-ve$ and

$\Delta S=+ve$ and hence, thermodynamically the process is spontaneous at all temperatures. But coal stored in coal depots does not burn automatically because of

0%
very high threshold energy barrier
0%
thermodynamically stability of coal
0%
lower energy of activation needed for burning
0%
low temperature in coal depots

Explanation

$\Delta G = \Delta H - T \Delta S$

$\Delta H$ = -ve

$\Delta S$ = +ve

since it is given the process is spontaneous

$\Delta G$ will be more -ve.

$\therefore \Delta G$ must be -ve.

$\Delta G = -nFE$

Hence it requires very high threshold energy barrier.

Which of the following curves represents a zero order reaction?

Which of the following curves represents a first order reaction?

Explanation

$t_{1/2}$ is independent of

$a$ for the first order reaction. So, it will be a straight line parallel to concentration axis.

The reaction

$A+B\to C+D;\ \Delta H=25\ kJ/ mole$ should have an activation energy :

0%
$-25\ kJ/ mole$
0%
$< 25\ kJ/ mole$
0%
$> 25\ kJ/ mole$
0%
either answer $(B)$ or $(C)$ depending upon experiment

Which of the following statements is wrong about reactions?

0%
There can be only three values of molecularity, that is $1, 2,$ and $3$ .
0%
There can be only four values of order, that is , $0, 1, 2,$ and $3$ .
0%
There can be infinite number of values for order.
0%
The order involves rate while molecularity does not

Explanation

The order can be anything

$-ve$ fractional,

$O$ or

$+ve$ . Its maximum value is

$3$ But as we know between any two real no. infinite no are possible & since order can be fractional / any real no. so it has infinite values.

$50\%$ of a zero order reaction completes in

$10$ minutes.

$100\%$ of the same reaction shall complete in:

0%
$5\ min$
0%
$10\ min$
0%
$20\ min$
0%
$\infty$ time

Explanation

For zeroth order reaction:

$At=Ao-Kt$ ;

After

$50\%$ completion, At

$=\dfrac{Ao}{2}$

$A/q, \dfrac{Ao}{2}=Ao-K\times 10\Rightarrow K=\dfrac{Ao}{20}$

Now for

$100\%$ completion

$At=0$

$0=Ao-\dfrac{Ao}{20}\times t\Rightarrow t=20\min$

The rate for a first order reaction is

$0.6932\times 10^{-2}\ mol\ L^{-1}\ min^{-1}$ and the initial concentration of the reactant is

$0.1\ M. t_{\tfrac{1}{2}}$ is equal to:

0%
$0.6932\times 10^{-2}\ min$
0%
$0.6932\times 10^{-3}\ min$
0%
$10\ min$
0%
$6.932\ min$

Explanation

Rate

$=0.6932\times 10^{-2}mole/lit-\min =KA_o$

$A_{0}=0.1\ M, t/1/2=\dfrac{0.693}{K}=\dfrac{0.693}{\dfrac{0.693\times 10^{2}}{10^{-1}}}=10\min$

Hence, option C is correct.

For a zero order reaction with the initial reactant concentration a, the time for completion of the reaction is

0%
$k/a$
0%
$a/k$
0%
$2k/a$
0%
$a/2k$

Which of the following curves represent(s) a zero-order reaction ?

In which of the following reactions of the following orders the molecularity and order can never be same?

0%
Zero order
0%
First order
0%
Second order
0%
Third order

A plot of reactant concentration versus time for a reaction is a straight line with a negative slope giving the rate constant, and the intercept, giving the initial concentration of the reactant. The order of the reactant is:

0%
$0$
0%
$1$
0%
$2$
0%
None of these

For a reaction of the order of

$0.5$ , when the concentration of the reactant is doubled the rate

0%
doubles
0%
increases four times
0%
decrease four times
0%
increases $\sqrt{2}$ times

Explanation

The expression for the rate of the reaction is

$r=k[A]^n$

now,

$r=k[A]^{1/2}......(1)$

When the concentration of A is doubled, then, rate:

$r′=k[2A]^{1/2}......(2)$

Dividing (2) by (1), we get

$\dfrac{k[2A]^{1/2}}{k[A]^{1/2}}=\sqrt2$

The rate constant of a first-order reaction is

$6.93 \times 10^{-2} min^-$ . The half-life of the reaction is

0%
$10 \min$
0%
$100 \min$
0%
$1 \min$
0%
$1000 \min$

Explanation

As we know that Half-life of first order reaction is given by:

$t_{1/2} = \cfrac{0.693}{k}$

$\implies t_{1/2} = \cfrac{0.693}{6.93 \times 10^{-2}}$

$\implies t_{1/2} = 10 \ min$

Hence, Option "A" is the correct answer.

For a first order reaction

$A\xrightarrow [ ]{ k } B$ , the degree of dissociation is equal to

0%
$e^{-kt}$
0%
$1-e^{-kt}$
0%
$e^{kt}$
0%
$1+e^{-kt}$

The order w. r. t A is :

0%
$1$
0%
$2$
0%
$3$
0%
$-1$

Explanation

$R = k [A]^m [B]^n$

$5 \times 10^{-4} = k[2.5 \times 10^{-4}]^m [3 \times 10^{-5}]^n$ .....(i)

$4 \times 10^{-3} = k[5 \times 10^{-4}]^m [6 \times 10^{-5}]^n$ .......(ii)

$1.6 \times 10^{-2} = k[1 \times 10^{-3}]^m [6 \times 10^{-5}]^n$ .......(iii)

Using equation (ii) and (iii) we get,

m = 2.

Option B is correct.

The time for the half-life period of a certain reaction

$A\to$ products is

$1$ hour. When the initial concentration of the reactant,

$A$ is

$2.0\ mol\ L^{-1}$ . How much time does it take initial concentration to make from

$0.50$ to

$0.25\ mol\ L^{-1}$ if it is a zero-order reaction?

0%
$4\ h$
0%
$0.5\ h$
0%
$0.25\ h$
0%
$1\ h$

Explanation

The initial concentration of A is

$\mathrm{2 \ M}$ .

If the reaction would be first order reaction, then

$\mathrm{k = \cfrac{a_o - 0.5a_o}{t} = 1}$

Time taken for the reactant concentration to decrease from 0.5 to 0.25;

$\mathrm{t = \cfrac{0.5 - 0.25}{k} = \cfrac{0.25}{1}}$

So, the time taken for reactant concentration to decrease from 0.5 M to 0.25 M is

$0.25 \ h$ .

Hence, Option "C" is the correct answer.

The number of collisions depend upon

0%
Pressure
0%
Concentration
0%
Temperature
0%
All the above

The rate constant is doubled when temperature increases from

$27^{o}C$ to

$37^{o}C$ . Activation energy in

$kJ$ is:

0%
$34$
0%
$53.6$
0%
$100$
0%
$50$

Explanation

$\log{\dfrac{K_2}{K_1}}=\dfrac{E_a}{2.303R}\left[\dfrac{1}{T_1}-\dfrac{1}{T_2}\right]$

$\dfrac{K_2}{K_1}=2$

$\log{2}=\dfrac{E_a}{2.303\times8.314}\left[\dfrac{1}{300}-\dfrac{1}{310}\right]$

${E}_{a}$ =

$.3010\times2.303\times8.314\left(\dfrac{300\times310}{10}\right)$

$53598.59 J{mol}^{-1}$

$53.6kJ$

The energy of activation is

0%
The energy associated with the activated molecules
0%
Threshold energy energy of normal molecules
0%
Threshold energy + energy of normal molecules
0%
Energy of products energy of reactants

Activation energy is given by the formula

0%
$\log{\frac{K_2}{K_1}}=\frac{E_a}{2.303R}\left[\frac{{T_2}-{T_1}}{{T_1}{T_2}}\right]$
0%
$\log{\frac{K_1}{K_2}}=-\frac{E_a}{2.303R}\left[\frac{{T_2}-{T_1}}{{T_1}{T_2}}\right]$
0%
$\log{\frac{K_1}{K_2}}=-\frac{E_a}{2.303R}\left[\frac{{T}_{1}-{T_2}}{{T_1}{T_2}}\right]$
0%
None of these

Explanation

$log_eK_2 = log_eA - \cfrac{E_A}{RT_2}$ ...........(1)

$log_eK_1 = log_eA - \cfrac{E_A}{RT_1}$ ..............(2)

Eqn. (1) - (2) we get,

$\log{\cfrac{K_2}{K_1}}=\cfrac{E_a}{2.303R}\left[\cfrac{{T_2}-{T_1}}{{T_1}{T_2}}\right]$

Option A is correct.

A graph between

$t_{1/2}$ and concentration for

$n^{th}-order$ reaction is a straight line. The reaction of this nature is completed

$50\%$ in

$10$ minutes when concentration is

$2\ mol. L^{−1}.$ This is decomposed

$50\%$ in t minutes at

$4$

$mol. L^{−1}$ . Then

$n$ and

$t$ are respectively:

0%
0, 20 min
0%
1, 10 min
0%
1, 20 min
0%
0, 5 min

Explanation

It can be observed from the given graph that

$t_{1/2}$ is constant and is independent of the concentration. We know that,

$t_{1/2}∝(1/a_o)^{n−1}$ ---- 1

where,

$a_o$ is the initial concentration of the reactant

n is the order of the reaction.

Now for first order reaction

$n=1$ and

$t_{1/2}$ will be independent of

$a_o$ (i.e. concentration) as depicted in the graph.

Hence, the reaction is of

$\text{first order.}$

As,

$t_{1/2}$ is independent of initial concentration, it will be the same whether the concentration is

$2\ mol /L$ or

$4\ mol /L$

So,

$n=1$ and

$t=10$ minutes (as already given in the question)

Option B is correct.

The energy of activation for reaction

$(KJ/mol)$ is :

0%
20.83
0%
13.83
0%
15.23
0%
10.23

8 gm of the radioactive isotope, cesium-137 were collected on February 1 and kept in a sealed tube. On July 1, it was found that only 0.25 gm of it remained. So the half-life period of the isotope is:

0%
37.5 days
0%
30 days
0%
25 days
0%
50 days

Explanation

t = Feb 1 to July 1 = 28 + 31 + 30 + 31 + 30 + 31 = 150 days

$\lambda =\dfrac{2.303}{150} log \dfrac{8}{0.25} = \dfrac{2.303}{150}log2^5 =\dfrac{0.693}{30}day^{-1}$

$t_{1/2} = \dfrac{0.693}{0.693/30} = 30 days$

For a zero order reaction

0%
The concentration of the reactant does not change during the reaction
0%
The concentration change only when the temperature changes
0%
The rate remains constant throughout
0%
The rate of the reaction is proportional to the concentration

Explanation

Rate of

$\textbf {zero order}$ reaction is independent of the concentration of the reactant and remains constant throughout the reaction.

hence option (C) is correct.

Half-life of a radioactive substance which disintegrates by

$75\%$ in 60 minutes, will be

0%
120 min
0%
30 min
0%
45 min
0%
20 min

Explanation

1st half- Life → 50% decay
2nd half- Life → 75% decay
3rd half- Life → 87.5% decay
4th half- Life → 93.75% decay

Thus, the half-life period of the radioactive element, if

$75$ % of it disintegrates in

$60$ min, is

$\dfrac{60}{2}$ =

$30 \;Minutes.$

Hence option (B) is correct.

A zero order reaction is one whose rate is independent of_________________________

0%
Temperature of the reaction
0%
The concentrations of the reactants
0%
The concentration of the products
0%
The material of the vessel in which the reaction is carried out

Explanation

The rate of zero order reaction is not depend on the concentration of the reactants.

$Rate = K[A]^0$

Certain bimolecular reactions which follow the first order kinetics are called_________________________

0%
First order reactions
0%
Unimolecular reactions
0%
Bimolecular reactions
0%
Pseudounimolecular reactions

Half-life of 10 gm of radioactive substance is 10 days. The half-life of 20 gm is:

0%
10 days
0%
20 days
0%
25 days
0%
Infinite

Explanation

Half-life period is independent of initial amount. Hence,

$t_{1/2}$ of 20 gm is 10 days.

The

$\Delta H$ value of the reaction

$H_2 + Cl_2 \rightleftharpoons 2HCl$ is

$-44.12 \,kcal$ . If

$E_1$ is the activation energy of the products, then for the above reaction

0%
$E_1 > E_2$
0%
$E_1 < E_2$
0%
$E_1 = E_2$
0%
$\Delta H$ is not related to $E_1$ and $E_2$
0%
None is correct

Explanation

$\Delta {H} = E_2- E_1= Negative$

$E_1$ is the activation energy of the products

$E_2$ is the activation energy of the reactants.

Thus ,

$E_1>E_2$

Option (A) is correct.

Half-life period of a zero order reaction is:

0%
Inversely proportional to the concentration
0%
Independent of the concentration
0%
Directly proportional to the initial concentration
0%
Directly proportional to the final concentration

Explanation

$t_{1/2}$ of zero-order reaction is independent of the concentration.

Hence, option B is correct.

If 2.0 g of a radioactive isotope has a half-life of 20 hr, the half-life of 0.5 g of the same substance is:

0%
20 hr
0%
80 hr
0%
5 hr
0%
10 hr

A radioactive isotope decays at such a rate that after 96 minutes only

$\dfrac{1}{8}th$ of the original amount remains. The half-life of this nuclide in minutes is:

0%
12
0%
24
0%
32
0%
48

Explanation

$\lambda = \dfrac{2.303}{t}log\dfrac{N}{N} = \dfrac{2.303}{96} log\dfrac{1}{1/8} = 0.0216$

$\therefore \ t_{1/2} = \dfrac{0.693}{\lambda} = 32.0 min$

Number of

$\alpha$ -particles emitted per second by a radioactive element falls to 1/32 of its original value in 50 days. The half-life-period of this elements is:

0%
5 days
0%
15 days
0%
10 days
0%
20 days

Explanation

$T = 50 \ days, t_{1/2} = ?, N_0 = 1, N = \dfrac{1}{32}$ ,

$N = N_0 \times \Big(\dfrac{1}{2}\Big)^n$ or

$\dfrac{1}{32} = 1 \times \Big(\dfrac{1}{2}\Big)^n$ ,

$\Big(\dfrac{1}{2}\Big)^5 = \Big(\dfrac{1}{2}\Big)^n$ or

$n = 5$

$T = t_{1/2} \times 2$ , or

$t_{1/2} = \dfrac{50}{5} = 10 \ days$ .

The radium and uranium atoms in a sample of uranium mineral are in the ratio of

$1:2.8 \times 10^6$ . If the half-life period of radium is 1620 years, the half-life period of uranium will be:

0%
$45.3 \times 10^9$ years
0%
$45.3 \times 10^{10}$ years
0%
$4.53 \times 10^9$ years
0%
$4.53 \times 10^{10}$ years

Explanation

According to radioactive equilibrium:

$\lambda_AN_A = \lambda_BN_B$

$\dfrac{0.693 \times N_A}{t_{1/2}(A)} = \dfrac{0.693 \times N_B}{t_{1/2}(A)} \Big[\lambda = \dfrac{0.693}{t_{1/2}}\Big]$

Where

$t_{1/2} (A)$ and

$t_{1/2} (B)$ are half periods of A and B respectively

$\therefore \dfrac{N_A}{t_{1/2}(A)} = \dfrac{N_B}{t_{1/2}(B)}$

$\therefore \dfrac{N_A}{t_{1/2}(A)} = \dfrac{N_B}{t_{1/2}(B)}$ or

$\dfrac{N_A}{N_B} = \dfrac{t_{1/2}(A)}{t_{1/2}(B)}$

$\therefore$ At equilibrium A and B are present in the ratio of their half lives

$\dfrac{1}{2.8 \times 10^6} = \dfrac{1620}{\text{Half Life of uranium}}$

$\therefore$ Halflife of uranium

$= 2.8 \times 10^6 \times 1620 = 4.53 \times 10^9$ years.

If the order of the reaction

$x+y\xrightarrow[]{hv}xy$ is zero, it means that the rate of ____

0%
Reaction is independent of temperature
0%
Formation of activated complex is zero
0%
Reaction is independent of the concentration of reacting species
0%
Decomposition of activated complex is zero

$8 \ gms$ of a radioactive substance is reduced to

$0.5 g$ after 1 hour. The

$t_{1 / 2}$ of the radioactive substance is:

0%
15 min
0%
30 min
0%
45 min
0%
10 min

Explanation

$N_0 = 8gms, N = 0.5 g$ and

$t = 1 hr. = 60 min.$

For

$t_{1/2}$ ,

$t = \dfrac{2.303 \times t_{1/2}}{0.693} log \dfrac{N_0}{N}$ .

$60 = \dfrac{2.303\times t_{1/2}}{0.693} log \dfrac{8}{0.5}$

$t_{1/2} = 15$ min

Hence, option A is correct.

CBSE Questions for Class 12 Engineering Chemistry Chemical Kinetics Quiz 15 - MCQExams.com

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Practice Class 12 Engineering Chemistry Quiz Questions and Answers

CBSE Questions for Class 12 Engineering Chemistry Chemical Kinetics Quiz 15 - MCQExams.com

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Practice Class 12 Engineering Chemistry Quiz Questions and Answers

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