MCQ Questions for Class 12 Engineering Chemistry Chemical Kinetics Quiz 2

For a given reaction which one is higher than the rest among the following?

0%
Average energy
0%
Threshold energy
0%
activation energy
0%
Normal energy

Explanation

Threshold enery

$\Rightarrow$ The collisions in which molecules collide with sufficient kinetic energy.

The energy to be possessed by the molecule participating in the reaction to give the products is:

0%
< activation energy
0%
threshold energy
0%
< average energy
0%
threshold energy + average energy

Which of the following is not a first order reaction?

0%
Decomposition of $H_{2}O_{2}$
0%
Decomposition of $N_{2}O_{5}$
0%
Decomposition of $N_{2}O$
0%
Decomposition of $SO_{2}Cl_{2}$

Explanation

Hint: The order of a reaction is experimentally determined.

Correct answer: Option

$C$

Explanation for correct option :

$N_{2}O$

Decomposition of

$N_2O$ into

$N_2$ and

$O_2$ in the presence of gaseous argon follows second-order kinetics

Explanation for incorrect options:

(A)Decomposition of

$H_{2}O_{2}$

The decomposition reaction of hydrogen peroxide was found to be first-order

(B)Decomposition of

$N_{2}O_{5}$

$2 \mathrm{~N}_{2} \mathrm{O}_{5} \rightarrow 4 \mathrm{NO}_{2}+\mathrm{O}_{2}$

Rate,

$\mathbf{r}=\mathbf{k}\left[\mathrm{N}_{2} \mathrm{O}_{5}\right]$

It is a first-order reaction that is not equal to the stoichiometry of the reactants in a balanced chemical equation.

(D)Decomposition of

$SO_{2}Cl_{2}$

Decomposition of

$SO_{2}Cl_{2}$ is the first order reaction.

The reaction

$L\rightarrow M$ is started with

$10.0\ gm$ of

$L$ . After

$30$ and

$90$ minutes,

$5.0\ gm$ and

$1.25\ gm$ of

$L$ respectively are left. The order of the reaction is:

0%
0
0%
1
0%
2
0%
3

Explanation

After 30 minutes, the concentration is reduced to half.

After the next 60 minutes, the concentration is reduced to one fourth.

Thus, 30 minutes is the half-life period and is independent of the concentration of

$L$ .

Hence, it is a first-order reaction.

Option B is correct.

A substance having initial concentration (a) reacts according to zero order kinetics. How much time will the reaction take to reach the completion?

0%
$\dfrac{a}{K}$
0%
$\dfrac{K}{a}$
0%
$\dfrac{a}{2K}$
0%
$\dfrac{2K}{a}$

Explanation

For a zero order reaction,

$[A_{0}] - [A] = Kt$

When the reaction is complete,

$[A] =0$ .

$[A_{0}] - [A] =[A_{0}] -0=[A_{0}] = Kt_{completion}$

$t_{completion}=\dfrac {[A_{0}] } {K}=\dfrac{a}{K}$

At 400K, the half-life of a sample of a gaseous compound initially at 56.0

$kPa$ is 340s. When the pressure is 28.0

$kPa$ , the half-life is 170s. The order of the reaction is:

0%
0
0%
2
0%
1
0%
1/2

$N_{2}O_{2} (g)\rightarrow2NO$ is a first-order reaction interms of the concentration of

$N_{2}O_{2} (g)$ .Which of the following is valid,

$[N_{2}O_{2} ]$ being constant?

0%
$[NO]=[N_{2}O_{2}]_{0}e^{-kt}$
0%
$[NO]=[N_{2}O_{2}]_{0}(1-e^{kt})$
0%
$[NO]=[N_{2}O_{2}]_{0}(e^{-kt}-1)$
0%
$[NO]=[N_{2}O_{2}]_{0}(1-e^{-kt})$

Explanation

The reaction is

$N_{2}O_{2} (g)\rightarrow2NO$
The initial concentration of

$N_{2}O_{2}$ is

$[N_{2}O_{2}]_0$ .. Assuming initially only

$N_{2}O_{2}$ is present and no

$NO$ is formed.
At time t, the concentrtion of

$N_{2}O_{2}$ and NO are

$[N_{2}O_{2}]$ and

$[NO]$ respectively.
The total concentration is equal to

$[N_{2}O_{2}]+[NO]=[N_{2}O_{2}]_0$ ......(1)
But from integrated rate law equation fo first order reaction ,

$[N_{2}O_{2}]=[N_{2}O_{2}]_0e^{-kt}$ ......(2)
Substituting equation (2) in equation (1)

$[N_{2}O_{2}]_0e^{-kt}+[NO]=[N_{2}O_{2}]_0$

$[NO]=[N_{2}O_{2}]_0-[N_{2}O_{2}]_0e^{-kt}=[N_{2}O_{2}]_0(1-e^{-kt})$

From the graph, pick out the correct one :

0%
$E*$ for the forward reaction is $E_{A}-E_{B}$
0%
$E*$ for the backward reaction is $E_{C}-E_{A}$
0%
$E*$ for reverse reaction is $\,> \,E*$ for forward reaction
0%
$E*$ for forward reaction $\,>\,E*$ for backward reaction

Explanation

$E^*\ for\ reverse = E^*\ for\ forward\ + \Delta H$

$so\ E^*\ reverse\ >\ E^*\ forward$

For a first order reaction, the half-life period is:

0%
inversely related to initial concentration
0%
directly related to initial concentration
0%
independent of initial concentration
0%
inversely related to the square of initial concentration

Explanation

$1^st\ order$

$t_{^1/_2}= \frac{0.0693}{K}$
so it is independent of initial concentration.

For a first order reaction, the half life is equal to:

0%
0.693 x k
0%
0.693/k
0%
k/0.693
0%
$\dfrac{2.303}{k}$

Explanation

$t= \cfrac{2.303}{k}\ log\ \left [ \cfrac{a}{a-x} \right ]$

$t\ ^{1/_2} = \cfrac{2.303}{k}\ log\ \left [ \cfrac{a}{^a/_2} \right ]$

$t\ ^{1/_2} = \cfrac{2.303\ log\ 2}{k}$

$= \cfrac{ln\ 2}{k}$

$= \cfrac{0.693}{k}$

Hence, the correct option is

$\text{B}$

If the initial concentration is reduced to 1/4th in a zero order reaction, then the time taken for half the reaction to complete:

0%
remains same
0%
becomes 4 times
0%
becomes one-fourth
0%
doubles

Explanation

For a zero order reaction,

$t_{1/2}=\dfrac{[A]_0}{2k}$

Let

$[A]_0=a$

$t_{1/2}=\dfrac{a}{2k}$ ......(1)

When the intial concentration is reduced to one fourth,

$[A]_0=\dfrac {1} {4} \times a.$

$t'_{1/2}=\dfrac{\dfrac {1} {4} \times a}{2k}$ ......(2)

By dividing equation (2) by (1), we get

$\dfrac {t'_{1/2}} {t_{1/2}} =\dfrac {\dfrac{\dfrac {1} {4} \times a}{2k}} {\dfrac{a}{2k}} =\dfrac {1} {4}$

Hence, option C is correct.

The amount left after completion of average life period in a first order reaction is :

0%
$\dfrac{a(e-1)}{e+1}$
0%
$\dfrac{a}{e-1}$
0%
$\dfrac{a(e-1)}{e}$
0%
$\dfrac{a}{e}$

Explanation

The average life period of a first order reaction is equal to the reciprocal of its rate constant.
Thus

$\lambda = \frac {1} {k}.$
The amount left after completion of average life period in a first order reaction is

$[A]=[A]_0e^{-kt}=[A]_0e^{-k \times \frac {1} {k}}=\frac{a}{e}.$

For the reaction A

$\rightarrow$ B it has been found that the order of the reaction is zero with respect to A. Which of the following expressions correctly describes the reaction?

0%
$K=\dfrac{2.303}{t}log\dfrac{[A_{0}]}{[A]}$
0%
$[A_{0}] - [A] = Kt$
0%
$t_{1/2}=\dfrac{0.693}{K}$
0%
$t_{1/2}\propto \dfrac{1}{[A_{0}]}$

Explanation

For a zero order reaction,

$d[A]=-Kdt.$

On integrating this equation between limits,

$[A]=[A]_0$ at

$t=0$

$[A]=[A]_t$ at

$t=t$ , we get

$[A_{0}] - [A] = Kt$

The following figure denotes the energy diagram for a reaction.
Then the activation energy for the reverse reaction is:

0%
2x
0%
2y
0%
x+ y
0%
y-x

$SO_{2}Cl_{2} \rightarrow SO_{2}+Cl_{2}$ is a first order gas reaction with

$k=2.2\times 10^{-5}\sec^{-1}$ at

$320^o C$ . The percentage of

$SO_{2}Cl_{2}$ decomposed on heating for 90 minutes is:

0%
1.118
0%
0.1118
0%
18.11
0%
11.18

Explanation

For a first order reaction,

$k=\dfrac{ln\left[\dfrac{Ro}{R} \right ]}{t}$

$ln\left[\dfrac{Ro}{R} \right ]=2.2\times10^{-5}\times90\times60$

$ln\dfrac{Ro}{R}=0.1188$

$\dfrac{Ro}{R}=1.126$

$R = 0.88 79 Ro$

Percentage of decomposition =

$R_0-R\ \Rightarrow Ro-0.8879\ Ro$

$\Rightarrow 11.18$

$Ro$ %

Hence, option D is correct.

Which equation represents the time to complete 90% of first order reaction?

0%
$\dfrac{k}{2.303}log(\frac{4}{3})$
0%
$\dfrac{2.303}{k}log(\frac{3}{4})$
0%
$\dfrac{2.303}{k}$
0%
$\dfrac{2.303}{k}log(3)$

Explanation

$t = \cfrac{2.303}{k}\ log \left [ \cfrac{Ro}{R} \right ]$

$t =\cfrac{2.303}{k}\ log\ \left [ \cfrac{a}{a-0.9a} \right ]$

$t = \cfrac{2.303}{k}\ log \left [ \cfrac{1}{0.1} \right ]$

$= \cfrac{2.303}{k}$

Hence, the correct option is

$\text{C}$

$A_{(g)} \rightarrow B_{(g)}$ is a first order reaction. The initial concentration of A is 0.2mol

$l^{-1}$ . After 10 minutes, the concentration of B is found to be 0.18mol

$l^{-1}$ . Calculate rate constant of the reaction?

0%
0.2303
0%
2.303
0%
0.693
0%
0.01

Explanation

At t=0,

$[A]=0.2$ and

$[B]=0.$
At t=10 min,

$[A]=0.2-0.18=0.02$ ,

$[B]=0.18.$

$k=\frac{1}{t} ln \frac{[C_o]}{[C_o - C_t]}=\frac{1}{10} ln\frac{[0.2]}{[0.02]} =0.2302.$

At room temperature the reaction between NO and

$O_{2}$ to give

$NO_{2}$ is fast while that of between CO and

$O_{2}$ is slow. It is because:

0%
CO is smaller in size than that of NO
0%
CO is poisonous
0%
the activation energy for the reaction 2NO+ $O_{2}$ $\rightarrow$ 2 $NO_{2}$ is less than 2CO+ $O_{2}$ $\rightarrow$ 2 $CO_{2}$
0%
the activation energy for the reaction 2CO+ $O_{2}$ $\rightarrow$ 2 $CO_{2}$ is less than 2NO+ $O_{2}$ $\rightarrow$ 2 $NO_{2}$

The concentration of reaction decreases from 0.2M to 0.05M in 5 minutes. The rate of reaction in

$mol. lit^{-1}.s^{-1}$ is:

0%
$8.3 \times 10^{-4}$
0%
$0.05$
0%
$0.0005$
0%
$0.15$

Explanation

The expression for the rate of the reaction is

$Rate=\displaystyle \frac{change\ in\ concentration}{time\ interval}.$

Substitute values in the above expression.

$Rate=\displaystyle \frac{0.2 - 0.05}{{5 \times 60}}=0.0005 \ mol.lit^{-1}.s^{-1}.$

A first order reaction was commenced with 0.2 M solution of the reactants. If the molarity of the solution falls to

$0.02M$ after 100 minutes the rate constant of the reaction is:

0%
$2\times 10^{-2}min^{-1}$
0%
$2.3\times 10^{-2}min^{-1}$
0%
$4.6\times 10^{-2}min^{-1}$
0%
$2.3\times 10^{-1}min^{-1}$

Explanation

$k=\frac{1}{t}ln\left [ \frac{Co}{Ct} \right ]$

Where,

$Co$ =initial concentration and

$Ct$ =change in concentration

$k=\frac{1}{100}ln\left [ \frac{0.2}{0.02} \right ]$

$=\frac{ln\ 10}{100}$

$=2.3\times 10^{-2}\ \ min^{-1}$

After how many seconds will the concentration of the reactant in a first order reaction be halved, if the rate constant is 1.155x10

$^{-3} sec^{-1}$ ?

0%
600
0%
100
0%
60
0%
10

Explanation

$t\ ^1/_2=\frac{0.693}{k}$

$4=\frac{0.693}{1.155\times 10^{-3}}$

$0.6 \times 10^{3}$
= 600

With respect to the figure which of the following statement is correct?

0%
$E_{A}$ for forward reaction is $E_{A}-E_{B}$
0%
$E_{A}$ for forward reaction is $E_{C}-E_{A}$
0%
$E_{A}$ for reverse reaction is greater than $E_{A}$ for forward reaction
0%
$E_{A}$ for forward reaction is greater than $E_{A}$ for backward reaction

For a first order reaction

$t_{0.75}$ is 1368 seconds.Therefore, the specific rate constant in

$sec^{-1}$ is:

0%
$10^{-3}$
0%
$10^{-2}$
0%
$10^{-9}$
0%
$10^{-5}$

Explanation

For first order reaction,

$t_{0.75}=2t_{0.50}=1368 sec.$

Hence,

$t_{0.50} = 684\ sec.$

Also

$k= \dfrac {0.693} {t_{0.50}}=\dfrac {0.693} {684}\approx 10^{-3} sec^{-1}.$

Hence, option A is correct.

A first-order reaction is carried out with an initial concentration of

$10$ mole per litre and

$80$ % of the reactant changes into the product. Now if the same reaction is carried out with an initial concentration of

$5$ mol per litre, the percentage of the reactant changing to the product is:

0%
$40$
0%
$80$
0%
$160$
0%
$\text{cannot be calculated}$

Explanation

$a_{0}=10\ M$

$80\%$ changed into product

$\therefore a_{t}=10\dfrac{10\times 80}{100}=2\ M$
If

$a_{0}=5\ M$ , In the same time duration same percentage of initial concentration changes into products because in first order reaction is independent from initial concentration.

Statement - I : The reactions

$2NO+O_{2}\rightarrow 2NO_{2}$ and

$2CO+O_{2}\rightarrow 2CO_{2}$ proceed at the same rate because they are similar.
Statement - II Above two reactions will have different activation energies.

0%
Both statements are true Statement - II is correct explantion of Statement - I
0%
Both statements are true but Statement -II is not correct explantion of Statement - I
0%
Statement - I is true but Statement - II is false.
0%
Statement - I is false but Statement - II is true.

The most common molecular collisions are in between:

0%
2 molecules
0%
3 molecules
0%
4 molecules
0%
5 molecules

The half life of first order reaction

$PCl_{5} \rightarrow PCl_{3}+Cl_{2}$ is 10 minutes at a certain temperature. The time in which the concentration of

$PCl_{5}$ would be reduced to 10% of the initial concentration is:

0%
26 min
0%
33 min
0%
71 min
0%
90 min

Explanation

The relationship between the rate constant and half life period is

$k=\displaystyle \frac{0.693}{t}=\displaystyle \frac{0.693}{10}=0.0693$ .
The integrated rate law expression is

$k=\displaystyle \frac{2.303}{t}\log\frac{a}{a-x}.$
Substitute values in the above expression.

$0.0693=\displaystyle \frac{2.303}{t}\log\frac{100}{10}.$

$t=33 min$ .

The rate constant of a first order reaction is

$0.0693$

$min^{-1}$ .

Time (in minutes) required for reducing an initial concentration of

$20mol$

$lit^{-1}$ to

$2.5mol$

$lit^{-1}$ is :

0%
$40$
0%
$30$
0%
$20$
0%
$10$

Explanation

For a

$1^{st}$ order reaction,

$t=\dfrac{2.303}{k}\ log\left [ \dfrac{R_o}{R} \right ]$

$t=\dfrac{2.303}{0.0693}\ log\left [ \dfrac{20}{2.5} \right ]$

t

$= 30.01 min$

Option B is correct.

For the gas phase decomposition

$A\rightarrow2B$ , the rate constant is

$6.93\times 10^{-3}$

$min^{-1}$ at 300K. The percentage of A remaining at the end of 300 minutes is :

0%
75
0%
50
0%
25
0%
12.5

Explanation

$ln\left [ \frac{Ao}{A} \right ]=kt$

$=6.93\times 10^{-3}\times 300$

$\frac{Ao}{A}=8$

$\frac{A}{Ao}=.1250$

$\frac{A}{Ao}\times 100$ =12.5%

From the graph pick out the correct one:

0%
$\Delta E$ for forward reaction is $B-A$
0%
$\Delta E$ for the forward reaction $C-A$
0%
$\Delta E$ reverse is greater than forward
0%
$\Delta E$ for forward reaction is $A + B$

Explanation

The change in energy of a reaction is equal to the energy of products minus the energy of reactants.

$\Delta E=E_{C}-E_{A}$ .

The rate of a certain reaction at different times are as follows:

Time	rate ( $mole$ $lit^{-1}\sec^{-1}$ )
I) 0	a) $2.8 \times 10^{-2}$
II) 10	b) $2.78\times 10^{-2}$
III) 20	c) $2.81\times 10^{-2}$
IV) 30	d) $2.79\times 10^{-2}$

The correct matching is:

0%
zero order
0%
first order
0%
second order
0%
third order

For a zero-order reaction

$A$

$\rightarrow$ product, the rate constant is

$10^{-2}\ mol\ L^{-1}\ s^{-1}$ . Starting with 10 moles of

$A$ in a 1 L vessel, how many moles of

$A$ would be left unreacted after 10 minutes?

0%
5 moles
0%
6 moles
0%
4 moles
0%
10 moles

Explanation

$k=10^{-2}$

$kt = R_o - R$

$10^{-2}\times 10\times 60=\dfrac{10}{1}-R$

$R$

$= 6\ mol L^{-1}$
6 mole react unreacted mole

$= 10 -6$

$= 4 moles$

Which is the graphical representation for the zeroth order reaction?

Explanation

rate = k

$\dfrac{-dx}{dt}=k$

Where k is rate constant. Thus, the rate of reaction is independent of concentration of reactant.

$\dfrac{dx}{dt}$ is constant.

Hence graph will be constant straight line parallel to x axis.

An activation energy of the forward reaction is same as activation energy of the backward reaction.

0%
True
0%
False

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

$k = A.e^{E^{\ast }/RT}. Activation energy$ (E^{\ast })$$ of the reaction can be calculated by plotting

0%
K vs T
0%
K vs $\frac{1}{log T}$
0%
log K vs $\frac{1}{T}$
0%
log k vs $\frac{1}{log T}$

Identify the false statements among the following:

0%
Rate of reaction is directly proportional to activation energy
0%
Order of a reaction can be determined by experimental method only
0%
Rate law never coincides with the stoichiometry of the reaction
0%
Order of reaction cort different reactants may be different.

What is the hybridisation in $AsF_4^-$ ion :

0%
sp
0%
$sp^2$
0%
$sp^3$
0%
$sp^3 d$

Explanation

See-saw shape

Since there is 4 bp and 1 lp.

$\therefore sp^3 d$ hybridisation'

Half life period of

$C^{14}$ in years is:

0%
$500$
0%
$5000$
0%
$50$
0%
$5\times 10^4$

Explanation

The half-period of

$C^{14}$ is 5000 years. It is used in the determination of age of fossils by using technique called carbon dating.

Increase in temperature increases the activation energy.

State whether the given statement is true or false.

0%
True
0%
False

Explanation

Change in temperature has no effect on Activation Energy.

The only way to alter the activation energy is by adding catalysts.

Arrhenius equation gives the relation between the rate constant and the temperature.

$K=A{ e }^{ -Ea/RT }$

K increases on increasing temperature.

Thus, the given statement is false.

For an endothermic reaction, energy of activation is E

$_{a}$ and enthalpy of reaction is H (both of these in kJ/mol). Minimum value of E

$_{a}$ will be:

0%
less than H
0%
equal to H
0%
more than H
0%
equal to zero

Explanation

Activation energy is the minimum quantity of energy which the reacting species must possess in order to undergo a specified reaction.

(Image)

${ E }_{ a }$ -Activation energy

Here, energy of products is higher than that of reactants.

This graph is for an endothermic reaction.

$\triangle H$ for an endothermic reaction is positive

See figure above, when the reaction reaches point P, the reactants possess enough energy for the reaction to take place.

In further reaction, some energy is use up for conversion of reactants to products.

This used up energy must be less than

${ E }_{ a }$ , for an endothermic reaction

Since,

${ E }_{ a }$ - H = (used up energy)

$\therefore { E }_{ a }>H$

Hence, the activation energy must be more than H

The rate of reaction is directly proportional to the activation energy.

0%
True
0%
False

Statement 1:
In a zero order reaction, if the concentration of the reactant is doubled, the half-life period is also doubled.
Statement 2:
For a zero-order reaction, the rate of the reaction is independent of initial concentration.

0%
Statement 1 is True, statement 2 is True, statement 2 is a correct explanation of statement 1.
0%
Statement 1 is True, statement 2 is True, statement 2 is not a correct explanation of statement 1.
0%
Statement 1 is true, Statement 2 is False.
0%
Statement 1 is False, Statement 2 is True.

Explanation

For a zero order reaction, the expression for the half life period is

$t_{\frac{1}{2}}=\frac{[A_{o}]}{2k}.$
Thus half life period is directly proportional to the initial concentration. When the concentration of the reactant is doubled, the half-life period is also doubled. Hence, the statement 1 is true.
The rate of a zero order reaction is independent of the concentration of the reaction. The reaction proceeds at constant rate throughout. Hence, statement 2 is True.

The activation energy of the forward reaction and backward reaction decreases by

$30 kcal$ then which of the following statement is correct?

0%
$\Delta H$ will remain constant.
0%
$\Delta H$ will decrease.
0%
$\Delta H$ will increase.
0%
The reaction will be exothermic.

Explanation

$\Delta H=E_{af}-E_{ab}$ .
So if Activation energy of the forward reaction and backward reaction decreases by 30kcal then

$\Delta H=[E_{af}+30]-[E_{ab}+30]$

$\Delta H=E_{af}-E_{ab}$ .

$\Delta H$ will remain same.

What are effective collisions?

0%
Collisions leading to the transformation of reactants to products
0%
formation of activated complex
0%
collison between two reactant to decrease the activation energy
0%
collison between two reactant to overcome activation energy barrier

Which of the following is/are correct statement?

0%
Stoichiometry of a reaction tells about the order of the elementary reactions
0%
For a zero order reaction, rate and the rate constant are identical
0%
A zero order reaction is controlled by the factors other than concentration of reactants
0%
A zero order reaction is always an elementary reaction

Explanation

$\bullet$ For an elementary reaction, the order of reaction with respect to a reactant is equal to the stoichiometric coefficient of the reactant.

$\bullet$ For a zero order reaction,

rate =

$K{ \left[ A \right] }^{ 0 }$

$\Rightarrow$ rate = K.

Hence, rate and the rate constant are identical.

$\bullet$ Since, the rate of a zero order reaction is indepedent of the concentration of reactants, it is controlled by other factors.

$\bullet$ A zero order reaction is not necessarily an elementary reaction. RDS of a complex reaction may also have zero order.

The initial rate of zero-order reaction of the gaseous reaction

$A(g)\rightarrow2B(g)$ is

$10^{-2}M min^{-1}$ . If the initial concentration of A is 0.1 M, then the concentration of B after 60 sec would be:

0%
0.09 M
0%
0.01 M
0%
0.02 M
0%
0.002 M

Explanation

$\left[A \right ]_{t}$ = $\left [A \right ]_{0}$ – $kt$

$\left[A \right ]_{t}$ = $(0.1) - (\dfrac{ (10) ^ {-2}\times60}{60})$

$0.09$ M

$\left [B \right ]_{t}$ = ${2}\times(.0.045)$

= $0.09$ M

hence the correct answer is -

(a) $0.09$ M

Statement: The rate of instantaneous reactions can be determined experimentally.

State whether the given statement is true or false.

0%
True
0%
False

Explanation

Instantaneous rate: It can be determined graphically by using the concentrations of reactant or product at different time intervals.

Hence, the given statement is

$\text{true}$

In a chemical reaction, two reactants take part. The rate of reaction is directly proportional to the concentration of one of them and inversely proportional to the concentration of the other. The order of reaction is:

0%
0
0%
1
0%
2
0%
4

Explanation

$\text{So zero is the answer}$

In the following reaction,

$A\rightarrow B$ , rate constant is

$1.2\times10^{-2} Ms^{-1}$ . What is the concentration of

$B$ after

$10$ min, if we start with

$10\ M$ of

$A$ ?

0%
7.2 M
0%
4.5 M
0%
6.7 M
0%
7.0 M

Explanation

Unit of k indicates that it's a zero-order reaction.

According to zero-order reaction.

$Rate \ of \ reaction = k$

$Rate \ of \ formation \ of \ B = \frac{Concenteration\ of\ B}{time(\ in \ seconds)} = k$

$Concentration\ of\ B\ after\ 10 mins=kt = 1.2\times10^{-2}\times10\times60 = 7.2M$

For the first-order reaction

$A\rightarrow product$ s, the half-life is

$200$ s. The rate constant of the reaction is:

0%
$3.46\times 10^{-2} s^{-1}$
0%
$3.46\times 10^{-3} s^{-1}$
0%
$3.46\times 10^{-4} s^{-1}$
0%
$3.75\times 10^{-5} s^{-1}$

Explanation

$k=0.693/200=0.00346s^{-1}$

CBSE Questions for Class 12 Engineering Chemistry Chemical Kinetics Quiz 2 - 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 2 - MCQExams.com

0:0:1

Practice Class 12 Engineering Chemistry Quiz Questions and Answers

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