MCQ Questions for Class 12 Engineering Chemistry Chemical Kinetics Quiz 14

For a reaction

$A \longrightarrow B+C$ , it was found that at the end of 10 minutes from the start the total optical rotation of the system was

${ 50 }^{ o }$ and when the reaction is complete, it was

${ 100}^{ o }$ . Assuming that only B and C are optically active and dextrorotatory. Calculate the rate constant of this first order reaction.

0%
$0.693 mi{ n }^{ -1 }$
0%
$0.0693 se{ c }^{ -1 }$
0%
$0.0693 { min }^{ -1 }$
0%
$0.00693 { sec }^{ -1 }$

The rate constant of a reaction at temperature 200

$\mathrm{K}$ is10 times less than the rate constant at 400

$\mathrm{K}$ . What is theactivation energy

$\left(E_{\mathrm{a}}\right)$ of the reaction

$(R=$ gas constant)?

0%
1842.4 $R$
0%
921.2 $R$
0%
460.6 $R$
0%
230.3 $R$

for a zero order reaction-

0%
$t_{\frac{1}{2}}\propto a$
0%
$t_{\frac{1}{2}}\propto \dfrac{1}{a}$
0%
$t_{\frac{1}{2}}\propto a^2$
0%
$t_{\frac{1}{2}}\propto \dfrac{1}{a^2}$

For a first order reaction velocity constant is

$K={ 10 }^{ -3 }{ s }^{ -1 }$ ,Two third life for it would be

0%
1100 s
0%
2200 s
0%
3300 s
0%
4400 s

Which one of the following formula represents a first-order reaction?

0%
$K=\dfrac { x }{ t }$
0%
$K=\dfrac { 1 }{ 2t } \left[ \dfrac { 1 }{ \left( 1-x \right) ^{ 2 } } -\dfrac { 1 }{ { a }^{ 2 } } \right]$
0%
$K=\dfrac { 2.303 }{ t } { log }_{ 10 }\dfrac { a }{ \left( a-x \right) }$
0%
$K=\dfrac { 1 }{ t } \dfrac { x }{ \left( a-x \right) }$

A graph plotted between log

$t_{50\%}$ vs log of concentration is a straight line. What conclusion can you draw from the given graph?

0%
$n=1,\ { t }_{ 1/2 }=\dfrac { 1 }{ k-a }$
0%
$n=2,\ { t }_{ 1/2 }=1/a$
0%
$n=1,\ { t }_{ 1/2 }= 0.693/k$
0%
none of these

Which of the following expression is correct for first- order reaction? (CO) refers to initial concentration of reactant

0%
${ t }_{ 1/2 }\propto CO$
0%
${ t }_{ 1/2 }\propto { CO }^{ -1 }$
0%
${ t }_{ 1/2 }\propto { CO }^{ -2 }$
0%
${ t }_{ 1/2 }\propto { CO }^{ o }$

Explanation

$t_{1/2}\propto{\left(CO\right)}^{0}$ i.e. half life for

$1^{st}$ order is independent of initial concentration.

For the first order reaction

${ 2N }_{ 2 }{ O }_{ 5 }\left( g \right) \longrightarrow { 4NO }_{ 2 }\left( g \right) +{ O }_{ 2 }\left( g \right)$

0%
the concentration of the reactant decrease exponentially with time
0%
the half-life of the reaction decrease with increasing temperature.
0%
the half-life of the reaction depends on the initial concentration of the reactant.
0%
the reaction proceeds to 99.6% completion in eight half-life duration.

According to molecular collision theory, the reaction is subjected to:

0%
Number of molecular collisions of reactant
0%
Number of collisions between reactants and activated complex
0%
The collision rate between reactants and product molecules
0%
Number of effective molecular collisions of reactants

The reactions having very high values of energy of activation are generally________

0%
Very slow
0%
Very fast
0%
Spontaneous
0%
Medium fast.

Reaction

$A+B\rightarrow C+D$ follows the following rate law:

$Rate\ =k{ \left[ A \right] }^{ 1/2 }{ \left[ B \right] }^{ 1/2 }$

On starting with initial conc. of 1 M of A and B each, what is the time taken for the concentration of A to becomes 0.25 M?
[Given;

$k=2.303\times { 10 }^{ -3 }{ sec }^{ -1 }$ ]

0%
300 sec
0%
600 sec
0%
900 sec
0%
None of these

The order of a reaction and rate constant for a chemical change having following curve would be:

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

For a first order reaction, the value of rate constant for the reaction

$A_{(gas)}\longrightarrow 2B_{(gas)}+C_{(solid)}$

0%
$\dfrac{1}{t}ln\left( \frac { { P }_{ 0 } }{ { P }_{ 0 }-{ P }_{ 1 } } \right)$
0%
$\dfrac{1}{t}ln\left( \frac { { P }_{ 0 } }{ { 2P }_{ 0 }-{ P }_{ 1 } } \right)$
0%
$\dfrac{1}{t}ln\left( \frac { { 2P }_{ 0 } }{ { 3P }_{ 0 }-{ P }_{ 1 } } \right)$
0%
$\dfrac{1}{t}ln\left( \frac { { P }_{ 0 } }{ { 3P }_{ 0 }-{ P }_{ 1 } } \right)$

From the above figure, the activation energy for the reverse reaction would be:

0%
$-120\ kJ{ mol }^{ -1 }$
0%
$+152\ kJ{ mol }^{ -1 }$
0%
$-120\ kJ{ mol }^{ -1 }$
0%
$+1760\ kJ{ mol }^{ -1 }$

The half life period for catalytic decomposition of

$A B_{3}$ at

$50 \mathrm{mm}$ is found to be

$4 hrs$ and at

$100 \mathrm{mm}$ it is

$2hrs$ . The order of reaction is -

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

For a zero order reaction

$\mathrm{t}_{1 / 2}$ is :

0%
$\frac{[\mathrm{A}]_{0}}{\mathrm{K}}$
0%
$\frac{2[\mathrm{A}]_{0}}{\mathrm{K}}$
0%
$\frac{[\mathrm{A}]_{0}}{2 \mathrm{K}}$
0%
$\frac{1}{[\mathrm{A}]_{0} \mathrm{K}}$

Which of the following represents threshold energy:

0%
Activation energy + energy of the reactant
0%
Activation energy - energy of the reactant
0%
Activation energy
0%
Energy of the reactants

Which of the following is correct with respect to a first order reaction?

0%
A plot of rate and concentration is a positive straight line graph whose slope given the value of rate constant
0%
A plot of log [A] and time is a straight line graph whose slope is $-\cfrac { k }{ 2.303 }$
0%
A plot of log $\cfrac { { [A] }_{ 0 } }{ [A] }$ and time is a positive straight line graph whose slope is $-\cfrac { k }{ 2.303 }$
0%
All of the above

For a given reaction of first order it takes 20 minutes for the concentration to drop from 1 M to 0.6 M. The time required for the concentration to drop from 0.6 M to 0.36 M will be:

0%
More than 20 min
0%
Less than 20 min
0%
Equal to 20 min
0%
Infinity

A reagent undergoes

$90 \%$ decomposition in 366 min. according to the first-order reaction. Its half-life is :-

0%
$366 \times 100\left(\dfrac{\ln 2}{90}\right)$
0%
$366\left(\dfrac{\ln 2}{\ln 10}\right)$
0%
$\dfrac{1}{366}$
0%
183

Which of the following graph(s) are correct for zero order reactions?

0%
0%
0%
0%
All of the above

The time taken for a first-order reaction to reduce the initial concentration by 4 times is 10 minutes. If the concentration is reduced 16 times, then the time required is:

0%
10 minutes
0%
96.33 minutes
0%
40 minutes
0%
80 minutes

Which among the following reaction is an example of a zero order reaction?

0%
$C_{12}H_{22}O_{11(aq)}+H_2O_{(l)}$ $\rightarrow C_6H_{12}O_{6(aq)}+C_6H_{12}O_{6(aq)}$
0%
$2NH_{3(g)}\overset{Pt}{\rightarrow}N_{2(g)}+3H_{2(g)}$
0%
$2H_2O_{2(l)}\rightarrow 2H_2O_{(l)}+O_{2(g)}$
0%
$H_{2(g)}+I_{2(g)}\rightarrow 2HI_{(g)}$

Explanation

The decomposition of ammonia

$(NH_3)$ on the surface of

$Pt$ is zero-order reaction.

Here, Rate of reaction

$r=k[NH_3]^0$

Here,

$Pt$ metal will act as a catalyst.

Hence, option

$B$ is correct.

$75\%$ of a zero order reaction complete in 4 hr and

$87.5\%$ of the same reaction completes in

0%
$6 h$
0%
$4.64 h$
0%
$8 h$
0%
$2 h$

Which option is valid for zero order reaction?

0%
$t_{1/2} = \dfrac {3}{2} t_{1/4}$
0%
$t_{1/2} = \dfrac {4}{3} t_{1/4}$
0%
$t_{1/2} = 2t_{1/4}$
0%
$t_{1/4} = (t_{1/2})^{2}$

Explanation

For zero order

$A = A_{0} - kt$

$t_{\dfrac {1}{2}} = \dfrac {A_{0}}{2k}$

$t_{\dfrac {1}{4}} = \dfrac {A_{0}}{4k}\ =\dfrac {t_{\dfrac {1}{2}}}{t_{\dfrac {1}{4}}} = \dfrac {2}{1}$

The number of collisions of Ar atoms with the walls of container per unit time?

0%
Increases when the temperature increases
0%
Remains the same when $CO_2$ is added to the container at constant temperature
0%
Increases when $CO_2$ is added to the container at constant temperature
0%
Decreases, when the average kinetic energy per molecule is decreased

Explanation

Number of effective collisions,

$z=\dfrac{1}{4}N$

$u_{avg}$

also

$u_{avg}=\sqrt{\dfrac{8RT}{M}}$ ,

$N=\dfrac{pv}{RT}$

$z=\dfrac{1}{4}\dfrac{pv}{RT}\times\sqrt{\dfrac{8RT}{m}}$

$\Rightarrow z\propto \dfrac{1}{\sqrt{T}}$

So if temperature is constant, 'Z' will be same

Option B.

If X is the total number of collisions which a gas molecule registers with other molecules per unit time under particular conditions, then the collision frequency of the gas containing N molecules per unit volume is:

0%
$X/N$
0%
$NX$
0%
$2NX$
0%
$NX/2$

Explanation

The collision frequency

$Z_{AA}=$ Total collision

$\times \dfrac{N}{2}$

$Z_{AA}=X\times \dfrac{N}{2}$

$=\dfrac{XN}{2}$

Option D is correct.

Consider the given plot of enthalpy of the following reaction between

$A$ and

$B$ .

$A + B\rightarrow C + D$
Identify the incorrect statement.

0%
$C$ is the thermodynamically stable product
0%
Formation of $A$ and $B$ from $C$ has highest enthalpy of activation
0%
$D$ is kinetically stable product
0%
Activation enthalpy of form $C$ is $5\ kJ\ mol^{-1}$ less than that to form $D$

Explanation

$A + B\rightarrow C + D$

Activation enthalpy for

$C = 20 - 5 = 15\ kJ/mol$

Activation enthalpy for

$D = 15 - 5 = 10\ kJ/ mol$ .

1247710_1614270_ans_81a9a4a3841b4abfa88bc0d970d57072.PNG

The order of a reaction is zero. It will be definitely

0%
exothermic
0%
endothermic
0%
elementary
0%
complex

Which of the following is incorrect 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 factors other than the concentration of reactants
0%
A zero-order reaction is always elementary reaction

Flask with greater mean free path for the molecules.(

$r_{H_2}:r_{CH_4}:r_{O_2}=1:2:2$ )

0%
A
0%
B
0%
C
0%
Same for all

Explanation

$\lambda$ (mean free path)

$=\dfrac{kT_A}{\sqrt{2\pi}\sigma^2_Ap_A}$

$\dfrac{\lambda_A}{\lambda_B}=\left(\dfrac{T_A}{T_B}\right)\left(\dfrac{\sigma^2_B}{\sigma^2_A}\right)\left(\dfrac{P_B}{P_A}\right)=\left(\dfrac{300}{600}\right)\left(\dfrac{2}{1}\right)^2\left(\dfrac{1}{4}\right)$

$\dfrac{\lambda_A}{\lambda_B}=\dfrac{1}{2}$

similarly,

$\dfrac{\lambda_B}{\lambda_C}=\dfrac{1}{4}$

$\therefore \lambda_C > \lambda_B > \lambda_A$

$\therefore$ Mean free path of C is highest

Option C is the correct answer.

For a zero-order reaction,

0%
the reaction rate is doubled when the initial concentration is doubled
0%
the time for half change is half the time taken for completion of the reaction
0%
the time for half change is independent of the initial concentration
0%
the time for completion of the reaction is independent of the initial concentration

Flask with greater collision number of the molecules.

0%
A
0%
B
0%
C
0%
Same in all

Explanation

Flask with greater collision number of molecules

$Z_A=\dfrac{u_{av}}{\lambda}$

$Z_A=\dfrac{(u_{av})_A}{\lambda_A}$

$Z_B=\dfrac{(u_{av})_B}{\lambda_B}$

$\dfrac{Z_A}{Z_B}=\dfrac{(u_{av})_A}{(u_{av})_B}\times \dfrac{\lambda_B}{\lambda_A}=2\times 2=4$

Similarly

$Z_B > Z_C$

$\therefore Z_A > Z_B > Z_C$

$Z_A$ is greatest

Option A is the correct answer.

Flask with greater number of collisions with the walls per unit area per unit time.

0%
A
0%
B
0%
C
0%
Same for all

Explanation

Effective no. of collision:

$z=\dfrac{1}{4}\times N\times u_{avg}$

$=\dfrac{1}{4}\times N\times \sqrt{\dfrac{8RT}{nm}}$

$\Rightarrow Z\propto \dfrac{1}{\sqrt{M}}$

Lower mass means higher Z

$\therefore$ A

$(H_2)$ option A.

When the temperature of an ideal gas is increased at constant pressure?

0%
Collision number increases
0%
Collision frequency increases
0%
Mean free path increases
0%
Number of molecules per unit volume increases

Explanation

Collision frequency is exceed to

$z=\sigma \times \bar{c}ra\times \dfrac{PNA}{RT}$

$z=\sigma\times \bar{c}ra\times \dfrac{p}{RBT}$

so when temperature $\uparrow =\downarrow$

Mean path is equal to

$\lambda =\dfrac{RT}{\sqrt{2}na^2NAD}$

so it will depend upon temperature so it will increases

(D) No of the molecule will be the same collision
No, the collision number increases because proper orientation also will be required.

So, the answer is C.

In a closed flask of

$5$ litres

$1.0\ g$ of

$H_2$ is heated from

$300$ to

$600\ K$ . Which statement is not correct?

0%
Pressure of the gas increases
0%
The rate of collisions increases
0%
The number of moles of gas increases
0%
The energy of gas molecules increases

One mole of helium and one mole of neon are taken in a vessel. Which of the following statements are correct?

0%
Molecules of helium strike the wall of vessel more frequently
0%
Moles of neon apply more average force per collision on the wall of vessel
0%
Molecules of helium have greater average molecular speed
0%
Helium exerts higher pressure than neon

Explanation

Helium gas has lower molecular mass and hence it moves faster than neon and strikes the wall more frequently.

$μ_{av}=\sqrt{\dfrac{8RT}{\pi M}}$

$μ_{av}∝(\dfrac1M)^{1/2}$

$\text{Hence. He has higher speed than neon}$

A container (cylindrical, base area

$=821cm^2$ ) fitted with frictionless, massless piston consists of five valves-I, II, III, IV, and V. The distance of valves from the initial position of the piston is

$15, 30, 40, 45$ and

$50$ cm, respectively. The initial height of the piston from the base of the container was

$60$ cm. These valves open automatically if pressure exceeds over

$1.5, 2.2, 2.5, 4.4$ , and

$4.8$ atm, respectively. Under the given conditions(shown in the diagram), the system is in a state of equilibrium. The piston is now compressed(moved downward) slowly and isothermally. Neglect the volumes of valve connectors. Select the correct option(s).

0%
Only the momentum of the molecules
0%
Only the frequency of collision of molecules
0%
Both momentum and frequency of collision of molecules
0%
Neither momentum nor frequency of collision of molecules

Explanation

As the piston (moved downward) slowly.

It does not change the pressure of the system.

$\therefore \Delta P=0$

$\therefore$ There is no change in momentum of molecule but as there is change in volume.

i.e., volume decreased.

$\therefore$ particles come closure & therefore particle collides with each other more frequently.

$\therefore$ As no. of collisions increases.

$\therefore$ Collision frequently changes

$\therefore$ There is change in frequency only.

Answer option B only.

The values of enthalpies of reactants and products are

$x$ and

$y$

$J/mol$ , respectively. If the activation energy for the backward reaction is

$z$

$J/mol$ , then the activation energy for forward reaction will be
(in J/mol)

0%
$x-y-z$
0%
$x-y+z$
0%
$y-x-z$
0%
$y-x+z$

Hydrolysis of ethyl acetate is catalysed by

$HCl$ . The rate is proportional to the concentration of both the ester and

$HCl$ . The rate constant is

$0.1{ M }^{ -1 }{ h }^{ -1 }$ . What is the half-life, if the initial concentrations are

$0.02M$ for the ester and

$0.01M$ for the catalysing acid?

0%
$347h$
0%
$519h$
0%
$836h$
0%
$693h$

The initial rate of a zero-order reaction:

$A(g)\rightarrow 2B(g)$ is

$0.01M\quad { min }^{ -1 }$ . If the initial concentration of

$A$ is

$0.1M$ , the concentration of

$B$ after

$60s$ is?

0%
$0.09M$
0%
$0.01M$
0%
$0.02M$
0%
$1.2M$

A zero-order reaction

$A\rightarrow B$ . At the end of

$1h$ ,

$A$ is

$75$ % reacted. How much of it will be left unreacted at the end of

$2h$ ?

0%
$12.5$ %
0%
$6.25$ %
0%
$3.12$ %
0%
$0$ %

A first-order reaction:

$A(g)\rightarrow nB(g)$ is started with

$A$ . The reaction takes place at constant temperature and pressure. If the initial pressure was

${P}_{0}$ and the rate constant of reaction is

$K$ , then at any time,

$t$ the total pressure of the reaction system will be

0%
${ P }_{ 0 }\left[ n+\left( 1-n \right) { e }^{ -kt } \right]$
0%
${ P }_{ 0 }\left( 1-n \right) { e }^{ -kt }$
0%
${ P }_{ 0 }.n{ e }^{ -kt }$
0%
${ P }_{ 0 }\left[ n-(1-n) \right] { e }^{ -kt }\quad$

For a zero-order reaction:

$2{ NH }_{ 3 }(g)\rightarrow { N }_{ 2 }(g)+3{ H }_{ 2 }(g)$ , the rate of reaction is

$0.1atm/s$ . Initially only

${ NH }_{ 3 }(g)$ was present at

$3atm$ and the reaction is performed at constant volume and temperature. The total pressure of gases after

$10s$ from the start of reaction will be

0%
$4atm$
0%
$5atm$
0%
$3.5atm$
0%
$4.5atm$

The half-life periods of two first-order reactions are in the ratio

$3:2$ . If

${t}_{1}$ is the time required for

$25$ % completion of the first reaction and

${t}_{2}$ is the time required for

$75$ % completion of the second reaction, then the ratio,

${ t }_{ 1 }:{ t }_{ 2 }$ is

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

0%
$3:10$
0%
$12:25$
0%
$3:5$
0%
$3:2$

At a certain temperature, the reaction between

$NO$ and

${O}_{2}$ is fast, while that between

$CO$ and

${O}_{2}$ is slow. It may be concluded that

0%
$NO$ is more reactive than $CO$
0%
$CO$ is smaller in size than $NO$
0%
activation energy for the reaction:
$2NO+{ O }_{ 2 }\rightarrow 2{ NO }_{ 2 }$ is less
0%
activation energy for the reaction:
$2NO+{ O }_{ 2 }\rightarrow 2{ NO }_{ 2 }$ is high

In general, the rate of a reaction can be increased by all the factors except

0%
increasing the temperature
0%
increasing the concentration of reactants
0%
increasing the activation energy
0%
using a catalyst

The time taken in

$75$ % completion of zero-order reaction is

$10h$ . In what time, the reaction will be

$90$ % completed?

0%
$12.0h$
0%
$16.6h$
0%
$10.0h$
0%
$20.0h$

Explanation

$75 %$ completed reaction means

Left amount

$A=25$

Initial amount

$A_∘ = 100$ and the reaction is completed in

$t=10\ hrs$

Similarly for

$90%,$

Left amount

$A = 10$

Initial amount

$A_∘ = 100$

$A = A_∘−kt$

$25=100−k×10$

$k×10=75$

$k=7.5mol \ l^{−1}s^{−1}$

$A=A_∘−kt$

$10=100−7.5×t$

$7.5t=90$

$t=12.00 hrs$

Hence, option is

$(A)$ is correct answer.

The rate expression for a reaction is

$\cfrac { -dC }{ dt } =\cfrac { \alpha C }{ 1+\beta C }$ , where

$\alpha$ and

$\beta$ are constants and

$C$ is the concentration of reactant at time,

$t$ . The half-life for this reaction is

0%
$\cfrac { 1 }{ \alpha } \ln { 2 } +\cfrac { \beta { C }_{ o } }{ 2\alpha } \quad$
0%
$\cfrac { 1 }{ \beta } \ln { 2 } +\cfrac { \beta { C }_{ o } }{ 2\alpha }$
0%
$\cfrac { \beta \ln { 2 } }{ \alpha }$
0%
$\cfrac { \alpha }{ \beta }$

Thermal decomposition of dibromosuccinic acid (DBSA) taking place according to the given equation, obeys first-order kinetics. The progress of reaction may be followed by means of alkali titration of the solution (definite volume of reaction mixture) at various time intervals. If

${T}_{0}$ and

${T}_{t}$ be the ml of alkali solution at zero time and at any time

$t$ , respectively and

$a$ and

$(a-x)$ be the concentrations of DBSA at zero time and at any time

$t$ , respectively, then the value of

$\cfrac { a }{ a-x }$ is

0%
$\cfrac { { T }_{ o } }{ 3{ T }_{ 0 }-2{ T }_{ t } }$
0%
$\cfrac { { T }_{ o } }{ { T }_{ t } }$
0%
$\cfrac { { T }_{ o } }{ { T }_{ o }-{ T }_{ t } }$
0%
$\cfrac { { T }_{ o } }{ { 2T }_{ o }-{ T }_{ t } }$

CBSE Questions for Class 12 Engineering Chemistry Chemical Kinetics Quiz 14 - MCQExams.com

0:0:1

Practice Class 12 Engineering Chemistry Quiz Questions and Answers

CBSE Questions for Class 12 Engineering Chemistry Chemical Kinetics Quiz 14 - MCQExams.com

0:0:1

Practice Class 12 Engineering Chemistry Quiz Questions and Answers

Support mcqexams.com by disabling your adblocker.