The concentration of $$N_{2}O_{5}$$ in liquid bromine varied with time as follows: $$t/s$$ $$0$$ $$100$$ $$300$$ $$500$$ $$[N_{2}O_{5}]M$$ $$0.12$$ $$0.088$$ $$0.049$$ $$0.026$$ What will be the rate constant and order of reaction?

$$k=2\times 10^{-3}s^{-1}$$, order$$=1st$$

$$k=3.06\times 10^{-3}s^{-1}$$, zero

$$k=4\times 10^{-3}s^{-1}$$, zero

MCQ Questions for Class 12 Engineering Chemistry Chemical Kinetics Quiz 13

A reaction requires 60 minutes and 20 minutes to complete 50% at

$27^o C$ and

$47^oC$ respectively. calculate activation energy of the reaction.

0%
$45678$ cal
0%
$26789$ cal
0%
$10480$ cal
0%
None of these

What will be the initial rate of a reaction if its rate constant is

${10}^{-3}{min}^{-1}$ and the concentration of reactant is

$0.2\ mol .{dm}^{-1}$ . Also, the amount of reactant converted into products in 200 minutes is:

0%
$2\times{10}^{-4} mol. {dm}^{-3}{min}^{-1}$ , $81.97\%$
0%
$2\times{10}^{-4} mol. {dm}^{-3}{min}^{-1}$ , $18.03\%$
0%
$2\times{10}^{-4} mol. {dm}^{-3}{min}^{-1}$ , $76\%$
0%
$2\times{10}^{-4} mol. {dm}^{-3}{min}^{-1}$ , $24\%$

$75\%$ of a first-order reaction is completed in 30 minutes. What is the time required for

$93.75\%$ of the reaction (in minutes) ?

0%
$45$
0%
$120$
0%
$90$
0%
$60$

When the initial concentration of the reactant doubled, the half-life period of the reaction is also doubled. Hence the order of the reaction is

0%
one
0%
two
0%
fraction
0%
zero

A reaction

$P \rightarrow Q$ is

$25\%$ complete in

$25 min,$

$50\%$ complete in

$25 min$ if

$[ P ]$ is halved. The reaction is

$25 \%$ completed in

$50 min$ if

$[ P ]$ is doubled. The order of reaction is

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

The initial concentration of

$N_2O_5$ in the following first order reaction

$N_2O_{5(g)} \rightarrow 2NO_2(g) + \dfrac{1}{2} O_2 (g)$ was

$1.24 \times 10^{-2} \ molL^{-1}$ at 318 K. The concentration of

$N_2O_5$ after 60 min was

$0.20 \times 10^{-2} \ molL^{-1}$ . Calculate the rate constant of the reaction at 318 K.

0%
$0.0104 min^{-1}$
0%
$0.0204 min^{-1}$
0%
$0.0304 min^{-1}$
0%
$0.0404 min^{-1}$

Reaction

$A\rightarrow B$ follows second order kinetics. Doubling the concentration of A will increase the rate of formation of B by a factor of:

0%
$1/4$
0%
$1/2$
0%
$2$
0%
$4$

The half-life of a first-order reaction is 10 min. If the initial amount is 0.08 mol/L and the concentration at some instant is 0.01mol/L then time will be:

0%
10 min
0%
30 min
0%
20 min
0%
40 min

Consider the following first order competing reaction :

$X\xrightarrow{{{K_1}}}A + B$ and

$Y\xrightarrow{{{K_2}}}C + D$

$50\%$ of the reaction of

$X$ was completed when

$96\%$ of the reaction of

$Y$ was completed , the ratio of their rate constant

$\left( {{K_2}/{K_1}} \right)$ is:

0%
$4.06$
0%
$0.215$
0%
$1.1$
0%
$4.65$

The curves

$M$ and

$N$ represent the variation of energy with reaction coordinates for the reaction

$A(g)+B(g) \rightleftharpoons C(g)+D(g)$ in absence and presence of negative catalyst
Of the curves

$M$ and

$N$ , which represents the

$E_{a}$ for the backward reaction in the presence of catalyst

0%
$P$
0%
$Q$
0%
$S+P$
0%
$P+Q$

In a first order reaction,

$0.16$ moles of reactant decrease to

$0.02$ moles in

$144$ minutes, its half life is ?

0%
$24$
0%
$12$
0%
$72$
0%
$48$

For a zero order reaction:

0%
${ t }_{ \frac { 1 }{ 2 } }\propto a$
0%
${ t }_{ \frac { 1 }{ 2 } }\propto \cfrac { 1 }{ a }$
0%
${ t }_{ \frac { 1 }{ 2 } }\propto { a }^{ 2 }$
0%
${ t }_{ \frac { 1 }{ 2 } }\propto \cfrac { 1 }{ { a }^{ 2 } }$

A graph of

$log_{10}k$ is plotted against

$1/T$ for a reaction with order 2.If the activation energy is 10 kJ/mol, then magnitude of the slope will be:

0%
$4.34\times 10^3$
0%
$5.2\times 10^2$
0%
$1.2\times 10^4$
0%
$6.8\times 10^{-3}$

In a zero order reaction 47.5% of the reactant remains at the end of 2.5 hours. The amount of reactant consumed in one hour is

0%
10.5%
0%
32.0%
0%
52.6%
0%
21.0%

The decomposition of A follows first order kinetics by the following equation:

$4A (g) \xrightarrow{\Delta} B(g) + 2C(g)$

If initially, the total pressure was

$800$ mm of Hg and after

$10$ minutes it is found to be

$650$ mm of Hg. What is the half-life of A?
(Assume only A is present initially)

0%
$10$ minutes
0%
$5$ minutes
0%
$7.5$ minutes
0%
None of these

For a reaction

$2A+B\rightarrow C+D$ , the active mass of

$B$ is kept constant but that of

$A$ is tripled. The rate of reaction will -

0%
Decrease by $3$ times
0%
Increased by $9$ times
0%
Increase by $3$ times
0%
Unpredictable

Explanation

According to the given reaction,

$Rate = k[A]^{2}[B]^{1}$

When A becomes 3A i.e. triples, the rate will become:

$Rate^{'} = k[3A]^{2}[B]^{1} = 9k[A]^{2}[B]$

Thus, rate of reaction is increased by 9 times.

For a reaction: nA

→→ product, the rate constant and rate of reactions are equal, What is the order of the reaction ?

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

The time taken for the completion of 90% of a first order reaction is 't' min. What is the time (in sec) taken for the completion of 99% of the reaction ?

0%
$2t$
0%
$t/30$
0%
$120t$
0%
$60t$

For zero order reaction, A

$\to$ ,a graph of rate vs time has slope equal to:( where k

$=$ rate of reaction)

0%
k
0%
-k
0%
zero
0%
-2.303 k

For a homogeneous reaction

$A\rightarrow 3B$ , if pressure after sometime

$t$ was

$P_t$ and after completion of reaction, pressure was

$P_\infty$ .Then select the correct relation.

0%
$K \times t$ = ln $(\cfrac { 2P_\infty }{ 3(P_\infty -Pt) } )$
0%
$K \times t$ = ln $(\cfrac {7 P_\infty }{ 3(P_\infty -Pt) } )$
0%
$K \times t$ = ln $(\cfrac {3P_\infty }{ 2(P_\infty -Pt) } )$
0%
$K \times t$ = ln $(\cfrac {5P_\infty }{ 2(P_\infty -Pt) } )$

The general integrated rate equation for first-order reaction is given by:

0%
$k=\cfrac{2.303}{t}ln\cfrac{[A]_{0}}{[A]_{t}}$
0%
$k=\cfrac{1}{t}ln\cfrac{[A]_{0}}{[A]_{t}}$
0%
$k=\cfrac{2.303}{t}log_{10}\cfrac{[A]_{0}}{[A]_{t}}$
0%
$k=\cfrac{1}{t}ln\cfrac{[A]_{0}}{[A]_{t}}$

The first order reaction has half-life of

$18$ hrs. What percentage of the reactant will remain after

$24$ hrs

$\left( {\log 2 = 0.3,\log 2.5 = 0.4} \right)$

0%
$20\%$
0%
$40\%$
0%
$45.5\%$
0%
$68.2\%$

The rates of most reactions doubled when their temperature is raised from

$298\ K$ to

$308\ K$ , calculate their activation energy?

0%
$42.0\ kJ$
0%
$5.29\ kJ$
0%
$4.2\ kJ$
0%
$52.9\ kJ$

A first order reaction goes upto 8% in 10 min. What time will it take to complete 99%

0%
555.55min
0%
305.52 min
0%
158.32 min
0%
155.55 min

The rate constant of a reaction increased by

$5%$ when temperature is raised from

$27^{o}$ to

$28^{o}\ C$ . The activation energy of the reaction is

$(\log\ 7=0.8450)$

0%
$36.5\ KJ$
0%
$16.5\ KJ$
0%
$47.5\ KJ$
0%
$27.5\ KJ$

The catalysts decrease the Ea from

$100 KJmol^-1$ to

$80 KJmol^-1$ . At what temperature the rate of reaction in the absence of catalyst at 500 K will be equal to the rate of reaction in the presence of a catalyst .

0%
$400K$
0%
$625K$
0%
$-1000K$
0%
$+1000K$

At least how half-lives should elapse for a I order reaction

$A\rightarrow$ products so that the reaction is at least

$95%$ completed ?

$(\log 2=0.3)$

0%
$6$
0%
$5$
0%
$4$
0%
$7$

The reaction,

$X \rightarrow Y$ is an exothermic reaction. Activation energy of the reaction for conversation of

$x$ into

$Y$ is

$150kJ mol^{-1}$ . Enthalpy is

$135 kJ mol^{-1}$ . The activation energy for the reverse reaction,

$Y \rightarrow X$ will be

0%
$280 kJ mol^{-1}$
0%
$285 kJ mol^{-1}$
0%
$270 kJ mol^{-1}$
0%
$15kJ mol^{-1}$

The rate constant of the first-order reaction may be described by following equation:

log K =

$14.34 - \cfrac{1.25 \times 10^4}{T}$ , calculate energy of activation.

0%
239.34kJ
0%
239.34 kcal
0%
239.34J
0%
239.34 cal.

Explanation

As we know that :

$lnK=ln\ A-\dfrac{E_a}{RT}$ -------(1)

We have given

$logK=14.34-\dfrac{1.25\times 10^4}{T}$ -----(2)

Comparing (1) and (2) we will come to know that

$\dfrac{E_a}{2.303R}=1.25\times 10^4$

$E_a=2.303\times 1.25\times 10^4\times R$

$E_a=2.303\times 1.25\times 10^4\times 8.314$

$E_a=239.34kJ$

Option A is correct.

Which one of the following plots is correct for a first order reaction:

Explanation

The plot shown in option C is correct for a first-order reaction, as for first order reaction:

$log(a-x)=-\dfrac{kt}{2.303}+loga$

It is similar to straight-line equation $y=mx+c$

where intercept is equal to $loga$ and slope equal to $-\dfrac{kt}{2.303}$

Negative slope shows Option C is correct answer.

If the concentration of reactant is reduced by n times, then the value of rate constant of the first order will

0%
Increase by n times
0%
Decrease by n times
0%
Not change
0%
Increase by 2n times

The initial concentration of cane sugar in presence of an acid was reduced from

$0.20$ to

$0.10\ M$ in

$5$ hours and to

$0.05\ M$ in

$10$ hours, what is the value of

$K$ ?

$(in\ hr^{-1})$

0%
$0.0693$
0%
$1.386$
0%
$0.1386$
0%
$3.465$

For a first order reaction, the ratio of time for the completion of 99.9% and half of the reaction is

0%
$8$
0%
$2$
0%
$6.6$
0%
$10$

If a reaction follows the Arrhenius equation, the plot

$lnK$ vs

$\dfrac{1}{(RT)}$ gives a straight line with a gradient

$(-y)$ unit. The energy required to activate the reactant is :

0%
y unit
0%
-y unit
0%
yR unit
0%
y/R unit

Explanation

We have,

$K=Ae^\frac{-E_a}{RT}$

$\therefore lnK=ln(Ae^\frac{-E_a}{RT})$

$\therefore lnK=lnA -Ea \dfrac{1}{RT}$

Compare it with

$y=mx+c$ we get slope,

$m=-E_a=-y\ (Given)$

$\therefore E_a=y$

The concentration of $N_{2}O_{5}$ in liquid bromine varied with time as follows:

$t/s$	$0$	$100$	$300$	$500$
$[N_{2}O_{5}]M$	$0.12$	$0.088$	$0.049$	$0.026$

What will be the rate constant and order of reaction?

0%
$k=2\times 10^{-3}s^{-1}$ , order $=1st$
0%
$k=3.06\times 10^{-3}s^{-1}$ , zero
0%
$k=4\times 10^{-3}s^{-1}$ , zero
0%
None of these

The rate law of the reaction A + 2B

$\rightarrow$ product is given by

$\frac{d(product)}{dt} = k[A]^2[B]$ . If

$A$ is taken in large excess, the order of the reaction will be:

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

Consider the following in respect of zero-order reaction
i.

${ t }_{\frac{1}{2} }$ is directly proportional to the initial concentration
ii. Time taken for the completion of the reaction is twice its

${ t }_{ 1/2 }$
iii. Concentration of the reactant decreases linearly with time
Which of the statements given above are correct?

0%
I&II only
0%
I&III only
0%
II & III only
0%
I, II & III

Activation energy of forward and backward process of reaction are 60 kJ and 40

$kJ \ mol^{-1}$ respectively . Which of the following are true statements ?

0%
It is endothermic reaction.
0%
It is exothermic reaction.
0%
Heat of reaction is $+20\ kJ \ mol^{−1}$
0%

Thershold enregy of reaction is $100 \ kJ\,mo{l^{ - 1}}$

If the values of enthalpies of reactants and products are p and q J/mol respectively. If the activation energy for the backward reaction is r J/mol, then the activation energy for forward reaction will be in (J/mol) (take only magnitudes)

0%
p-q-r
0%
p-q+r
0%
q-p-r
0%
q-p+r

Activation energy of the backward reaction is 10 kcal and

$\Delta H=$ 2.813 kcal.

Activation energy of the forward reaction is ___________ kcal.

0%
$128.13$
0%
$12.813$
0%
$12813$
0%
none of these

Decomposition of HI (g) on Gold surface is zero order reaction.Initially few moles of

$H_2$ are present in container then which of the following graph is correct?

For the decomposition of

${ N }_{ 2 }{ O }_{ 5 }(g)$ it is given that -

${ 2N }_{ 2 }{ O }_{ 5 }(g)\longrightarrow 4{ NO }_{ 2 }(g)$ Activation energy = Ea,

${ N }_{ 2 }{ O }_{ 5 }(g)\longrightarrow { 2NO }_{ 2(g) }+\frac { 1 }{ 2 } { O }_{ 2 }(g)$ activation energy =

${ E }a^{ ' }$ then

0%
$Ea=2{ E }a$
0%
$Ea>{ E }a^{ ' }$
0%
$Ea<{ E }a^{ ' }$
0%
$Ea={ E }a^{ ' }$

For a particular gaseous reaction a graph was plotted as shown. It shows that the reaction of

$A$ is

0%
Zero order w.r.t $A$
0%
1st order w.r.t $A$
0%
Second order w.r.t $A$
0%
a non-integer order w.r.t $A$

For a first order reaction,

${\text{A}} \to {\text{P,}}{{\text{t}}_{{\text{1/2}}}}\left( {{\text{half -}}\,{\text{life}}} \right)$ is

$10$ days. The time required for

$\dfrac{1}{4}^{th}$ conversion of

$A$ (in days) is :

$\left( {{\text{ln}}\,{\text{2 = 0}}{\text{.693, ln}}\,{\text{3 = 1}}{\text{.1}}} \right)$

0%
$4.1$
0%
$5$
0%
$2.5$
0%
$3.2$

For the chemical reaction,

$A \to$ products, the following

$\log \,K = 16.398 - \cfrac{{2800}}{T}$
Arrhenius factor for the reaction is______________

0%
$2.5 \times {10^{16}}$
0%
$5 \times {10^{16}}$
0%
$10^9$
0%
None of these

According to the Arrhenius equation,

0%
a high activation energy usually implies a fast reaction
0%
rate constant increases with increase in temperature. This is due to a greater number of collisions whose energy exceeds the activation energy
0%
higher the magnitude of activation energy, stronger is the temperature dependence of the rate constant.
0%
the pre-exponential factor is a measure of the rate at which collisions occure, irrespective of their energy.

For a gaseous equilibrium :

$2A(g)\rightleftharpoons 2B(g)+C(g)$

$K_{p}$ has a value 1.8 at 700 K. The value of

$K_{c}$ for the equilibrium :

$2B(g+C(g)\rightleftharpoons 2A(g)$ at that temperature is about _________.

0%
0.031
0%
32
0%
57.4
0%
103.3

In the following first order competing reaction:
A+Reagent

$\longrightarrow$ Product B+Reagent

$\longrightarrow$ 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:

0%
4.06
0%
0.246
0%
2.06
0%
0.06

$A\rightarrow B$ first order reaction

$A$ is optical active and

$B$ is optically inactive, a series of experiment were conducted on a solution of

$A$

Time	0	60 min	$\infty$
optical rotation	${82}^{o}$	${77}^{o}$	${2}^{o}$

Assume some impurity present calculate the optical rotation after

$5$ hours
(Given in

$1.066=0.064,{e}^{0.16}=1.17$ )

0%
60
0%
30
0%
20
0%
120

For an exothermic reaction

$Ea, Ea^{'}$ are the activation energies of the forward and backward reactions respectively. Then which is always false:

0%
$E_{a} < {E_a}^{'}$
0%
$E_{a} > \triangle H$
0%
$E_{a} < \triangle H$
0%
${E_a}^{'} < \triangle H$

CBSE Questions for Class 12 Engineering Chemistry Chemical Kinetics Quiz 13 - MCQExams.com

The plot shown in option C is correct for a first-order reaction, as for first order reaction:

$log(a-x)=-\dfrac{kt}{2.303}+loga$

It is similar to straight-line equation $y=mx+c$

where intercept is equal to $loga$ and slope equal to $-\dfrac{kt}{2.303}$

Negative slope shows Option C is correct answer.

0:0:1

Practice Class 12 Engineering Chemistry Quiz Questions and Answers

CBSE Questions for Class 12 Engineering Chemistry Chemical Kinetics Quiz 13 - MCQExams.com

The plot shown in option C is correct for a first-order reaction, as for first order reaction:

log(a−x)=−kt2.303+logalog(a-x)=-\dfrac{kt}{2.303}+loga

It is similar to straight-line equation y=mx+cy=mx+c

where intercept is equal to logaloga and slope equal to −kt2.303-\dfrac{kt}{2.303}

Negative slope shows Option C is correct answer.

0:0:1

Practice Class 12 Engineering Chemistry Quiz Questions and Answers

Support mcqexams.com by disabling your adblocker.

$log(a-x)=-\dfrac{kt}{2.303}+loga$

It is similar to straight-line equation $y=mx+c$

where intercept is equal to $loga$ and slope equal to $-\dfrac{kt}{2.303}$