Consider the three statements about reaction energy diagrams and the relative magnitudes of the activation energy, $$E_{a}$$, and the enthalpy of reaction, $$\Delta H$$. One or more of the statements is true. Identify the correct statement or combination of statements from the four choices below. Statement I For an endothermic reaction, the magnitude of $$E_{a}$$ is always greater than $$\triangle H$$. II For an exothermic reaction, the magnitude of $$E_{a}$$ is always greater than $$\triangle H$$. III For an exothermic reaction, adding a catalyst will decrease the magnitude of $$\triangle H$$.

Both statements I and II are true.

Both statements I and III are true.

MCQ Questions for Class 12 Engineering Chemistry Chemical Kinetics Quiz 11

The decomposition of

$Cl_2O_7$ at 400 K in gaseous phase to

$Cl_2$ and

$O_2$ is 1st order reaction. After 55 sec at 400 K, the pressure of reaction mixture increases from 0.62 to 1.88 atm. Calculate the rate constant of the reaction. Also, calculate the pressure of reaction mixture after 100 seconds:

0%
$1.58\times 10^{-2}, 2.33 atm$
0%
$1.58\times 10^{-3}, 2.33 atm$
0%
$15.8\times 10^{-2}, 23.3 atm$
0%
None of these

Explanation

$\displaystyle Cl_2O_7 (g) \rightarrow Cl_2(g) + 3.5 O_2 (g)$

Initial pressure of

$\displaystyle Cl_2O_7 = 0.62$ atm
Equilibrium pressure of

$\displaystyle Cl_2O_7 = 0.62 - x$ atm
Equilibrium pressure of

$\displaystyle Cl_2 = x$ atm
Equilibrium pressure of

$\displaystyle O_2 = 3.5 x$ atm.

Total equilibrium pressure

$\displaystyle = 0.62 - x + x + 3.5 x = 0.62 + 3.5 x = 1.88$ atm

$\displaystyle 3.5 x = 1.26$

$\displaystyle x = 0.36$ atm

Hence the equilibrium pressure of

$\displaystyle Cl_2O_7 = 0.62 -x = 0.62 - 0.36 = 0.26$ atm.

The rate constant of the reaction

$\displaystyle k = \dfrac {2.303}{t}log \dfrac {a}{a-x}$

$\displaystyle k = \dfrac {2.303}{55 \: sec} log \dfrac {0.62}{0.26} = 1.58 \times 10^{-2} /sec$

After 100 sec,

$\displaystyle 1.58 \times 10^{-2} = \dfrac {2.303}{100} log \dfrac {0.62}{a-x}$

$\displaystyle 0.6861 = log \dfrac {0.62}{a-x}$

$\displaystyle 4.854 = \dfrac {0.62}{a-x}$

$\displaystyle a-x = 0.1277$

$\displaystyle x = a - 0.1277 = 0.62 - 0.1277 = 0.4923$

Total equilibrium pressure

$\displaystyle = 0.62 + 3.5 x = 0.62 + 3.5 (0.4923) = 2.33$ atm

In the thermal decomposition of

$N_2O$ at 830 K, the time required to decompose half of the reactant was 263 sec at the initial pressure 290 mm. It takes 212 sec to decompose half of the reactant if initial pressure was 360 mm. What is the order of the reaction? Also calculate

$t_{1/2}$ for

$N_2O$ decomposition if initial pressure of

$N_2O$ is 1 atm.

0%
$O.R=2$ , Half life $=100.2 sec$
0%
$O.R=1$ , Half life $=10.02 sec$
0%
$O.R=4$ , Half life $=1.002 sec$
0%
None of these

Explanation

$\displaystyle \dfrac {t_{1/2,1}}{t_{1/2, 2}} = \dfrac {263}{212} = 1.24$

$\displaystyle \dfrac {P_2}{P_1} = \dfrac {360}{290} = 1.24$

Thus

$\displaystyle \dfrac {t_{1/2, 1}}{t_{1/2, 2}} = \dfrac {P_2}{P_1}$

$\displaystyle t_{1/2} \propto \dfrac {1}{P_i}$

Thus, the half life period is inversely proportional to the initial pressure. This is characteristic of second order reaction.

$\displaystyle \dfrac {t_{1/2, }}{263} = \dfrac {290}{760}$ (because 1 atm = 760 mm Hg)

$\displaystyle t_{1/2} = 100.2 sec$

A substance reacts according to 1

$^{st}$ order kinetic and rate constant for the reaction is

$1\times 10^{-2} sec^{-1}$ . If the initial concentration is 1 M. Rate of the reaction after 1 minute is:

0%
$5.49\times 10^{-3} mol\ litre^{-1} sec^{-1}$
0%
$5.55\times 10^{-3} mol\ litre^{-1} sec^{-1}$
0%
$7.49\times 10^{-3} mol\ litre^{-1} sec^{-1}$
0%
None of these

Explanation

$K=\dfrac {2.303}{t} log \dfrac {a}{(a-X)}$

$\therefore 10^{-2}=\dfrac {2.303}{1\times 60} log \dfrac {1}{(1-X)} (\because t=1 minute = 60 sec)$

$\therefore (1-X)=0.549$

Therefore, rate after 1 minute

$=K[1-X]^1$

$=10^{-2}\times (0.549)=5.49\times 10^{-3} mol\ L^{-1} sec^{-1}$

Hence, the correct option is

$A$

Mathematical expression for

$t_{1/4}$ i.e., when

$(1/4)^{th}$ reaction is over following first order kinetics can be given by:

0%
$t_{1/4}=\frac {2.303}{K}log 4$
0%
$t_{1/4}=\frac {2.303}{K}log 1/4$
0%
$t_{1/4}=\frac {2.303}{K}log 2$
0%
$t_{1/4}=\frac {2.303}{K}log \frac {4}{3}$

Explanation

As we know, for a first order reaction,

$t=\dfrac {2.303}{K}log \dfrac {a}{(a-x)}$

$t=t_{1/4}; x=\dfrac a4$

$\therefore t_{1/4}=\dfrac {2.303}{K} log \dfrac {a}{a-a/4}$

$=\dfrac {2.303}{K} log\dfrac {4}{3}$

Catalytic decomposition of nitrous oxide by gold at

$900^oC$ at an initial pressure of

$200$ mm, was

$50$ % in

$53$ minutes and

$73$ % in

$100$ minutes.

Velocity constant of the reaction is:

[Note: assume decomposition as the first order.]

0%
$1.308\times 10^{-2}$
0%
$2.317\times 10^{-2}$
0%
$3.208\times 10^{-3}$
0%
none of these

Explanation

Taking decomposition as first order:

$\because K=\dfrac {2.303}{t}log \dfrac {a}{(a-X)}$

Case

$I$ :

$a\propto\ 200\ mm$ ;

$X\propto 200\times (50/100)\ mm$ and

$t_{1/2}=53\ minutes$

$\therefore K_1=\dfrac {2.303}{53} log \dfrac {200}{200-100}$

$\quad \quad=1.308\times 10^{-2}\ min^{-1}$

Case

$II$ :

$a\propto 200\ mm; X\propto 200\times (73/ 100)\propto 146 mm$

$\therefore K_2=\dfrac {2.303}{100}log \dfrac {200}{200-146}$

$\quad \quad =1.309\times 10^{-2} min^{-1}$

$K=\dfrac {K_1+K_2}{2}=\dfrac {(1.308+1.309)\times 10^{-2}}{2}=1.308\times 10^{-2}$

Hence, the correct option is

$A$

According to the collision theory, most molecular collisions do not lead to a reaction. Which of the following is(are) necessary for collisions to successfully lead to the reaction?

0%
The total kinetic energy of the collision must be greater than some minimum value.
0%
A catalyst must be present at the collision.
0%
The colliding particles must be properly oriented in space when they collide.
0%
None of the above.

A 1st order reaction is 50% complete in 30 minute at

$27^oC$ and in 10 minute at

$47^oC$ . The sum of rate constants for reaction at

$27^oC$ and

$47^oC$ is:

0%
$9.24\times 10^{-2} min^{-1}$
0%
$12.24\times 10^{-2} min^{-1}$
0%
$8.24\times 10^{-2} min^{-1}$
0%
None of these

Explanation

For a

$1^{st}$ order reaction constant,

$K=\dfrac{0.693}{t_{1/2}}$
At

$27^oC: K_1=(0.693/30)=2.31\times 10^{-2} min^{-1}$
At

$47^oC: K_2=(0.693/10)=6.93\times 10^{-2} min^{-1}$

Hence, the correct option is (A).

Reaction

$A+B\longrightarrow C+D$ follow's following rate law rate

$=k={ \left[ A \right] }^{ \frac { 1 }{ 2 } }{ \left[ B \right] }^{ \frac { 1 }{ 2 } }$ . Starting with initial conc. of one mole of A and B each, what is the time taken for amount of A of become 0.25 mole. Given

$k = 2.31\times 10^{-3} sec^{-1}$ .

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

Explanation

$A+B\longrightarrow C+D$

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

$\displaystyle \frac { dx }{ dt } =k\sqrt { (a-x)(a-x) }$

$\displaystyle \frac { dx }{ dt } =k(a-x)$

$\Rightarrow t= \displaystyle \frac { 2.303 }{ k } \log_ { 10 } \left( \displaystyle \frac { a }{ a-x } \right)$

$t= \displaystyle \frac { 2.303 }{ 2.31\times 10^{ -3 } } \log_ { 10 } \left( \displaystyle \frac { 1 }{ 0.25 } \right)$

$t=600\ sec.$

Hence, the correct option is

$\text{B}$

The gas phase decomposition of dimethyl ether follows first order kinetics.

$CH_3OCH_3(g)\, \rightarrow\, CH_4(g)\, +\, H_2(g)\, +\, CO(g)$
The reaction is carried out in a constant volume container at

$500^\circ C$ and has a half life of 14.5 min. Initially, only dimethyl ether is present at a pressure of 0.40 atm. What is the total pressure of the system after 12 min? Assume ideal gas behaviour.

0%
0.74 atm
0%
7.4 atm
0%
74 atm
0%
none

Explanation

For first order reaction,
Half life =

$\displaystyle \frac{0.693}{k}$
or 14.5 =

$\displaystyle \frac{0.693}{k}$
or k =

$\displaystyle \frac{0.693}{k}\, =\, 4.78\, \times\, 10^{-2}\, min^{-1}$
Also,

$\displaystyle k\, =\, \frac{2.303}{t} log \frac{A_0}{A_t}\, =\, \frac{2.303}{t} log \frac{P_0}{P}$

or

$\displaystyle \frac{0.693}{14.5}\, =\, \frac{2.303}{12} log \frac{0.40}{(0.40\, -\, x)}$
or x = 0.1746 atm
Total pressure of container after 12 min is
(0.40 - x) + x + x + x = 0.4 + 2x
= 0.4 + 2

$\times$ 0.1746
= 0.7492 atm

Catalytic decomposition of nitrous oxide by gold at

$900^{0}C$ at an initial pressure of 200 mm was 50 % in 53 min and 73 % in 100 min, the value of velocity constant is :

0%
$1.308 \, \times\, 10^{-2}\, min^{-1}$
0%
$1.432 \, \times\, 10^{-2}\, min^{-1}$
0%
$2.318 \, \times\, 10^{-2}\, min^{-1}$
0%
None of these

Explanation

Case I:

$a\, \propto\, 200 mm$

$x\, \propto\, 200\,\times\, \displaystyle \frac{50}{100}$
or

$x\, \propto\, 100 mm.$

$t_{1/2}\,=\, 53 min.$

$k_1\,=\, \displaystyle \frac{2.303}{t}\,log\,\frac{a}{a\,-\,x}$

$k_1=\displaystyle \frac{2.303}{53} \,log\,\frac{200}{200 \,- \,100}\,=\, 1.308\,\times\,10^{-2}\,min^{-1}$

Case II :
a

$\propto$ 200 mm

$x$

$\propto\, 200\,\times\,\displaystyle \frac{73}{100}$
or,

$x\, \propto\, 146 mm$

$t_{73 \%}$ = 100 min

$k_2\,=\, \displaystyle \frac{2.303}{100}\, log\, \left ( \displaystyle \frac{200}{200\, -\, 146} \right )\, =\, 1.309\,\times\, 10^{-2}\, min^{-1}$

(b)

$k\,=\, \displaystyle \frac{k_1\,+\,k_2}{2}\,=\, \frac{(1.308\,+\, 1.309)\, \times\, 10^{-2}}{2}$

$k=1.308 \, \times\, 10^{-2}\, min^{-1}$

A first-order reaction: A

$\rightarrow$ Products and a second order reaction: 2R

$\rightarrow$ Products both have a half time of 20 min when they are carried out taking 4 mol

$L^{-1}$ of their respective reactants. The number of a mole per litre of A and R remaining unreached after 60 min from the start of the reaction, respectively, will be:

0%
1 and 0.5 M
0%
0.5 M and negligible
0%
0.5 and 1 M
0%
1 and 0.25 M

Explanation

In the case of first order reaction

$t_{1/2}$ will remain constant independent of initial concentration. So,

$\displaystyle 4\, mol\, L^{-1}\, \overset{20\, min}{\rightarrow}\, 2\, mol\, L^{-1}\, \overset{20\, min}{\rightarrow}\, 1\, mol\, L^{-1}\, \overset{20\, min}{\rightarrow}\, 0.5\, mol\, L^{-1}.$

That is, after 60 min 0.5 mol

$L^{-1}$ of A will be left unreached.
In the case of second order reaction,

$t_{1/2}$ in inversely proportional to the initial concentration of reactant, i.e.,

$t_{1/2}$ will go on doubling as the concentration of reactant will go on getting half, i.e.,

$t_{1/2}a$ will be constant. So,

$\displaystyle 4\, mol\, L^{-1}\, \overset{20\, min}{\rightarrow}\, 2\, mol\, L^{-1}\, \overset{40\, min}{\rightarrow}\, 2\, mol\, L^{-1}.$
That is, after 60 min, the concentration of R remaining unreached will be 1 mol

$L^{-1}$ .

In a first order reaction,

$75\%$ of the reactants disappeared in 1.386 hr. What is the rate constant?

0%
$3.6\, \times\, 10^{-3}\, s^{-1}$
0%
$2.7\, \times\, 10^{-4}\, s^{-1}$
0%
$72\, \times\, 10^{-3}\, s^{-1}$
0%
$1.8\, \times\, 10^{-3}\, s^{-1}$

Explanation

$t_{75\%}\, =\, 2t_{1/2}$

$\displaystyle \therefore\, t_{1/2}\, =\, \frac{1.386\, hr}{2}\, =\, 0.693\, hr$

$\displaystyle k\, =\, \frac{0.693}{t_{1/2}}\,= \frac{0.693}{0.693}\, =\, 1\, hr^{-1}\, =\, \frac{1}{3600s}$

$=\, 2.7\, \times\, 10^{-4}\, s^{-1}$

Which of the following isomerization reactions is/are of the first order ?

0%
Cyclopropane $\rightarrow$ Propane
0%
cis-But-2-ene $\rightarrow$ Trans-but-2-ene
0%
Vinyl allyl ether $\rightarrow$ Pent-4-enal
0%
$CH_4NC\, \rightarrow\, CH_3CN$

$A\rightleftharpoons B+C$
Time

$t$

$\infty$
Total pressure of

$(B + C)$

$P_2$

$P_3$
Find equilibrium constant

$k$ .

0%
$\displaystyle k=\frac{1}{t}ln\frac{P_{3}}{2(P_{3}-P_{2})}$
0%
$\displaystyle k=\frac{1}{t}ln\frac{P_{2}}{2(P_{3}-P_{2})}$
0%
$\displaystyle k=\frac{1}{t}ln\frac{P_{3}}{2(P_{3}+P_{2})}$
0%
None of these

Explanation

$a=P_{3}$

$x=2P_{2}-P_{3}$

$x=P_{3}-2(P_{3}-P_{2})$

$a-x=2(P_{3}-P_{2})$

$\displaystyle k=\frac{1}{t}ln\frac{a}{a-x}$

$\displaystyle k=\frac{1}{t}ln\frac{P_{3}}{2(P_{3}-P_{2})}$

$A \rightarrow Product, [A]_{0} = 2M$ . After

$10\ min$ reaction is

$10\%$ completed. If

$\dfrac {d[A]}{dt} = k[A]$ , then

$t_{1/2}$ is approximately.

0%
$0.693 min$
0%
$69.3 min$
0%
$66.0 min$
0%
$0.0693 min$

Explanation

$z=3,n=6$

$ln\quad A=-kt+ln\quad A$

$ln(0.9)=-k(600)+ln1$

$-0.10536=-k\times 600$

$\Longrightarrow k=0.0001756$

$ln\quad 0.5=-(0.0001756)\times t\times ln(1)$

$-0.693=-0.0001756\times t$

$\Longrightarrow t=3946.46$

$t{ 1 }/{ 2 }=\cfrac { (ln2) }{ k }$

$\Longrightarrow t\frac { 1 }{ 2 } =3947\quad secs=66mins$

80% of a first order reaction was completed in 70 min. How much it will take for 90% completion of a reaction ?

0%
114 min
0%
100 min
0%
140 min
0%
70 min

Explanation

According to given conditions,

$\displaystyle t_{80\%}\, \propto \, log\, \frac{100}{100\, -\, 80}\, =\, log\, \frac{10}{2}$

$= log\ 5 = 0.69$

$\approx$

$0.7$

$\displaystyle t_{80\%}\, \propto \, log\, \frac{100}{100\, -\, 90}\, =\, log\, 10\, =\, 1$

$\displaystyle \frac{t_{80\%}}{t_{90\%}}\, =\, \frac{0.7}{1}\, \Rightarrow\, \frac{70}{t_{90\%}}\, =\, \frac{0.7}{1}$

$\displaystyle \therefore t_{90\%}\, =\, \frac{70}{0.7}\, =\, \frac{700}{7}\, =\, 100\, min$

In a first order reaction, the initial conc. of the reactant was

$M/10.$ After 8 minutes 20 seconds the conc. becomes

$\,M/100.\,$
What is the rate constant?

0%
$\;5\times10^{-3}\,second^{-1}$
0%
$\;2.303\times10^{-5}\,second^{-1}$
0%
$\;2.303\times10^{-4}\,second^{-1}$
0%
$\;4.606\times10^{-3}\,second^{-1}$

Explanation

As we know,

$K\,=\,\displaystyle\frac{2.303}{t}log\left[\displaystyle\frac{a_0}{a} \right]$

$K=\dfrac{2.303}{500} \log \left (\dfrac{\dfrac{1}{10}}{\dfrac{1}{100}}\right )$

$=\,\displaystyle\frac{2.303}{500}log\,(10)\,=\,4.606\times10^{-3}\,sec^{-1}$

The decomposition of a compound P, at temperature T according to the equation

$2P_{(g)}\rightarrow4Q_{(g)}+R_{(g)}+S_{(l)}$ is the first order reaction. After 30 minutes from the start of decomposition in a closed vessel, the total pressure developed is found to be 317 mm Hg and after a long period of time the total pressure observed to be 617 mm Hg. Calculate the total pressure of the vessel after 75 minute, if volume of liquid S is supposed to be negligible. Also calculate the time fraction

$t_{7/8}$ ?
Given : Vapour pressure of S(l) at temperature T = 32.5 mm Hg.

0%
$P_{t}=37.955\:mm\:Hg,\:t_{7/8}=39.996\:min$
0%
$P_{t}=379.55\:mm\:Hg,\:t_{7/8}=399.96\:min$
0%
$P_{t}=179.55\:mm\:Hg,\:t_{7/8}=199.96\:min$
0%
None of these

Explanation

$2P_{(g)}\rightarrow 4Q_{(g)}+R_{(g)}+S_{(l)}$

$t=0$

$P_0$

$t=30\:min.$

$P_{0}-P$ 2PP/2

$t=\infty$ -

$2P_0$

$P_0/2$
so

$P_{0}-P+2P+P/2=317-32.5$
i.e.

$P_{0}+1.5\:P=284.5$ .........(i)
&

$2.5\:P_{0}=617-32.5=584.5$
so

$P_{0}=233.8$

$P=33.8$

$\displaystyle k\times 30=\ln \frac{233.8}{200}\Rightarrow k=0.0052$
At

$t=75\:min$

$\displaystyle 0.0052\times 75=\ln \frac{233.8}{P^{\circ}-P}$

$P^{\circ}-P=158.23\Rightarrow P=75.57$

$P_{T}=32.5+P_0+1.5P=347.155+32.5$

$P_{T}=379.65\:mm\:Hg$
(ii)

$0.0052\times t=\ln \:8$

$t=399.89\:min.$

$60$ % of the first-order reaction was completed in

$60$ min. The time taken for reactants to decompose to half of their original amount will be:

0%
$\approx\, 30\, min$
0%
$\approx\, 60\, min$
0%
$\approx\, 90\, min$
0%
$\approx\, 45\, min$

Explanation

$\displaystyle t_{1/2}\, =\, \frac{0.69}{k}\, =\, \frac{0.3\, \times\, 2.3}{k}$

$\displaystyle t_{60\%}\, =\, \frac{2.3}{k}\, log\, \frac{100}{100\, -\, 60}\, =\, \frac{2.3}{k}\, log\, \frac{10}{4}$

$\displaystyle k\, =\, \frac{2.3}{60}\, log\, \frac{10}{4}$

$\displaystyle \frac{0.3\, \times\, 2.3}{t_{1/2}}\, =\, \frac{2.3}{60}\, log\, \frac{10}{4}\, =\, \frac{1}{60}\, [1\, -\, 2\, log\, 2]\, =\, \frac{1}{60}\, [1\, -\, 0.6]$

$\displaystyle \frac{0.3}{t_{1/2}}\, =\, \frac{0.4}{60}$

$\therefore t_{1/2}\, =\, \frac{60\, \times\, 0.3}{0.4}\, =\, 45\, min$
Use direct relation

$\displaystyle t_{1/2}\, =\, 0.3,\, t_{x\%}\, =\, \left ( log\, \frac{100}{100\, -\, x}\right )$

$t_{60\%}\, =\, log\, \frac{10}{4}\, =\, (0.4)$

$\displaystyle \frac{t_{1/2}}{t_{60\%}}\, =\, \frac{0.3}{0.4}$

$\displaystyle \therefore t_{1/2}\, =\, t_{60\%}\, \times\, \frac{0.3}{0.4}\, =\, \frac{60\, \times\, 3}{4}\, =\, 45\, min$

A vessel contains dimethyl ether at a pressure of 0.4 atm. Dimethyl ether decomposes as

$CH_{3}OCH_{3}(g)\rightarrow CH_{4}(g)+CO(g)+H_{2}(g)$ . The rate constant of decomposition is

$4.78\times10^{-3}\:min^{-1}$ . Calculate the ratio to initial rate of diffusion of rate of diffusion after 4.5 hour of initiation of decomposition. Assume the composition of gas present and composition of gas diffusing to be same.

0%
$0.26 : 1$
0%
$0.13 : 1$
0%
$1: 0.26$
0%
$1 : 0.13$

Explanation

$CH_{3}OCH_{3}\rightarrow CH_{4}+CO+H_{2}$

$t = 00.4$

$t = 4.5 hr.$ 0.110.290.290.29

$\displaystyle kt=\ln \:\frac{P_{0}}{P}\Rightarrow 4.78\times 10^{-3}\times 4.5\times 60=\ln\frac{0.4}{P}$

$P=0.11\:atm$
at

$t=0\:P_{0}=0.4$

$M_0=46$
at

$t=4.5\:hr.\:P=0.11+0.29\times 3=0.98\:atm$

$\displaystyle M=\frac{0.11\times 46+0.29(16+28+2)}{0.98}=18.77$

$\displaystyle \frac{r_{0}}{r}=\frac{P_{0}}{P}\sqrt{\frac{M}{M_{0}}}=\frac{0.4}{0.98}\sqrt{\frac{18.77}{46}}=0.26$

In this case we have

$A\rightarrow B+C$
Time t

$\infty$
Total

$press^{r}$

$P_2$

$P_3$
Find k?

0%
$\displaystyle k=\frac{1}{t}ln\frac{P_{3}}{2(P_{3}-P_{2})}$
0%
$\displaystyle k=\frac{1}{t}ln\frac{P_{3}}{2(P_{3}+P_{2})}$
0%
$\displaystyle k=\frac{1}{t}ln\frac{P_{2}}{2(P_{3}-P_{2})}$
0%
$\displaystyle k=\frac{1}{t}ln\frac{P_{2}}{2(P_{3}+P_{2})}$

Explanation

Let P be the initial pressure of A and x be the change in the pressure of A to reach equilibrium.

When time

$\displaystyle t = \infty P = \dfrac {P_3}{2}$

When time is t, the total pressure

$\displaystyle P-x+x+x=P+x = P_2 \\ x = P_2 - P=P_2 - \dfrac {P_3}{2} \\ P-x = \dfrac {P_3}{2}-[P_2 - \dfrac {P_3}{2} ]= P_3 - P_2$

$\displaystyle k = \dfrac {1}{t}ln \dfrac {P}{P-x}$

$\displaystyle k = \dfrac {1}{t} ln \dfrac {\dfrac {P_3}{2}}{(P_3 - P_2)}$

$\displaystyle \displaystyle k=\frac{1}{t}ln\frac{P_{3}}{2(P_{3}-P_{2})}$

In the following reaction,

$A\rightarrow B$ , rate constant is

$1.2\times10^{-2}\:M\:s^{-1}\:$ . What is concentration of B after 10 min, if we start with 10 M of A?

0%
7.2 M
0%
14.4 M
0%
3.6 M
0%
None of these

Explanation

$k=1. 2 \times 10^{–2}Ms^{–1}$

Since unit of rate constant is

$Ms^{–1}$ , it is zero order

reaction

Rate law

$[B] = kt$

At t = 10 min

$[B] = 1. 2 \times 10^{–2} \times 10 \times 60\ M$

$[B] = 7. 2\ M$

90% of a first order reaction was completed in 100 min. What is the half life of the reaction ?

0%
63.3 min
0%
53.3 min
0%
43.3 min
0%
30 min

Explanation

$\bf{Hint:}$ Half-life of first-order reaction inversely proportional to rate constant.

$\bf{Explanation:}$

$\bf{Formula \ used:}$

$k=\dfrac{2.303}{t}log\dfrac{a}{a-x}$

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

Here,

$t_{1/2}$ is half life,

$k$ is rate constant

$t$ is time,

$a$ is initial concentration and

$(a-x)$ concentration at time

$t$

$\bf{Step:1}$ To find rate constant

$k$ .

Given that,

$a=100$

$x=90$

$t=100$

On substituting these values in the formula

$k=\dfrac{2.303}{t}log\dfrac{a}{a-x}$

We get,

$k=\dfrac{2.303}{100}log\dfrac{100}{100-90}$

$k=\dfrac{2.303}{100}log\dfrac{100}{10}$

$k={0.02303}$

$\bf{Step:2}$ To find half-life

$t_{1/2}$ .

On substituting rate constant value in the formula

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

We get,

$t_{1/2}=\dfrac{0.693}{0.02303} = 30 \ minutes$

$\bf{Final \ answer:}$ Option

$D$

In a certain reaction, 10% of the reactant decomposes in one hour, 20% in two hours, 30% in three hours, and so on. The dimension of the velocity constant (rate constant) is:

0%
$Hr^{-1}$
0%
$Mol\, L^{-1}\, hr^{-1}$
0%
$L\, mol^{-1}\, s^{-1}$
0%
$Mol\, s^{-1}$

Explanation

For zero-order reaction rate is independent of concentration. Hence, it is a zero-order reaction. So, dimension of

$k =$

$mol L^{-1}\, hr^{-1}$

For zero order: t

$=\displaystyle \frac{x}{k}\, or\, k\, =\, \frac{x}{t}$

If t

$=t_{10\%}\, =\, 60\, min,\, x\, =\, 10,$

Then, r

$=\displaystyle \frac{-10}{1}\, =\, 10\, mol\, L^{-1}\, hr^{-1}$

If t

$=t_{20\%}\, =\, 120\, min,\, x\, =\, 20,$

Then, r

$=\displaystyle \frac{20}{2}\, =\, 10\, mol\, L^{-1}\, hr^{-1}.$

Similarly, for 30% and so on.

Thus, the reaction is of zero order.

In a first order reaction, the initial concentrated of the reactant was

$M/10.$ After

$8$ minutes

$20$ seconds the concentrated becomes

$\,M/100.\,$ . What is the rate constant?

0%
$\;5\times10^{-3}\,sec^{-1}$
0%
$\;2.303\times10^{-5}\,sec^{-1}$
0%
$\;2.303\times10^{-4}\,sec^{-1}$
0%
$\;4.606\times10^{-3}\,sec^{-1}$

Explanation

$K\,=\,\displaystyle\frac{2.303}{t}log\left[\displaystyle\frac{a_0}{a} \right]$

$K\,=\,\displaystyle\frac{2.303}{500}log\displaystyle\frac{\displaystyle\frac{1}{10}}{\displaystyle\frac{1}{100}}$

$=\,\displaystyle\frac{2.303}{500}log\,10\,=\,4.606\times10^{-3}\,sec^{-1}$

Unit of frequency factor (A) is:

0%
mol/L
0%
mol/L.s
0%
depend upon order of reaction
0%
it does not have any unit

For a first-order reaction, the concentration changes from

$0.8\ M$ to

$0.4\ M$ in

$15$ mins. The time taken for the concentration to change from

$0.1\ M$ to

$0.025\ M$ is:

0%
$30$ mins.
0%
$15$ mins.
0%
$7.5$ mins.
0%
$60$ mins.

The rate constant of the reaction

$A\rightarrow B$ is

$0.6\times 10^{-3} \ moleL^{-1}s^{-1}$ . If the concentration of

$A$ is

$5 \ M$ , then concentration of

$B$ after

$20 \ minutes$ is:

0%
$0.36 M$
0%
$0.72 M$
0%
$1.08 M$
0%
$3.60 M$

Explanation

The rate constant has unit mole per second which indicates zero order reaction.

x = kt
The concentration of

$B$ after 20 minutes is

$\displaystyle 20 \times 60 \times 0.6\times 10^{-3} = 0.72 \: M$

_________ increases effective collisions without increasing average energy.

0%
An increase in the reactant concentration
0%
An increase in the temperature
0%
A decrease in pressure
0%
Catalysts
0%
$\displaystyle pH$

What is the activation energy of the reverse reaction by this diagram?

0%
$E$
0%
$D$
0%
$C$
0%
$B$

Explanation

The activation energy of the reverse reaction by this diagram is given by

$E$ .
Activation energy is the amount of energy required to raise reactants to transition state.

Which of the following choice is correct regarding to increase the rate of a reaction?

0%
Decreasing the temperature
0%
Increasing the volume of the reaction vessel
0%
Reducing the activation energy
0%
Decreasing the concentration of the reactant in the reaction vessel

Explanation

The relation between specific rate (k), temperature (T) and activation energy (E_a) can be given by Arrhenius equation as follows,

$k = A e^{-E_a/RT}$

$k = p.Z.e^{-Ea/RT}$

Where,

k = specific rate (or) rate constant

A = pZ = Frequency factor

p = probability factor or steric factor or orientation factor

Z = no. of binary collisions per unit time

E_a = Activation energy

R = Universal Gas constant

T = Absolute Temperature

It is clear that the rate of a reaction is proportional to the temperature but inversely proportional to the activation energy.

Which statements describe the condition(s) required for a successful formation of a product in a reaction ?

0%
The collision must involve a sufficient amount of energy, provided from the motion of the particles, to overcome the activation energy
0%
The relative orientation of the particles has little or no effect on the formation of the product
0%
The relative orientation of the particles has an effect only if the kinetic energy of the particles is below some minimum value
0%
The relative orientation of the particles must allow for formation of the new bonds in the product
0%
The energy of the incoming particles must be above a certain minimum value and the relative orientation of the particles must allow for formation of new bonds in the product

$50$ % of the reactant is converted into a product in a first order reaction in 25 minutes, how much of it would react in 100 minutes?

0%
$93.75$ %
0%
$87.5$ %
0%
$75$ %
0%
$100$ %

Explanation

According to first order reaction kinetics:

$K=\dfrac{ln\ 2}{25}min^{-1}$

$[A]_t=[A]_oe^{-4\ ln2}$

$[A]_t=\dfrac{[A]_o}{2^4}=0.0625[A]_o$

Hence, the percentage will be

$100-6.25=93.75$ %.

Why does the probability curve become narrower when gas particles are more massive?

0%
The gas particles can travel at higher speeds
0%
The gas particles are less likely to travel at slower speeds
0%
The gas particles are more likely to travel at higher speeds
0%
The gas particles can not travel at higher speeds
0%
There are more collisions between gas particles

Consider the three statements about reaction energy diagrams and the relative magnitudes of the activation energy,

$E_{a}$ , and the enthalpy of reaction,

$\Delta H$ .
One or more of the statements is true.
Identify the correct statement or combination of statements from the four choices below.

	Statement
I	For an endothermic reaction, the magnitude of $E_{a}$ is always greater than $\triangle H$ .
II	For an exothermic reaction, the magnitude of $E_{a}$ is always greater than $\triangle H$ .
III	For an exothermic reaction, adding a catalyst will decrease the magnitude of $\triangle H$ .

0%
Statement I only is true.
0%
Statement II only is true.
0%
Both statements I and II are true.
0%
Both statements I and III are true.

Explanation

For endothermic reaction, $\Delta H={ H }_{ P }-{ H }_{ R }=positive.$

For exothermic reaction, $\Delta H=negative$ From the above figures, it is confirmed that

(i) For an endothermic reaction, $E_a$ is always greater than $\Delta H.$

(ii) For an exothermic reaction, $E_a$ may or may not be greater than $\Delta H.$ While catalysts decrease the activation energy only and does not affect the $\Delta H$ of the reaction.

Therefore, option $(A)$ is the correct answer.

Among the following which will decrease the rate of the reaction?
i. Using highly concentrated reactants
ii. Decreasing the temperature by

$25\ K$
iii. Stirring the reactants

0%
i only
0%
ii only
0%
i and iii only
0%
ii and iii only
0%
i, ii, and iii

Explanation

$\bullet$ Using highly concentrated reactants will increase the rate of the reaction as rate is directly proportional to the

concentration of reactants.

$\bullet$ Decreasing the temperature by 25 K will decrease the rate constant and hence the rate of reaction will decrease.

$\bullet$ Stirring the reactants increases the rate of interaction between the reactants and hence the rate of reaction increases.

$\therefore$ The correct answer is (ii) only.

The rate of reaction increases with rise in temperature because of :

0%
Increase in number of activated molecules
0%
Increase in energy of activation
0%
Decrease in energy of activation
0%
Increase in the number of effective collisions

Explanation

The rate of reaction increases when the rate constant increases which is given by the formula

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

When the temperature increases the number of effective collisions increases which is due to the increase in the number of activated molecules having energy greater than threshold energy.

$2N_2O_5(g) \rightarrow 4NO_2(g) + O_2(g)$
What is the ratio of the rate of decomposition of

$N_2O_5$ to rate of formation of

$NO_2$ ?

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

What does it mean when a collision is elastic?

0%
No energy is gained or lost.
0%
Energy is gained.
0%
Energy is lost.
0%
The particles can stretch out.
0%
The particles slow down.

Consider the Arrhenius equation,

$k = Ae^{-E_{a}/(RT)}$ , which of the following would not increase the rate constant of the reaction?

0%
A new value of the constant $A$ was calculated, this new value being higher than the previous one
0%
The temperature of the reaction is increased
0%
Experiments have revealed that the previously accepted activation energy of the reaction was calculated incorrectly. The new value is larger.
0%
The universal gas constant was redetermined to a smaller value

Explanation

$k={ A.e }^{ { -E }_{ a }/RT }$

(i) If the value of $A$ increases, the value of $k$ increases.

(ii) If the value of $T$ increases, ${ E }_{ a }/RT$ decreases. Therefore ${ e }^{ { -E }_{ a }/RT }$ increases. Therefore, $k$ increases.

(iii) If ${ E }_{ a }$ increases, ${ e }^{ { -E }_{ a }/RT }$ decreases. Therefore, $k$ decreases.

(iv) If $R$ increases, ${ e }^{ { -E }_{ a }/RT }$ increases. Therefore, $k$ increases.

Here is another look at the reaction of crystal violet with sodium hydroxide, a first-order reaction (In

$A$ v time):
What is the significance of the slope?

0%
The slope represents the change in concentration over time.
0%
The slope represents the inverse of change in concentration over time.
0%
The slope represents the negative value of the rate constant.
0%
The slope represents the negative rate of product concentration over time.

Explanation

By graph,

$\ln { [{ A }_{ t }] } =-Kt+\ln { [{ A }_{ o }] }$

$\therefore \ln { \left[ \cfrac { { A }_{ t } }{ { A }_{ o } } \right] } =-Kt\longrightarrow (1)$

$K\longrightarrow$ Rate constant=slop of graph

$[{ A }_{ t }],[{ A }_{ o }]\rightarrow$ concentrations of reactant at

$t=0,t=5$ respectively.

It is a first order reaction.

$\therefore$ The slop represents change in concentration over time. (from (1))

In a zero-order reaction, if the initial concentration of the reactant is doubled, the time required for half the reactant to be consumed:

0%
increases two-fold
0%
increases four-fold
0%
decreases by half
0%
does not change

Explanation

For a zero order reaction,

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

Time increases twice as half-life is directly proportional to initial concentration

A zero-order reaction,

$A \rightarrow$ Product, with an initial concentration

${ \left[ A \right] }_{ 0 }$ has a half-life of

$0.2 s$ . If one starts with the concentration

$2{ \left[ A \right] }_{ 0 }$ , then the half-life is:

0%
$0.1 s$
0%
$0.4 s$
0%
$0.2 s$
0%
$0.8 s$

Explanation

For zero-order reaction, half-life is

$\dfrac{[A]_0}{2k}$

"or"

$k=\dfrac{[A]_0}{2\ t_{1/2}}=\dfrac{[A]_0}{0.4}$

If initial concentration is

$2[A] _0$ then,

$t_{1/2}=\dfrac{2[A]_0}{2\times \dfrac{[A]_0}{0.4}}=0.4\ sec$

Decomposition of

$NH_3$ on gold surface follows zero order kinetics. If rate constant

$K$ is

$5\times 10^{-4}M$

$s^{-1}$ , rate of formation of

$N_2$ will be:

0%
$10^{-3}M\ s^{-1}$
0%
$2.5\times 10^{-4}M \ s^{-1}$
0%
$5\times 10^{-4}M \ s^{-1}$
0%
zero

Explanation

$N_{2} + 3H_{2} \rightarrow 2NH_{3}$

Rate

$= \dfrac{-1}{2} \dfrac{d [NH_{3}]}{dT} = \dfrac{d[N_{2}]}{dT} = \dfrac{1}{3} \dfrac {d[NH_{3}]}{dT} = k = 5 \times 10^{-4} M \ s^{-1}$

Rate of formation of

$N_{2}$

$= \dfrac{d[N_{2}]}{dT} = k = 5 \times 10^{-4} M \ s^{-1}$

Acid hydrolysis of ester is first order reaction and rate constant is given by

$k=\dfrac { 2.303 }{ t } \log { \dfrac { { V }_{ \infty }-{ V }_{ 0 } }{ { V }_{ \infty }-{ V }_{ t } } }$ where,

${ V }_{ 0 }$ ,

${ V }_{ t }$ and

${ V }_{ \infty }$ are the volume of standard

$NaOH$ required to neutralise acid present at a given time, if ester is

$50$ % neutralised then :

0%
${ V }_{ \infty }={ V }_{ t }$
0%
${ V }_{ \infty }=\left( { V }_{ t }-{ V }_{ 0 } \right)$
0%
${ V }_{ \infty }=2{ V }_{ t }-{ V }_{ 0 }$
0%
${ V }_{ \infty }=2{ V }_{ t }+{ V }_{ 0 }$

Explanation

$RCOO{ R }^{ \prime }+{ H }_{ 2 }O\xrightarrow { \quad { H }^{ + }\quad } RCOOH+{ R }^{ \prime }OH$

At $t=0$ ,	$a$	$0$	$0$
At time $t$ ,	$a-x$	$x$	$x$
At time $\infty$ ,	$a-a$	$a$	$a$

$t=0$ ,

${ V }_{ 0 }=$ volume of

$NaOH$ due to

${ H }^{ + }$ (catalyst)

${ V }_{ t }=x+{ V }_{ 0 }$

${ V }_{ \infty }=a+{ V }_{ 0 }$
If ester is

$50$ % hydrolysed then,

$x=\dfrac { a }{ 2 }$

${ V }_{ t }=\dfrac { a }{ 2 } +{ V }_{ 0 }$

$a=2{ V }_{ t }-2{ V }_{ 0 }$

$\therefore { V }_{ \infty }=2{ V }_{ t }-2{ V }_{ 0 }+{ V }_{ 0 }$

$=2{V}_{t}-{V}_{0}$

The rate of a particular reaction triples when temperature changes from

$50^o$ C to

$100^o$ C. What is the activation energy of the reaction (in

$J.mol^{-1}$ )?

$(log 3=0.4771, R=8.314K^{-1}mol^{-1})$

0%
$24.012\times 10^{-3}$
0%
$24.012\times 10^3$
0%
$22.012\times 10^{-3}$
0%
$22.012\times 10^3$

Explanation

From Arrhenius equation we have,

$2.303\ log\dfrac{K_2}{K_1}=\dfrac{E_a}{R}(\frac{1}{T_1}-\frac{1}{T_2})$

$2.303\ log\dfrac{3}{1}=\dfrac{E_a}{8.314}(\frac{1}{323}-\frac{1}{373})$

$\Rightarrow E_a=22.012\times 10^3\ J$

The formation of

$H_2O_2$ in the upper atmosphere follows the mechanism:

$H_2O+O\rightarrow 2OH \rightarrow H_2O_2$ ;

$\Delta H=72\ kJ mol^{-1}$ ,

$E_a=77\ kJ$

$mol^{-1}$ .

Then,

$E_a$ for the backward reaction will be (per mol):

0%
$-149$ kJ
0%
$+149$ kJ
0%
$-5$ kJ
0%
$+5$ kJ

Explanation

Relation used;

$\Delta H=E_{a(for.)}-E_{a(back)}$

$E_{a(back)}=E_{a(for)}-\Delta H$

$E_{a(b)}=77-72=+5\ kJ$ .

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%
The intrinsic energy of the reaction $2NO+O_2\rightleftharpoons 2NO_2$ is less
0%
$CO$ is smaller in size than that of $NO$
0%
$CO$ is poisonous
0%
The activation energy for the reaction $2NO+O_2\rightleftharpoons 2NO_2$ is less

$K_{p}$ of a reaction at

$300\ K$ is

$6\ atm$ and

$2\ atm$ at

$450\ K$ . Which of the following statements is incorrect about this reaction, if

$\triangle n_{g} = 1$ ?

0%
The reaction is exothermic
0%
The rate of backward reaction increases more than that of forward reaction with increase in temperate
0%
$E_{a}$ for the forward reaction is more than that of backward reaction
0%
The difference between heat of reaction at constant pressure and that at constant volume is $RT$

Explanation

Since the value of the equilibrium constant,

$K_{p}$ decreases with rise in temperature, it is an exothermic reaction. For an exothermic reaction, the rate of backward reaction increases more than that of forward reaction with the increase in temperature. For such reactions,

$(E_{a})_{f} < (E_{a})_{b}$

We know that,

$\triangle H = \triangle E + \triangle n_{g}RT$

$\therefore \triangle H - \triangle E = RT = constant\ [\because \triangle n_{g} = 1(given)]$ .

What is the formula to find the value of

$t_{1/2}$ for a zero order reaction?

0%
$\dfrac {k}{[R]_{0}}$
0%
$\dfrac {2k}{[R]_{0}}$
0%
$\dfrac {[R]_{0}}{2k}$
0%
$\dfrac {0.693}{k}$

Explanation

For zero order reaction:

$R_0-R=kt$ or

$t=\dfrac{R_0-R}{k}$ ------- 1

$t_{1/2} = t_{50\%}\ \Rightarrow R = \dfrac {R_0}{2}$

$t_{1/2} = \dfrac{R_0-\dfrac{R_0}{2}}{k}=\dfrac{R_0}{2k}$

Therefore, the formula of

$t_{1/2}$ for a zero order reaction is

$\dfrac {[R]_{0}}{2k}$ .

CBSE Questions for Class 12 Engineering Chemistry Chemical Kinetics Quiz 11 - 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 11 - MCQExams.com

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Practice Class 12 Engineering Chemistry Quiz Questions and Answers

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