MCQ Questions for Class 12 Engineering Chemistry Chemical Kinetics Quiz 6

Which of the following graphs is correct for the following reaction?

$C{ H }_{ 3 }-C{ H }_{ 2 }-CH=C{ H }_{ 2 }\xrightarrow [ { 300 }^{ o }C ]{ \quad { { H }_{ 2 } }/{ Ni }\quad } C{ H }_{ 3 }-C{ H }_{ 2 }-C{ H }_{ 2 }-C{ H }_{ 3 }$

Explanation

Reaction involves adsorption on the surface of

$Ni$ catalyst; hence it is a zero order reaction.

$a=2{ t }_{ { 1 }/{ 2 } }k$ (for zero order)

$\log { a } =\log { { t }_{ { 1 }/{ 2 } } } +\log { 2k }$

$Y=MX+C$
Here, slope

$M=1$ i.e.,

$\tan { \theta } =1$

$\therefore \theta ={ 45 }^{ o }$
Intercept

$C=\log { 2k }$

The rate of a reaction at

$10$ sec intervals are as follows:

Time $\left( sec \right)$	Rate $\left( mol\ { L }^{ -1 }{ sec }^{ -1 } \right)$
$0$	$4.8\times { 10 }^{ -2 }$
$10$	$4.79\times { 10 }^{ -2 }$
$20$	$4.78\times { 10 }^{ -2 }$
$30$	$4.81\times { 10 }^{ -2 }$

What will be the order of the reaction?

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

$A$ and

$B$ are two different chemical species undergoing first order decompositions with half-life periods as

$3$ and

$4.5$ minutes respectively. If the initial concentrations of

$A$ and

$B$ are in the ratio

$1 : 2$ , The ratio

$C_{t_{(A)}}/ C_{t_{(B)}}$ after three half-lives of

$A$ would be:

0%
$3 : 4$
0%
$1 : 1$
0%
$1 : 4$
0%
$4 : 3$

Explanation

For the first order reaction, $C_t=C_o\left (\cfrac {1}{2}\right)^n$

where, $n=\cfrac {t}{t_{1/2}}$ , $\quad C_o=$ Initial concentration

After $3$ half- lives of $A$ , ${C_t}_{(A)}={C_o}_{(A)}\left(\cfrac {1}{2}\right)^\cfrac {3t_{1/2}}{t_{1/2}}=\cfrac {{C_o}_{(A)}}{8}$

After $3$ half -lives of $A$ , for $B$ , $\quad n=\cfrac {3\times 3}{4.5}=2$

$\therefore {C_t}_{(B)}={C_o}_{(B)}\left(\cfrac {1}{2}\right)^2=\cfrac {{C_o}_{(B)}}{4}$

$\therefore \cfrac {{C_t}_{(A)}}{{C_t}_{(B)}}=\left(\cfrac {{C_o}_{(A)}}{{C_o}_{(B)}}\right)\times \cfrac {4}{8}=\cfrac {1}{2}\times \cfrac {4}{8}=1:4$

The half-life period of a first order process is

$1.6$ min. It will be

$90\%$ complete in :

0%
$5.3$ min
0%
$10.6$ min
0%
$43.3$ min
0%
$99.7$ min

Explanation

For a first order reaction,

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

At half-time,

$x=\displaystyle\frac{a}{2}$

$\therefore k=\displaystyle\frac{2.303}{1.6}\times log\frac{a}{\displaystyle a-\frac{a}{2}}$

$=\displaystyle\frac{2.303}{1.6}\times log 2$

$=\displaystyle\frac{2.303}{1.6}\times 0.3010$

$=0.4332$

For

$90\%$ completion,

$x=0.9$ a

$\therefore k=\displaystyle\frac{2.303}{t}log\frac{a}{a-0.9a}$

$0.4332=\displaystyle\frac{2.303}{t}log\frac{a}{0.1a}$

$=\displaystyle\frac{2.303}{t}log10$

$=\displaystyle\frac{2.303}{t}\times 1$

$t=\displaystyle\frac{2.303}{0.4332}$

$=5.3$ min.

Graph between

$\log { k }$ and

$\cfrac{1}{T}$ is a straight line with

$OX=5,\tan{\theta}=\left( \cfrac { 1 }{ 2.303 } \right)$ . Hence

${E}_{a}$ will be:

0%
$2.303\times 2$ cal
0%
$\cfrac { 5 }{ 2.303 } cal$
0%
$-2$
0%
None of these

Explanation

$\quad \log { k } =\log { A } -\cfrac { { E }_{ a } }{ 2.303RT }$

Slope

$=\cfrac { -{ E }_{ a } }{ 2.303R } =\cfrac { 1 }{ 2.303 } (given)\quad$

$\quad { E }_{ a }=-2.303\times R\times slope=R=-2\ cal$

Hence, the correct option is

$\text{C}$ .

The activation energy for the forward reaction

$X\rightarrow Y$ is

$60KJ$

${mol}^{-1}$ and

$\Delta H$ is

$-20KJ$

${mol}^{-1}$ . The activation energy for the backward reaction

$Y\rightarrow X$ is:

0%
$80KJ$ ${mol}^{-1}$
0%
$40KJ$ ${mol}^{-1}$
0%
$60KJ$ ${mol}^{-1}$
0%
$20KJ$ ${mol}^{-1}$

Explanation

$\implies\Delta H = E_a(f) - E_a(b)$
$\implies -20 = 60 - E_a(b)$
$\implies E_a(b) = 60 + 20$
$\implies 80 \space KJ mol^{-1}$

$N_2(g)+3H_2(g)\rightleftharpoons 2NH_3(g)+22$ kcal. The activation energy for the forward reaction

$50$ kcal. What is the activation energy for the backward reaction?

0%
$72$ kcal
0%
$28$ kcal
0%
$-72$ kcal
0%
$-28$ kcal

Explanation

$N_2(g)+3H_2(g)\rightleftharpoons 2NH_3(g)+22$ kcal

$\Rightarrow \Delta H = -22\ kcal$

$\because$ Activation energy for the forward reaction

$E_a=50$ kcal

$\therefore$ Activation energy for the backward reaction

$E_b=E_a-\Delta H=50+22=72$ kcal.

The unit ,

$\\mol L^{-1} s^{-1}$ is meant for the rate constant of the reaction having the order :

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

Explanation

General unit for rate constant

$= \left( \dfrac {mol}{L} \right)^{1-n} s^{-1}$
(where,

$n =$ order of reaction)

Given, unit for rate constant

$= \left( \dfrac {mol}{L} \right) s^{-1}$

On comparing,

$\left( \dfrac {mol}{L} \right)^{1-n} s^{-1} = \left( \dfrac {mol}{L} \right) s^{-1}$

$1 - n = 1$

$\Rightarrow n = 0$

Hence, the order of the reaction is zero.

If for a reaction in which

$A(g)$ converts to

$B(g)$ the reaction carried out at const.

$V$ and

$T$ results into the following graph.

0%
Then the reaction must be $A(g) \rightarrow 3B(g)$ and is a first order reaction
0%
Then the reaction must be $A(g) \rightarrow 3B(g)$ and is a second order reaction
0%
Then the reaction must be $A(g) \rightarrow 3B(g)$ and is a zero order reaction
0%
Then the reaction must be $A(g) \leftrightarrow 3B(g)$ and is a first order reaction

A first order reaction is

$75 \%$ completed in

$100$ minutes. How long time will it take for it's

$87.5 \%$ completion?

0%
$125\ min$
0%
$150\ min$
0%
$175\ min$
0%
$200\ min$

Explanation

Explanation:

From the above chart, we can conclude that after $2$ half-lives $75$ % of reaction is completed.

For completion of $87.5$ % reaction $3$ half lives are required.

For $2$ half lives $100$ $mins$ required.

Therefore, for $1$ half life $50$ $mins$ required.

For $3$ half lives $3\times 50$ = $150$ $mins$ required.

Hence, the correct answer is option $B$ .

In a reaction carried out at

$400\ K, 0.0001$ % of the total number of collisions are effective. The energy of activation of the reaction is:

0%
zero
0%
$7.37\ kcal/mol$
0%
$9.212\ kcal/mol$
0%
$11.05\ k cal/mol$

Explanation

The fraction of molecules having energy equal to or greater than $E_a$ is :
$x = \dfrac{n}{N} = e^{-E_a/RT} (x = \frac{n}{N} = 0.0001\% = \frac{0.0001}{100} = 10^{-6})$

$\therefore \log x = \dfrac{-E_a}{2.3 \times RT}$
$\implies \ log 10^{-6} = \dfrac{-E_a}{2.3 \times 2 \times 400}$
$\implies E_a = 6 \times 2.3 \times 800$
$\implies E_a = 11.05 \space kcal/mol$

For the zero order reaction

$A \rightarrow B + C ;$ initial concentration of

$A$ is

$0.1 M.$ If

$A = 0.08 M$ after

$10$ minutes, then it's half-life and completion time are respectively :

0%
$10$ min $; 20$ min
0%
$2 \times 10^{-3}$ min $; 4 \times 10^{-3}$ min
0%
$25$ min $; 50$ min
0%
$250$ min $; 500$ min

Explanation

Its a Zero Order Reaction.

$k = \dfrac{[{A}_{0}] - [A]}{t}$

$k = \dfrac{[0.1] - [0.08]}{10}$ =

$2 \times {10}^{-3}\space M \ {min}^{-1}$

${t}_{1/2} = \dfrac{[{A}_{0}]}{2k}$

${t}_{1/2} = 25\space min$

$t = \dfrac{[{A}_{0}]}{k}$

$t = 50\space min$

In a first order reaction, the concentration of the reactant decreases from 0.8 M to 0.4 in 15 minutes. The time taken for the concentration to change from 0.1M to 0.025 M is:

0%
30 min
0%
40 min
0%
35 min
0%
25 min

Explanation

Order = 1 (given)

$\because t_{75} = 2 \times t_{50}$

$\implies 2 \times 15 = 30$

$k = \frac{0.693 }{t_{50}} = \frac{0.693}{15}$

$a = 0.1 \space M$

$(a-x) = 0.025 \space M$

For first order reaction $k = \frac{2.303}{t} \log(\frac{a}{a-x})$

$\implies \frac{2.303 \times \log2}{15} = \frac{2.303}{t} \log(\frac{0.1}{0.025})$

$\implies t= 30 \space minutes$

A reaction takes place in three steps. The rate constants are

$k_{1}, k_{2}$ and

$k_{3}$ . The over all rate constant

$k = \dfrac {k_{1}k_{3}}{k_{2}}$ . If (energy of activation)

$E_{1}, E_{2}$ and

$E_{3}$ are

$60, 30$ and

$10\ kJ$ , the overall energy of activation is:

0%
$40$
0%
$30$
0%
$400$
0%
$60$

Explanation

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

$\therefore k=\dfrac{k_{1}k_{2}}{k_{2}}=\dfrac{A_{1}A_{3}}{A_{2}} \times e^{-\dfrac{E_a^{(1)}+E_a^{(3)}-E_a^{(2)}}{RT}}$

$E_{a}=E_a^{(1)}+E_a^{(3)}-E_a^{(2)}$

$=40KJ$

In bimolecular reaction, the steric factor

$P$ was experimentally determined to be

$4.5$ . The correct option(s) among the following is (are):

0%
the activation energy of the reaction is unaffected by the value of the steric factor
0%
experimentally determined value of frequency factor is higher than that predicted by Arrheneius equation
0%
since $P=4.5$ , the reaction will not proceed unless an effective catalyst is used
0%
tThe value of frequency factor predicted by Arrhenius equation is higher than that determined experimentally

Explanation

Activation Energy does not depend on the steric factor.

The steric factor which is 4.5 is much higher than Arrhenius factor for the frequency factor because

$P=\frac{Frequency Factor}{Arrhenius Factor}$

Which graph represents the zero order reaction?

$A(g) \rightarrow B(g)$ .

Explanation

For zero order reaction:
Rate of formation of product i.e.

$\dfrac {dB}{dt}$ increases with time

$(t)$
Hence,

$\dfrac {dB}{dt}\propto t$ .

A first order reaction has

$k = 1.5\times 10^{-6}$ per second at

$200^0C$ . If the reaction is allowed to run for 10 hrs, what percentage of the initial concentration would have changed into the product?

0%
5.213%
0%
10.21%
0%
20.3%
0%
15.2%

Explanation

Given,

$K=1.5\times 10^{-6} s^{-1}$

We know for a first order reaction,

$t=\dfrac{2.303}{K}log(\dfrac{N_0}{N})$

So on substituting given values,

$\dfrac{N_0}{N}=1.0555$

$\implies N=0.9475N_0$

Now product formed

$=0.9475N_0-N_0=0.0521N_0=$ 5.213%

The half-life of a radioisotope is four hours. If the initial mass of the isotope was 200 g, the mass remaining after 24 hours undecayed is:

0%
3.125 g
0%
2.084 g
0%
1.042 g
0%
4.167 g

Explanation

$k=\dfrac{0.693}{t_{\dfrac{1}{2}}}=\dfrac{0.693}{4}=0.17325$

$A=A_{0}e^{-kt}=200 \times e^{-0.17325 \times 24}=3.1277$ g

Select incorrect statements :

0%
Pre exponential factor for zero order reaction is a unitless quantity
0%
If $t_{1/4}$ = 30 sec then $t_{1/2}$ = 60 sec for first order reaction
0%
If $t_{1/3}$ = 30 sec then $t_{2/3}$ = 90 sec for sec-order reaction
0%
If $t_{1/5}$ = 30 sec then $t_{3/5}$ = 90 sec for zero order reaction.

Explanation

Arrhenius Equation is given by:-

$K=A$

$e^{-Ea/RT}$

where,

$K$ =rate constant of the reaction

$A$ = pre exponential factor

$Ea$ = Activation energy

$R$ = Ideal gas constant

$T$ = Temperature in Kelvin

Now,

$e^{-Ea/RT}$ is a dimensionless quantity.

So, unit of

$A$ will be same at that of

$K$

For zero order reaction we have,

$K=\cfrac {1}{t} \left([A]_o-[A]_t\right)$

so unit of

$K$ is

$mol L^{-1} time^{-1}$ , so it will be unit of pre exponential factor as well and it will not be an unitless quantity.

First order reaction completes

$20\%$ in

$5$ min. How much time it will take for

$60\%$ completion?

0%
$26.5$ min
0%
$20.5$ min
0%
$19.5$ min
0%
$18$ min

Explanation

In 5 minutes, the reaction is 20% complete.

$\displaystyle [A]_0=100$ and

$\displaystyle [A]=100-20=80$

$\displaystyle k= \dfrac {2.303}{t} log_{10} \dfrac {[A]_0}{[A]}$

$\displaystyle k= \dfrac {2.303}{5 \ min} log_{10} \dfrac { 100}{80}$

$\displaystyle k=0.0446 \ min^{-1}$

Now, the reaction is 60% complete.

$\displaystyle [A]_0=100$ and

$\displaystyle [A]=100-60=40$

$\displaystyle t= \dfrac {2.303}{k} log_{10} \dfrac {[A]_0}{[A]}$

$\displaystyle t= \dfrac {2.303}{0.0446 \ min^{-1}} log_{10} \dfrac { 100}{40}$

$\displaystyle t=20.5 \ min$

Cyclopropane rearranges to form propene:

$\triangle \rightarrow CH_{2} - CH = CH_{2}$

This follows first-order kinetics. The rate constant is

$2.714\times 10^{-3} sec^{-1}$ . The initial concentration of cyclopropane is

$0.29\ M$ . What will be the concentration of cyclopropane after

$100\ sec$ ?

0%
$0.035\ M$
0%
$0.22\ M$
0%
$0.145\ M$
0%
$0.0018\ M$

Explanation

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

$(a - x)$ is the concentration left after

$100\ sec$ .

$2.7\times 10^{-3} = \dfrac {2.303}{100} \log \dfrac {0.29}{(a - x)}$

$\Rightarrow \dfrac {0.27}{2.303} = \log \dfrac {0.29}{(a - x)} \Rightarrow 0.117 = \log \dfrac {0.29}{(a -x)}$

$\Rightarrow (a - x) = 0.22\ M$ .

Hence, the correct answer is option

$\text{B}$ .

The reaction

$2X \rightarrow Y + Z$ would be zero order reaction when:

0%
rate remains unchanged at any concentration of $Y$ and $Z$
0%
rate of reaction doubles if concentration of $Y$ is doubled
0%
rate of reaction remains same at any concentration of $X$
0%
rate of reaction of directly proportional to square of concentration of $X$

In a zero order reaction, 20% of the reaction complete in 10 s. How much time it will take to complete 50% of the reaction ?

0%
20 s
0%
25 s
0%
30 s
0%
40 s

Explanation

$20\%\quad completed\quad reaction\quad means\quad \\ left\quad amount\quad A=80\\ initial\quad amount\quad { A }_{ \circ }=100\quad and\quad the\quad reaction\quad is\quad completed\quad in\quad t=10s\\ Similarly\quad for\quad 50\%,\\ left\quad amount\quad A=50\\ initial\quad amount\quad { A }_{ \circ }=100\\ A={ A }_{ \circ }-kt\\ 80=100-k\times 10\\ k\times 10=20\\ k=2mol\quad { l }^{ -1 }{ s }^{ -1 }\quad \\ A={ A }_{ \circ }-kt\\ 50=100-2\times t\\ 2t=50\\ t=25s$

For the first order reaction,

$2H_2O_2\rightarrow 2H_2\,+\,O_2 \,\log k\,=\,14.34\,-\,1.25\,\times\,10^4\,K/T$ . The energy of activation for the above reaction if its half life period is 256 min.

0%
400 KJ / mole
0%
230 KJ / mole
0%
480 KJ / mole
0%
239 KJ / mole

Explanation

Given order of the reaction= First order

$t_{1/2}=256min$

$K= \cfrac {0.693}{t_{1/2}}=\cfrac {0.693}{256\times 60}=8.67 \times 10^{-8}$

Given,

$\log K= 14.34- \cfrac {1.25 \times 10^4}{T}K \longrightarrow (1)$

By Arrhenius equation,

$\log K= \log A- \cfrac {E_a}{2.303RT} \longrightarrow (2)$

(1) & (2)

$\longrightarrow$

$\Rightarrow \cfrac {E_a}{2.303RT}= \cfrac {1.25 \times 10^4}{T}$

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

$= 2.39 \times 10^5$

$=239 kJ/mole$

At low pressure, the fraction of the surface covered follows:

0%
zero-order kinetics
0%
first order kinetics
0%
second order kinetics
0%
fractional order kinetics

Explanation

At low pressure $(P_A \rightarrow 0)$ , $\theta_A$ is very small and proportional to the pressure.
The rate becomes first order rate (low P) = $k_2K_AP_A$

If a substance is adsorbed on surfaces.

It is the first order reaction.

The following data were obtained during the first order thermal decomposition of

$SO_2Cl_2$ at constant volume

$SO_2Cl_{2(g)} \rightarrow SO_{2(g)}\,+\,Cl_{2(g)}$

Expt	Time / $S^{-1}$	Total pressure/ atm
1	0	0.5
2	100	0.6

when the total pressure is 0.65 atm then the rate constant of reaction is ..............
if a = pi, and a - x = 2pi - Pt

0%
2.0 x $10^{-3}\,S^{-1}$
0%
2.12 x $10^{-3}\,S^{-1}$
0%
2.23 x $10^{-3}\,S^{-1}$
0%
2.34 x $10^{-3}\,S^{-1}$

Explanation

First order thermal decomposition

$SO_2 \, Cl_2 (g) \rightarrow SO_2 (g) + Cl_2 (g)$

$\begin{matrix} at\, t=0 & P_{ 0 } & 0 & 0 \\ at\, t=100 & P_{ 0 }-P & P & P \end{matrix}$

$P_0 = 0.5\ atm$

$P_0 + P = 0.6$

$P = 0.6 - 0.5 = 0.1$

$k = \dfrac{1}{t} \, \ln \, \dfrac{P_0}{P_0 - P}$

$k = \dfrac{1}{100} \, \ln \, \dfrac{0.5}{0.4}$

$k = \dfrac{1}{100} \, \ln \, 1.25$

$= 0.223 \times 10^{-2} = 2.23 \times 10^{-3}$

option

$C$ is correct option

The rate constant of a first order reaction is

$15\times 10^{-3} s^{-1}$ . How long will

$5.0\ g$ of this reactant take to reduce to

$3.0\ g$ ?

0%
$34.07\ s$
0%
$7.57\ s$
0%
$10.10\ s$
0%
$15\ s$

Explanation

$t = \dfrac {2.303}{k}\log \left(\dfrac {a}{(a - x)}\right)$

$t = \dfrac {2.303}{15\times 10^{-3}} \log\left( \dfrac {5}{3}\right) = 34.00\ s$

A first order reaction takes

$40$ min for

$30$ % decomposition. What will be

$t_{1/2}$ ?

0%
$77.7\ min.$
0%
$52.5\ min.$
0%
$46.2\ min.$
0%
$22.7\ min.$

Explanation

$30$ % decomposition means

$X = 30$ % of

$R_{0}$ or

$R = R_{0} - 0.3R_{0} = 0.7R_{0}$

For first order,

$k = \dfrac {2.303}{t}\log \dfrac {[R_{0}]}{[R]}$

$= \dfrac {2.303}{40}\log \dfrac {10}{7} min^{-1} = 8.918\times 10^{-3} min.^{-1}$

$t_{1/2} = \dfrac {0.693}{k} = \dfrac {0.693}{8.918\times 10^{-3} min.^{-1}} = 77.7\ min/$

The half-life of the reaction

$X\rightarrow Y$ , following first-order kinetics, when the initial concentration of

$X$ is

$0.01$ mol L

$^{-1}$ and the initial rate is

$0.00352$ mol L

$^{-1}$ min

$^{-1}$ will be :

0%
$19.60\ min$
0%
$1.969\ min$
0%
$7.75\ min$
0%
$77.5\ min$

Explanation

For a first order reaction,

$\dfrac {dx}{dt} = k[X]$

$0.00352 = k\times 0.01$

$\Rightarrow k = 0.352$

$t_{1/2} = \dfrac {0.693}{k} = \dfrac {0.693}{0.352} = 1.969\ min$

Hence, the correct answer is option

$\text{B}$ .

The potential energy diagram for a reaction

$X\rightarrow Y$ is given.

$A$ and

$C$ in the graph corresponds to:

0%
$A \rightarrow$ activation energy, $C\rightarrow \triangle H^{\circ}$
0%
$A\rightarrow$ energy of reactants, $C \rightarrow$ energy of products
0%
$A\rightarrow \triangle H^{\circ}, C\rightarrow$ activation energy
0%
$A\rightarrow$ activation energy, $C\rightarrow$ threshold energy

Explanation

Activation energy is defined as the least possible amount of energy (minimum) which is required to start a reaction or the amount of energy available in a chemical system for a reaction to take place

$\to A$ .

$C \to \triangle { H }^{ o }$ is the difference of the forward and backward activation energies.

Hence, the correct answer is option

$\text{A}$ .

The decomposition of a substance follows first order kinetics. If its concentration is reduced to

$1/8$ of its initial value in

$12$ minutes, the rate constant of the decomposition system is ?

0%
$\left (\dfrac {2.303}{12}\log \dfrac {1}{8}\right ) min.^{-1}$
0%
$\left (\dfrac {2.303}{12}\log 8\right ) min.^{-1}$
0%
$\left (\dfrac {0.693}{12}\right ) min.^{-1}$
0%
$\left (\dfrac {1}{12}\log 8\right )min.^{-1}$

Explanation

$k = \dfrac {2.303}{t} \log\left( \dfrac {a}{a - x}\right)$ (for first order)

$k = \dfrac {2.303}{12}\log\left( \dfrac {1}{1/8}\right)= \left (\dfrac {2.303}{12}\log 8\right ) min.^{-1}$ .

Graph 9c) An endothermic reaction with high activation energy for the forward reaction can be shown by the figure:

What will be the half-life of the first order reaction for which the value of rate constant is

$200\ s^{-1}$ ?

0%
$3.46\times 10^{-2} s$
0%
$3.46\times 10^{-3} s$
0%
$4.26\times 10^{-2} s$
0%
$4.26\times 10^{-3} s$

Explanation

$Given:k=200s^{-1}$

$t_{1/2} = \dfrac {0.693}{k} = \dfrac {0.693}{200} = 3.46\times 10^{-3} s$ .

The activation energy in a chemical reaction is defined as ?

0%
The difference in energies of reactants and products
0%
The sum of energies of reactants and products
0%
The difference in energy of intermediate complex with the average energy of reactants and products
0%
The difference in energy of intermediate complex and the average energy of reactants

A first order reaction has a rate constant

$1.15\times 10^{-3}s^{-1}$ . How long will

$5\ g$ of this reactant take to reduce to

$3\ g$ ?

0%
$444\ s$
0%
$400\ s$
0%
$528\ s$
0%
$669\ s$

Explanation

$t = \dfrac {2.303}{k}\log \dfrac {[R_{0}]}{[R]} = \dfrac {2.303}{1.15\times 10^{-3}} \log\left( \dfrac {5}{3}\right)$

$= 2.00\times 10^{3} \log (1.667)$

$= 2\times 10^{3} \times 0.2219 = 444\ s$ .

The rate constant for a first order reaction is

$2\times 10^{-2} min^{-1}$ . The half-life period of reaction is ?

0%
$69.3\ min.$
0%
$34.65\ min.$
0%
$17.37\ min.$
0%
$3.46\ min.$

Explanation

$t_{1/2} = \dfrac {0.693}{k} = \dfrac {0.693}{2\times 10^{-3}} = 34.65\ min.$

In a first order reaction, the concentration of reactant is reduced to

$1/8$ of the initial concentration in

$75$ minutes at

$298\ K$ . What is the half-life period of the reaction in minutes?

0%
$50\ min$
0%
$15\ min$
0%
$30\ min$
0%
$25\ min$

Explanation

Given:

$a = 1, a - x = 1/8, t = 75\ min$

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

$= \dfrac {2.303\times 0.903}{75} \ min.^{-1}$

For first order reaction,

$t_{1/2} = \dfrac {0.693}{k} = \dfrac {0.693\times 75}{2.303\times 0.903} = 25\ min.$

Hence, the correct answer is option

$\text{D}$ .

Activation energy of a chemical reaction can be determined by _________.

0%
determining the rate constant at standard temperature
0%
determining the rate constants at two temperatures
0%
determining probability of collision
0%
using catalyst

Explanation

We know that,

$\log { \dfrac { { k }_{ 2 } }{ { k }_{ 1 } } } =\dfrac { { E }_{ a } }{ 2.303R } \left[ \dfrac { { T }_{ 2 }-{ T }_{ 1 } }{ { T }_{ 1 }{ T }_{ 2 } } \right]$

From this equation, we observe that

${ E }_{ a }$ at temperature

${ T }_{ 1 }$ and

${ T }_{ 2 }$ can be determined by determining rate of constant

${ K }_{ 1 }$ and

${ K }_{ 2 }$

Option B is correct.

The rate constant is given by the equation

$k = P.Ze^{-E_{a}/RT}$ . Which factor should register

$a$ decrease for the reaction to proceed more rapidly:

0%
$T$
0%
$Z$
0%
$E$
0%
$P$

Explanation

$K=P.ze^{-E_a/RT}$

$\Rightarrow log \ k= log (Pze^{-E_a/RT})$

$\Rightarrow log \ k= log \ (Pz) + log \ (e^{-E_a/RT})$

$\Rightarrow log \ k= log \ (Pz) - \cfrac {E_a}{RT} (log \ e)$

For

$k$ to increase,

$\cfrac {-E_a}{RT}$ shold be small and that can happen only if

$E_a$ is less or

$T$ is more. So, reaction proceeds rapidly if and only if

$E_a$ decreases.

Fill up the following with suitable terms.

(i) Activation energy

$=$ Threshold energy

$-$ _______.

(ii) Half-life period of zero order reaction

$=$ ________.

(iii) Average rate of reaction

$=$ _______.

(iv) Instantaneous rate of reaction

$=$ ______.

0%
Potential energy, $\dfrac {0.693}{k}, \dfrac {dx}{dt}, \dfrac {\triangle [A]}{\triangle t}$
0%
Energy of reactants, $\dfrac {1}{k}, \dfrac {\triangle [A]}{\triangle t}, \dfrac {dx}{dt}$
0%
Energy of reaction, $\dfrac {\log k}{t}, \dfrac {\triangle [A]}{\triangle t}, \dfrac {dx}{dt}$
0%
Average kinetic energy of reactants, $\dfrac {a}{2k}, \dfrac {\triangle [A]}{\triangle t}, \dfrac {dx}{dt}$

Explanation

a) Activation energy = Threshold energy

$-$ the Average kinetic energy of molecules(by definition)

b) half-life of a zero order reaction =

${ t }_{ \frac { 1 }{ 2 } } =\frac { a }{ 2k }$

c) Average rate of reaction =

$\dfrac {\displaystyle \text{ Total change concentration reactants }}{ \displaystyle \text{Total time taken} }$ =

$\dfrac { \triangle \left[ A \right] }{ \triangle t }$

d) Instantaneous rate of reaction =

$\dfrac { \displaystyle \text {Instantaneous change in concentration} }{\displaystyle \text { Instantaneous time} }$ =

$\dfrac { dx }{ dt }$

A first order reaction is

$50$ % complete in

$30$ minutes at

$27^{\circ}C$ and in

$10$ minutes at

$47^{\circ}C$ . The reaction rate constant at

$27^{\circ}C$ and the energy of activation of the reaction are respectively.

0%
$k = 0.0231\ min^{-1}, E_{a} = 43.848\ kJ\ mol^{-1}$
0%
$k = 0.017\ min^{-1}, E_{a} = 52.54\ kJ\ mol^{-1}$
0%
$k = 0.0693\ min^{-1}, E_{a} = 43.848\ kJ\ mol^{-1}$
0%
$k = 0.0231\ min^{-1}, E_{a} = 28.92\ kJ\ mol^{-1}$

Explanation

Since

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

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

Given:

$t_{1/2} = 30\ min$ at

$27^{\circ}C$ and

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

$47^{\circ}C$

$\therefore k_{27^{\circ}C} = \dfrac {0.693}{30}min^{-1} = 0.0231\ min^{-1}$

and

$k_{47^{\circ}C} = \dfrac {0.693}{10} = 0.0693\ min^{-1}$

We know that

$\log \dfrac {k_{47^{\circ}C}}{k_{27^{\circ}C}} = \dfrac {E_{a}}{2.303R}\times \dfrac {(T_{2} - T_{1})}{T_{2}\times T_{1}}$

$\therefore E_{a} = \dfrac {2.303R\times T_{1}\times T_{2}}{(T_{2} - T_{1})} \log \dfrac {k_{47^{\circ}C}}{k_{27^{\circ}C}}$

$E_{a} = \dfrac {2.303\times 8.314\times 10^{-3} \times 300\times 320}{(320 - 300)} \log \dfrac {0.0693}{0.0231}$

$= 43.848\ kJ\ mol^{-1}$ .

Hence, the correct answer is option

$\text{A}$ .

Threshold energy is equal to:

0%
Activation energy
0%
Activation energy $-$ energy of molecules
0%
Activation energy $+$ energy of molecules
0%
None of these

Explanation

According to the collision theory, only a small fraction of collisions is effective in bringing about the chemical reaction and the rest of the collisions are ineffective.

For effective collision (to yield product), the colliding molecules must have more than or equal to a certain minimum amount of energy called threshold energy.

Threshold energy

$=$ Activation energy

$+$ Average kinetic energy of the molecules.

Hence, the correct answer is option

$\text{C}$ .

If hydrogen and oxygen are mixed and kept in the same vessel at room temperature, the reaction does not take place to form water because:

0%
activation energy for the reaction is very high at room temperature
0%
molecules have no proper orientation to react to form water
0%
the frequency of collisions is not high enough for the reaction to take place
0%
no catalyst is present in the reaction mixture

The decomposition of a hydrocarbon follows the equation

$k = (4.5\times 10^{11}s^{-1})e^{-28000/ K/T}$ . What will be the value of activation energy?

0%
$669\ kJ\ mol^{-1}$
0%
$232.79\ kJ\ mol^{-1}$
0%
$4.5\times 10^{11}\ kJ\ mol^{-1}$
0%
$28000\ kJ\ mol^{-1}$

Explanation

Arrhenius equation,

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

Given equation is

$k = (4.5\times 10^{11}s^{-1})e^{-28000\ K/T}$

Comparing both the equations, we get

$-\dfrac {E_{a}}{RT} = -\dfrac {28000\ K}{T}$

$E_{a} = 28000\ K\times R = 28000\ K\times 8.314\ J\ K^{-1} mol^{-1}$

$= 232.79\ kJ\ mol^{-1}$ .

The half-life period of an 1st order reaction is

$60$ minutes. What percentage will be left over after

$240$ minutes?

0%
$6.25\%$
0%
$4.25\%$
0%
$5\%$
0%
$6\%$

Explanation

Let us consider the Amount of the substance left after one-half life is

$\dfrac{{{A_0}}}{2}$ .

Similarly,

Amount of the substance left after two half-lives is

$\dfrac{{{A_0}}}{{{2^2}}}$

Amount of the substance left after Three half-lives is

$\dfrac{{{A_0}}}{{{2^3}}}$ .

In general word,

Amount of the substance left after n half-life =

${A_0}/{2^n}$

The half-life period of a first-order reaction is

$60$ minutes.

Number half lives

$= \dfrac{{240}}{{60}} = 4$ i.e.

$n = 4$

Amount of the substance left after 4 half life

$= \dfrac{{{A_0}}}{{{2^4}}} = \dfrac{{{A_0}}}{{16}}$

Hence the solution is

$= 0.0625{\text{ }}$ of

${\text{ }}{A_0} = 6.25{\text{ }}\%$ .

Hence, option A is correct.

What is the half-life of a radioactive substance if

$87.5\%$ of any given amount of the substance disintegrate in

$40$ minutes?

0%
$160$ min
0%
$10$ min
0%
$20$ min
0%
$13$ min

Explanation

Radioactive reacting have order one so

$\displaystyle Kt = 2.303 \, log \frac{[R]_0}{[R]_t}$

$K\times40 =2.303\, log\frac{[R]_0}{12.5[R]_0} \times100$

$K=0.052 \ min^{-1}$

$\displaystyle t_{50\%}=\frac{0.693}{K}$

$\displaystyle t_{50\%}=\frac{0.693}{0.052}$

$t_{50\%}=13.32 \,min$

For an elementary bi-molecular reaction of the type

$A + B \rightarrow C$ activation energy is equal to 20 Kcal. If specific rate of reaction at 500 K is 4.2

$\times$

$10^{-3}M^{-1}sec^{-1}$ , then identify the incorrect option(s). [Given : ln (

$2.1 \times$

$10^{-9}$ ) = - 20]

0%
% of activated molecules at 500 K is $2.1 \times$ $10^{-7}$ %
0%
Maximum possible rate of reaction when $[A]$ and $[B]$ are both equal to 1 is $2 \times$ $10^6 M sec^{-1}$
0%
% of activated molecules at 300 K is $2 \times 10^{-6}$ %
0%
At some temperature rate of reaction can be $4 \times 10^3 M sec^{-1}$ if $[A] =\ 2M,\ [B] = 0.1 M$

The activation energy for the forward reaction X Y is

$60$ KJ

$mol^{-1}$ and

$\Delta H$ is

$-20$ KJ

$mol^{-1}$ . The activation energy for the backward reaction Y X is:

0%
$80$ kJ $mol^{-1}$
0%
$40$ kJ $mol^{-1}$
0%
$60$ kJ $mol^{-1}$
0%
$20$ kJ $mol^{-1}$

Explanation

We know that,

$\Delta H = E_f - E_b$

$-20=60 - E_b$

$E_b=(60+20) KJ/mol$

$=80KJmol^{-1}$

The half life for the first order reaction,

$N_2O_5\rightarrow 2NO_2+1/2O_2$ is

$24$ hours at

$30^o$ C. Starting with

$10$ g of

$N_2O_5$ , how many grams of

$N_2O_5$ will remain after a period of

$96$ hours?

0%
$1.25$ g
0%
$0.63$ g
0%
$1.77$ g
0%
$0.5$ g

Explanation

Given

$t_{1/2}=24hrs$ thus, we can re write

$k=\cfrac{0.693}{t_{1/2}}=0.02875hr^{-1}$

so using

$ln\left(\cfrac{A_o}{A}\right)=kt$ where

$A=$ Unknown

we get

$A=A_oe^{-kt}$

$\Rightarrow A=10.e^{-0.02875\times96}$

$=0.6329g$

Following reaction are of first order

$A\xrightarrow [ ]{ { k }_{ 1 }=2{ sec }^{ -1 } } 2B$

$A\xrightarrow [ ]{ { k }_{ 2 }=5{ sec }^{ -1 } } 3C$
Which of the following statement

$(s)$ is/are correct?

0%
$\dfrac { { \left[ B \right] }_{ t } }{ { \left[ C \right] }_{ t } } =\dfrac { 2 }{ 5 }$ at any time during reaction.
0%
After completion of reaction $\dfrac { { \left[ B \right] } }{ { \left[ C \right] } } =\dfrac { 2 }{ 15 }$
0%
$\dfrac { { \left[ B \right] }_{ t } }{ { \left[ C \right] }_{ t } } =\dfrac { 4 }{ 15 }$ at any time during reaction
0%
None of these

Explanation

$A \xrightarrow []{K_1} 2B$

$K_1= 2 sec^{-1}$

For first order,

$R_1=K_1(B)^1_t$

$A \xrightarrow []{K_2} 3C$

$K_2=5sec^{-1}$

$R_2= K_2(C)^1_t$

As the rate of the above reactions are not equal and depend upon concentration of (B) & (C) respectively.

Thus,

$\cfrac {(B)_t}{(C)_t}\neq \cfrac {2}{5}$ at any time.

So, none of the statement is correct.

CBSE Questions for Class 12 Engineering Chemistry Chemical Kinetics Quiz 6 - 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 6 - MCQExams.com

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Practice Class 12 Engineering Chemistry Quiz Questions and Answers

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