MCQ Questions for Class 12 Engineering Chemistry Chemical Kinetics Quiz 7

For certain first-order reaction, 75% of the reaction complete in 30 min. How much time did it require to complete 99.9% of the reaction?

0%
150 min
0%
100 min
0%
90 min
0%
300 min

Explanation

Its a 1st order reaction, So

$\dfrac{{t}_{x\text%}}{{t}_{y\text%}} = \dfrac {log{\dfrac{100}{100 - x}}}{log{\dfrac{100}{100 - y}}}$

Here x% = 75% and y% = 99.9%

Substituting the values, We get

$\dfrac{30}{{t}_{y\text%}} = \dfrac {log{\dfrac{100}{100 - 75}}}{log{\dfrac{100}{100 - 99.9}}}$

${t}_{y\text%} = \dfrac{30}{0.2}$ = 150 min

For the reaction system:

$2NO(g)+O_2(g)\rightarrow 2NO_2(g)$ , volume is suddenly reduced to half its value by increasing the pressure on it. If the reaction is of first order with respect to

$O_2$ and second order with respect to

$NO$ , the rate of reaction will:

0%
diminish to one-eighth of its initial value
0%
increase to eight times of its initial value
0%
increase to four times of its initial value
0%
diminish to one-fourth of its initial value

Explanation

Order of reaction with respect to $O_2: first$

Order of reaction with respect to $NO: second$

$Rate = k [NO]^2[O]$

Volume is decreased to half, then concentration of each reactants gets doubled.

$Rate \ New= k[2NO]^2[2O]$

$\therefore$ Rate of reaction will increase to eight times of it’s initial value.

Which is correct about zero order reaction?

0%
Rate of reaction depends on decay constant
0%
Rate of reaction is independent of concentration
0%
Unit of rate constant is concentration
0%
Unit of rate onstant is concentration time

Explanation

For a zero order reaction,

Rate=

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

so, rate of reaction depends on K only. K is called rate constant or decay constant.

The rate of a particular reaction doubles when temperature change from

${ 27 }^{ 0 }$ C to

${ 37 }^{ 0 }$ C. Calculate the energy of the activation of such reaction.

0%
53582 KJ
0%
53.58 KJ
0%
29.86 KJ
0%
none of these

Explanation

We are given that:

When

$T_1 = 27 + 273 = 300 K$

Let

$K_1 = k$

When

$T_2 = 37 + 273 = 310 K$

$K_2 = 2 k$

Substituting these values the equation:

$log\dfrac {K_2}{ K_1}$

$\dfrac{Ea}{ 2.303 R}\times \dfrac{(T_2 – T_1)}{ T_1 T_2}$

We will get:

$log\dfrac {2 k }{ k}$

$= \dfrac{E_a}{ 2.303 \times 8.314}\times\dfrac {310 – 300} { 300 \times310}$

$log_ 2 = \dfrac{E_a} {2.303 \times 8.314} \times\dfrac {10}{ 300 \times 310}$

$E_a = 53598.6 J mol^{-1}$

$E_a = 53.6 kJ mol^{-1}$

Hence, the energy of activation of the reaction is

$53.6 kJ mol^{-1}$

The activation energy of a reaction is

$94.14$ kj/mole, and the value of rate constant at

$313$ K is

$1.8\times 10^{-1}sec^{-1}$ . Calculate the frequency factor A.

0%
$[A]=8.25\times 10^{10}s^{-4}$ .
0%
$[A]=7.923\times 10^{10}s^{-4}$ .
0%
$[A]=9.194\times 10^{10}s^{-4}$ .
0%
none of these

Explanation

By using arrhenius equation,
$\implies \log k = \frac{-E_a}{2.303\space RT} + \log A$

We get
$\implies \log A = \log(1.8 \times 10^{-5}) + \frac{94140}{2.303 \times 8.314 \times 313 }$
$\implies (\log 1.8) - 5 + 15.7082)$
$\implies 0.2553 - 5 + 15.7082 = 10.9635$

$\therefore \log A = (10.9634) = 9.914 \times 10^{10}$

Find the two third life

$(t_{2/3})$ of a first order reaction in which

$K=5.48\times 10^{-14}sec^{-1}$ .

0%
$t_{2/3}=5\times 10^{13}$ sec
0%
$t_{2/3}=2\times 10^{13}$ sec
0%
$t_{2/3}=1.25\times10^{13}$ sec
0%
none of these

Explanation

1st Order Reaction,

$t = \dfrac{2.303}{k} log \dfrac{[100]}{[100 - x]}$

Substituting the values, we get

$t_{2/3} = \dfrac{2.303}{5.48 \times {10}^{-14}}\times log{[3]}$

$t_{2/3} = 2 \times {10}^{13}\space sec$

What is the activation energy of a reaction if its rate doubles when temperatures is raised from

$20^o$ C to

$35^o$ C?(R

$=8.314$ J

$K^{-1}$

$mol^{-1}$ )

0%
$34.7$ kJ $mol^{-1}$
0%
$15.1$ kJ $mol^{-1}$
0%
$342$ kJ $mol^{-1}$
0%
$269$ kJ $mol^{-1}$

Explanation

Let us consider the Arrhenius equation,

We have

$\log\dfrac{K_2}{K_1}=-\dfrac{E_a}{2.303R}[\dfrac{T_1-T_2}{T_1T_2}]$

Given that

Initial temperature

$T_1=20+273=293\;K$

Final temperature

$T_2=35+273=308\;K$

$R=8.314\;J\;mol^{-1}K^{-1}$

Since, the rate becomes double on raising the temperature,

$\therefore r_2=2r_1$

$\dfrac{r_2}{r_1}=2$

Hence the rate constant,

$K \propto r$

Since

$\frac{K_2}{K_1}=2$

putting the values

$\log2=-\dfrac{E_a}{2.303 \times 8.314}[\dfrac{293-308}{293 \times 308}]$

$E_a=34673.48\;J\;mol^{-1}$

$E_a=34.7\;KJ\;mol^{-1}$

Hence, the correct option is

$A$

The rate constant for the decomposition of a hydrocarbon is

$2.418\times 10^{-5}S^{-1}$ at

$546$ K. If the energy of activation is

$179.9$ kJ/mol. What will be the value of pre-exponential factor A?

0%
$[A]=9.245\times 10^{12}s^{-1}$ .
0%
$[A]=3.912\times 10^{12}s^{-1}$ .
0%
$[A]=7.83\times 10^{12}s^{-1}$ .
0%
None of these

Explanation

By using arrhenius equation,

$\implies \log k = \frac{-E_a}{2.303\space RT} + \log A$

We get
$\implies \log A = \log(2.418 \times 10^{-5}) + \frac{179900}{2.303 \times 8.314 \times 546 }$
$\implies (\log 2.418) - 5 + 15.7082)$
$\implies 0.3834 - 5 + 17.209 = 12.6$

$\therefore \log A = (12.6) = 3.9 \times 10^{12}$

Which one of is first order reaction ?

0%
$NH_4NO_2 \rightarrow N_2 + 2H_2O$
0%
$2HI\rightleftharpoons H_2 + I_2$
0%
$2NO_2 \rightarrow 2NO + O_2$
0%
$2NO + O_2 \rightarrow 2NO_2$

Explanation

${ NH }_{ 4 }{ NO }_{ 2 }\rightarrow { N }_{ 2 }+2{ H }_{ 2 }O$

Rate

$=K{ \left[ { NH }_{ 4 }{ NO }_{ 2 } \right] }^{ 1 }$

$\therefore \\$ order of this reaction is 1.

The energy of activation for a first order reaction is

$104 kJ mol^{-1}$ . The rate constant at

$25^oC$ is

$3.7\times 10^{-5} s^{-1}$ . What is the rate constant at

$30^oC$ ?

0%
$5.2\times 10^{-4}s^{-1}$
0%
$7.4 \times 10^{-5} s^{-1}$
0%
$7.72\times 10^{-5}s^{-1}$
0%
None of these

Explanation

$log \dfrac{{k}_{2}}{{k}_{1}}$ =

$\dfrac{{E}_{a}}{2.303 \times R} (\dfrac{1}{{T}_{1}} - \dfrac{1}{{T}_{2}})$

Substituting the values,

$log \dfrac{{k}_{2}}{3.7 \times {10}^{-5}}$ =

$\dfrac{104 \times {10}^{3}}{8.314 \times 2.303} (\dfrac{1}{298} - \dfrac{1}{303})$

${k}_{2} = 7.4 \times {10}^{-5}\space {s}^{-1}$

What is the activation energy for a reaction whose rate constant doubles when temperature changes from

$30^oC$ to

$40^oC$ ?

0%
$72.56\ KJmol^{-1}$
0%
$45.45\ KJmol^{-1}$
0%
$54.66\ kJ mol^{-1}$
0%
None of these

Explanation

Let

${ K }_{ 1 }={ K }$

then

${ K }_{ 2 }=2K$

${ T }_{ 1 }={ 30 }^{ 0 }C+273\\ =303K$

${ T }_{ 2 }={ 40 }^{ 0 }+273\\ =313K$

$\log { \left( \frac { { K }_{ 2 } }{ { K }_{ 1 } } \right) } =\frac { { E }_{ a } }{ 2.303R } \left[ \frac { { T }_{ 2 }-{ T }_{ 1 } }{ { T }_{ 1 }{ T }_{ 2 } } \right]$

$or,\quad \log { \left( \frac { { 2K } }{ { K } } \right) } =\frac { { E }_{ a } }{ 2.303X8.3 } \left[ \frac { 313-303 }{ 313X303 } \right]$

${ E }_{ a }=54571\quad KJ/mol\\ =54.5\quad KJ/mol\\ \approx \quad 54.66\quad KJ/mol$

Find out the percentage of the reactant molecules crossing over the activation energy barrier at 325K, given that

$\triangle { H }_{ 325 }=0.12kcal,\quad { E }_{ a(b) }=+0.02kcal$ .

0%
0.80%
0%
80.63%
0%
20%
0%
None of these

Explanation

$\Delta H={ E }_{ a(f) }-{ E }_{ a(b) }$

(activation energy of forward reaction-action energy of backward reaction)

$0.02+0.12={ E }_{ a(f) }$

${ E }_{ a(f) }= 0.14 kcal/mole.k= 140 cal/mol.K$

Only those molecules will cross the activation energy barrier which possess energy greater than

${ E }_{ a }.$

Formula:

Fraction of molecules with energy greater than

${ E }_{ a }=x={ e }^{ -\left( \cfrac { { E }_{ a } }{ RT } \right) }\\ x={ e }^{ -\left( \cfrac { 140 }{ 2\times 325 } \right) }$ Where

$R=1.98cal/molek,\quad R\approx 2cal/mol.k\\ x={ e }^{ -0.2154 }\\ \ln { x } =0.2154\\ \ln { x } =\cfrac { -0.2154 }{ 2.303 } =0.0935\\ x=antilog(0.0935)\\ x=0.8063\\ Percentage=80.63\%$

For a reversible reaction :

${ N }_{ 2 }+{ O }_{ 2 }\rightleftharpoons \quad 2NO$
Activation energy of the backward reaction is lower than that of forward reaction. The slope of k verse 1/T graph will be:

0%
zero
0%
$-\frac { H }{ 2.303\quad R }$
0%
$\frac { H }{ 2.303\quad R }$
0%
$-\frac { \Delta H }{ \quad R }$

Explanation

For reactions whose forward direction has less activation energy, the graph has -ve slope, indicating, a faster reaction in the forward direction.

It is given in the question that the backward reaction has a lower

${ E }_{ A }$ than the forward reaction. this means graph has a positive slope.

Slope of graph

$=\cfrac { \Delta H }{ 2.303R }$

Hence answer is [C].

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.

State whether it is True or False:

0%
True
0%
False

The activation energy for a certain reaction is

$334.4 kJ mol^{-1}$ . How many times larger is the rate constant at

$610$ K than the rate constant at

$600$ K?

0%
Two times
0%
Three times
0%
Four times
0%
None of these

Explanation

Let

${ K }_{ 1 } and { K }_{ 2 }$ be rate constants.

$\log { \left( \cfrac { { K }_{ 2 } }{ { K }_{ 1 } } \right) } =\cfrac { { E }_{ a } }{ 2.303R } \left[ \frac { { T }_{ 2 }-{ T }_{ 1 } }{ { T }_{ 1 }{ T }_{ 2 } } \right]$

$\log { \left( \cfrac { { K }_{ 2 } }{ { K }_{ 1 } } \right) } =\left( \cfrac { 334.4 \times { 10 }^{ 3 } }{ 2.303\left( 8.316 \right) } \right) \left[ \cfrac { 610-600 }{ 600 \times 610 } \right]$

$\cfrac { { K }_{ 2 } }{ { K }_{ 1 } } =3$

$\therefore \quad { K }_{ 2 }$ is .greater by three times

In a first-order reaction the concentration of the reactant is decreased from $1.0{\text{ M to 0}}{\text{.25 M }}$ in 20 min.The rate constant of the reaction would be:

0%
$10{\text{ mi}}{{\text{n}}^{ - 1}}$
0%
$6.931{\text{ mi}}{{\text{n}}^{ - 1}}$
0%
$0.6931{\text{ mi}}{{\text{n}}^{ - 1}}$
0%
$0.06931{\text{ mi}}{{\text{n}}^{ - 1}}$

Explanation

Its First Order Reaction

$k = \dfrac{2.303}{t} log\space\dfrac {[A]}{[A - x]}$

Substituting the values,

$k = \dfrac{2.303}{20} log\space\dfrac {[1]}{[0.25]}$

$k = 0.06931\space {min}^{-1}$

A first order reaction is half-completed in

$45$ minutes. How long does it need for

$99.9\%$ of the reaction to be completed?

0%
$20$ hours
0%
$10$ hours
0%
$7\ \dfrac{3}{10}$ hours
0%
$5$ hours

Explanation

${ t }_{ 1/2 }=45$ minutes

For 1st order reaction

${ t }_{ 1/2 }=\cfrac { 0.693 }{ k } \\ \therefore k=\cfrac { 0.693 }{ 45 } { min }^{ -1 }=0.0154{ min }^{ -1 }$

Again for 1st order reaction we have

$k=\cfrac { 2.303 }{ t } \log { \left[ \cfrac { 100 }{ 100-99.9 } \right] }$

(considering original concentration to be 100)

$t=\cfrac { 2.303 }{ k } \log { \left[ \cfrac { 100 }{ 0.1 } \right] }$

$t=\cfrac { 2.303 }{ 0.0154 } \log { 1000 }$

$t=\cfrac { 2.303 }{ 0.0154 } \times 3=448.636\quad minutes$

$t=7 \dfrac{3}{10}\quad hours.$

Hence, the correct option is

$C$

For a first order reaction,

$(A)$

$\rightarrow$ products,

the concentration of

$A$ changes from

$0.1$ M to

$0.025$ M in

$40$ minutes. The rate of reaction when the concentration of

$A$ is

$0.01$ M is:

0%
$1.73\times 10^{-4}$ mol $dm^{-3} min^{-1}$
0%
$1.73\times 10^{-5}$ mol $dm^{-3} min^{-1}$
0%
$3.47\times 10^{-4}$ mol $dm^{-3} min^{-1}$
0%
$3.47\times 10^{-5}$ mol $dm^{-3} min^{-1}$

Explanation

For

$1^{st}$ order reaction,

rate constant,

$R=\cfrac {2.303}{t}\log\cfrac {[A]}{[A]_t}$

$\therefore R=\cfrac {2.303}{40}\log \cfrac {0.1}{0.025}=\cfrac {0.693}{20}$

For

$1^{st}$ order reaction,

rate=

$K[A]=\cfrac {0.693}{20}\times 0.01$

$=3.47 \times 10^{-4} mol\ dm^{-3} min^{-1}$

Hence, the correct option is

$C$

The reaction :

$X \rightarrow$ product, follows first-order kinetics in

$20$ minutes, the concentration of

$X$ changes from

$0.1 M$ to

$0.05\ M$ then rate of reaction when concentration of

$X$ is

$0.02\ mol/ L$ is:

0%
$1.73 \times 10^{-4} M/min$
0%
$3.47 \times 10^{-5} M/min$
0%
$6.94 \times 10^{-4} M/min$
0%
$1.73 \times 10^{-5} M/min$

Explanation

By first order kinetic rate constant,

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

$a = 0.1 \space M$

$(a-x) = 0.05 \space M$

$t = 20 \space min$

$\implies k = \cfrac{2.303}{20} \log(\cfrac{0.1}{0.05})$

$\implies k = 0.0347 \space min^{-1}$

$Rate= \cfrac{dx}{dt} = k[A]^1$

$\implies 0.0347 \times 0.02$

$\implies 6.94 \times 10^{-4} \times M \space min^{-1}$

For zero order reactions, the linear plot was obtained for

$[A]$ vs t. The slope of the line is equal to:

0%
$k_{0}$
0%
$-k_{0}$
0%
$\dfrac {0.693}{K_{o}}$
0%
$-\dfrac {K_{o}}{2.303}$

Explanation

For a Zero Order Reaction,

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

Integrating on both sides and finding the value of 'c'(Constant of proportionality), we get

('c' can be found out by substituting [A] =

$[{A}_{0}]$ at t = 0)

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

Where,

$-{k}_{0}$ is rate constant,

$[{A}_{0}]$ is Initial Concentration.

So, the Slope of the line is

$-{k}_{0}$

Rate of formation of

${ SO }_{ 3 }$ according to the reaction

$2{ SO }_{ 2 }+O_{ 2 }\rightarrow 2{ SO }_{ 3 }\quad is\quad 1.6\times { 10 }^{ -3 }kg\quad min^{ -1 }$ . Hence rate at which

$SO_{ 2 }$ reacts is:

0%
$1.6\times { 10 }^{ -3 }kg\quad { min }^{ -1 }$
0%
$8.0\times { 10 }^{ -4 }kg\quad { min }^{ -1 }$
0%
$3.2\times { 10 }^{ -3 }kg\quad { min }^{ -1 }$
0%
$1.28\times { 10 }^{ -3 }kg\quad { min }^{ -1 }$

Explanation

$2 SO_2 + O_2 \rightleftharpoons 2 SO_3$

$\dfrac{1}{2}\dfrac{d[SO_3]}{dt} = 1.6 \times 10^{-3} kg/min = \dfrac{1.6 \times 10^{-3}}{80} mol/min$

$[\because M_{SO_3}= 80 \times 10^{-3} Kg/mol]$

$\dfrac{-1}{2}\dfrac{d[SO_2]}{dt} = \dfrac{+1}{2}\dfrac{d[SO_3]}{dt}$

$= -\dfrac{1.6 \times 10^{-3}}{80 \times 10}$ mol/min

$= -\dfrac{1}{50} \times 64 \times 10^{-3} = -1.28 \times 10^{-3}$ kg/min

An endothermic reaction

$A\rightarrow B$ has an activation energy

$15kcal/mole$ and the heat of reaction is

$5kcal/mole$ . The activation energy of reaction

$B\rightarrow A$ is:

0%
$20kcal/mole$
0%
$15kcal/mole$
0%
$10kcal/mole$
0%
zero

Explanation

Activation energy=

$15 kcal/mole$

Heat of reaction=

$5 kcal/mole$

Total

${ E }_{ a }=20kcal/mol$

For

$A\rightarrow B$

For any reaction the activation energy in the forward direction is numerically equal to activation energy of reaction in backward direction.

Therefore

${ E }_{ a }$ for

$B\rightarrow A$ is 20 kcal/mole and option [A] is correct answer.

The rate of chemical reaction is directly proportional to the equilibrium constant.

In which of the following process reaction will be completed first?

0%
$K=10$
0%
$K=1$
0%
$K={ 10 }^{ 3 }$
0%
$K={ 10 }^{ -2 }$

Explanation

The rate of chemical reaction is directly proportional to the equilibrium constant.

$\therefore$ For

$K={ 10 }^{ 3 }$ , the reaction will complete fastest.

In a reaction involving one single reactant, the fraction of the reactant consumed may be defined as

$f=\left(1-\dfrac {C}{C_{0}}\right)$ where

$C_{0}$ and

$C$ are the concentrations of the reactant at the after time, t. For a first order reaction:

0%
$\dfrac {df}{dt}=k(1-f)$
0%
$-\dfrac {df}{dt}=kf$
0%
$-\dfrac {df}{dt}=k(1-f)$
0%
$\dfrac {df}{dt}=kf$

Explanation

Integrated rate equation for 1st order reaction

$K=\dfrac{2.303}{t}\log\dfrac{[A]_1}{[A]_2}$

Where

${[A]_1}$ is the initial concentration and

${[A]_2}$ is the concentration of A at time t

Given initial concentration is

$C_0$ and concentration at time t is C

Integrating option A we get

$\ln(1-f)=-kt$

given

$f=(1-\dfrac{C}{C_0})$

$ln\dfrac{[C_0]}{[C]}=kt$

$K=\dfrac{2.303}{t}\log\dfrac{[C_0]}{[C]}$

In a first order reaction the amount of reactant decayed in three half lives (let

$a$ be is initial amount) would be:

0%
$7a/8$
0%
$a/8$
0%
$a/6$
0%
$5a/6$

Explanation

The amount of reactant left after n half lives for a 1st order reaction is

$({\dfrac{1}{2}})^{n}$ times the Initial concentration.

Here n =3 . So Amount of Reactant left is

$\dfrac{a}{8}$

So Amount of reactant decayed is

$\dfrac{7a}{8}$

If a first order reaction is completed to the extent of

$75$ % and

$50$ % in time interval,

${t}_{1}$ and

${t}_{2}$ , what is the ratio

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

0%
$\ln {2}$
0%
$\cfrac{\ln{(3/4)}}{\ln{2}}$
0%
$2$
0%
$1/2$

Explanation

For a first order reaction

$t=\cfrac { 2.303 }{ k } \log { \cfrac { { [A] }_{ o } }{ { [A] }_{ t } } }$

Assuming initial concentration as 100;

${ t }_{ 1 }(75\%)=\cfrac { 1 }{ k } \ln { \left[ \cfrac { 100 }{ 25 } \right] } =\cfrac { \ln { (4) } }{ k } \\ { t }_{ 2 }(50\%)=\cfrac { 1 }{ k } \ln { \left[ \cfrac { 100 }{ 25 } \right] } =\cfrac { \ln { (2) } }{ k } \\ \cfrac { { t }_{ 1 } }{ { t }_{ 2 } } =\cfrac { \ln { (4) } }{ \ln { (2) } } =2$

Hence answer answer is option [C]

For the consecutive unimolecular-type first order reaction

$A \xrightarrow{k_1} R \xrightarrow{k_2} S$ , the concentration of component

$R, C_R$ at any time

$t$ is given by

$: C_R = C_{AO} K_1 \left[\dfrac{e^{k_1t}}{(k_2 - k_1)} + \dfrac{e^{-k_2t}}{(k_1 - k_2)}\right]$ if

$C_A = C_{AO}, CR = C_{RO} = 0$ at

$t = 0$ the time at which the maximum concentration of

$R$ occurs is:

0%
$t_{max} = \dfrac{k_2 - k_1}{ln(k_2/k_1)}$
0%
$t_{max} = \dfrac{ln(k_2/k_1)}{k_2 - k_1}$
0%
$t_{max} = \dfrac{e^{k_2/k_1}}{k_2 - k_1}$
0%
$t_{max} = \dfrac{e^{k_2 - k_1}}{k_2 - k_1}$

The time for half-life period of a certain reacting

$A\rightarrow$ product is an hour. How much time does it take for its concentration to come from 0.50 to 0.25 mol

${ L }^{ -1 }$ if it is a zero order reaction?

0%
0.25h
0%
1h
0%
4h
0%
0.5h

Explanation

concerntration is halued.

$\therefore$ time must be

$t_{1/2}$

$\therefore t.025n$

The rate constant, the activation energy and the arrhenius parameter of a chemical reaction at

${ 25 }^{ 0 }C$ are

$3\times { 10 }^{ 14 }{ sec }^{ -1 }:\ 104.4J{ mol }^{ -1 }$ and

$6.0\times { 10 }^{ 14 }{ sec }^{ -1 }$ respectively, the value of the rate constant as

$T\rightarrow \infty$ is:

0%
$2\times { 10 }^{ 8 }{ sec }^{ -1 }$
0%
$6\times { 10 }^{ 14 }{ sec }^{ -1 }$
0%
$\infty$
0%
$3.6\times { 10 }^{ 30 }{ sec }^{ -1 }$

In the following first-order competing reactions.

$A+ \text{Reagent}\rightarrow \text{Product},\quad B+ \displaystyle{\text{Reagent}} \rightarrow \text{Product}$ .

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%
4.07
0%
0.246
0%
2.06
0%
0.06

Explanation

$A + Reagent \, \xrightarrow{k_1} P$

$B + Reagent \, \xrightarrow{k_2} P$

Let initial care be x and given time be

$t$ first order

$\therefore k_1 t = \ln \dfrac{[A_0]}{[A_0]}$

$\Rightarrow k_1 \times t = \ln \dfrac{[A_0]}{[A_0] - 0.94[A_0]}$

$\Rightarrow k_1 t = \ln \dfrac{[A]_0}{0.06 [A]_0}$

$\Rightarrow k_1 t = \ln \dfrac{50}{3}$ ....

$(I)$

Similarly

$k_2 t = \ln \dfrac{[B_0]}{[B_0]}$

$\Rightarrow k_2 t = \ln \dfrac{[B_0]}{0.5 [B_0]}$

$\Rightarrow k_2 t = \ln 2$ .....

$(II)$

$(I)/(II) \Rightarrow \dfrac{k_1}{k_2} = \dfrac{\ln(50/3)}{\ln 2} = \dfrac{2.81}{0.693}$

$= 4.072$

Hence, the correct option is

$A$

For a firt order reaction

$A\to P$ , the temperature (T) dependent rate constant (k) was found to follow the equation

$\log k = -(2000)\dfrac{1}{T}+6.0$ . The pre-exponential factor A and the activation energy

$E_a$ , respectively, are?

0%
$1.0\times 10^6s^{-1}$ and $9.2kJ\ mol^{-1}$
0%
$6.0^{-1}$ and $16.6kJ\ mol^{-1}$
0%
$1.0\times 10^6s^{-1}$ and $16.6kJ\ mol^{-1}$
0%
$1.0\times 10^6s^{-1}$ and $38.3kJ\ mol^{-1}$

Explanation

$log\space k = \dfrac{-2000}{T} + 6.0$

Multiply by

$2.303$

$ln\space \dfrac{k}{1 \times {10}^{6}} = \dfrac{-4606}{T}$

$k = 1 \times {10}^{6}\space {e}^{\dfrac{-38.3 \times {10}^{3}}{RT}}$

$A = 1 \times {10}^{6}\space {s}^{-1}$ and

${E}_{a} = 38.3\space kJ\space {mol}^{-1}$

$T (in \,K)$	$\dfrac{1}{T} (in \, K^{-1})$	$\log_{10} K$
$769$	$1.3\times 10^{-3}$	$2.9$
$667$	$1.5\times 10^{-3}$	$1.1$

From the following data: the activation energy for the reaction (cal/mol)

${ H }_{ 2 }+{ I }_{ 2 }\rightarrow 2HI$

0%
$4\times { 10 }^{ 4 }$
0%
$2\times { 10 }^{ 4 }$
0%
$8\times { 10 }^{ 4 }$
0%
$3\times { 10 }^{ 4 }$

Explanation

$\displaystyle \log\frac{K_{2}}{K_{1}}=\frac{E_{act}}{R}\left ( \frac{1}{T_{1}}-\frac{1}{T_{2}} \right )$

$\Rightarrow 2.9-1.1=\dfrac{E_{act}}{2}\times 0.2\times 10^{-3}$

$\Rightarrow E_{act}=\dfrac{2\times 1.8\times 10^{3}\times 10}{2}$

$=18000 \ cal$

$\approx 2\times 10^{4}cal$

For a reaction

$A \rightarrow B + C$ .. 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 was completed a was

$100^o$ Assuming that "

$A$ " a optically inactive. '

$B$ ' dextro rotatory and '

$C$ ' is laevo rotatory rate constant of this first order reaction is ?

0%
$0.069 \,min^{-1}$
0%
$0.69 \,min^{-1}$
0%
$2.303 \,min^{-1}$
0%
$4.6 \,min^{-1}$

$373\ K$ , a gaseous reaction

$A\rightarrow 2B+C$ is found to be first order. Starting with pure

$A$ , the total pressure at the end of

$10\ min$ . was

$176\ mm$ and after a long time when

$A$ was completely dissociated, it was

$270\ mm$ . The pressure of

$A$ at the end of

$10$ minutes was:

0%
$94\ mm$
0%
$47\ mm$
0%
$43\ mm$
0%
$90\ mm$

Explanation

A single gaseous molecule produces 3 molecules. The pressure of the system is 270 mm.

${p}_{a} = \dfrac{270}{3} = 90\space mm\space of\space Hg$

If x is the decrease in the partial pressure of A at t = 10 min, then partial pressure of B and C is 2x and x

Total Pressure at this stage would be

${p}_{b} + {p}_{c} + {p}_{a} = 176\space mm\space of\space Hg$

90 -x + 2x + x = 176 mm of hg

$x = 43 mm of Hg$

So Option C is correct

In the case of a zero-order reaction, the ratio of time required for

$75$ % completion to

$50$ % completion is:

0%
$ln 2$
0%
$2$
0%
$1.5$
0%
none

Explanation

Its a Zero Order Reaction,

Time required for 75% completion

${t}_{1} = \dfrac{100 -25}{2k}$ =

$\dfrac{75}{2k}$

Time required for 50% completion

${t}_{2} = \dfrac{100 -50}{2k}$ =

$\dfrac{50}{2k}$

The Ratio of times =

$1.5$

Which integrated equation is correct for the following

$1^{st}$ order reaction started with only

$A(g)$ in a closed rigid vessel?

$A(g)\rightarrow B(g) + C(g)+ D(g)$

where,

$P_i=$ initial pressure

$;\quad P_t=$ total pressure at time

$t$

0%
$K=\dfrac { 2.303 }{ t } \log _{ 10 } \left[\dfrac { P_i }{P_t } \right]$
0%
$K=\dfrac { 2.303 }{ t } \log _{ 10 } \left[\dfrac { P_t }{P_i } \right]$
0%
$K=\dfrac { 2.303 }{ t } \log _{ 10 } \left[\dfrac { 2P_i}{3P_i-P_t} \right]$
0%
$K=\dfrac { 2.303 }{ t } \log _{ 10 } \left[\dfrac { 3P_i}{2P_i-3P_t} \right]$

Explanation

$A\left( g \right) \longrightarrow B\left( g \right) +C\left( g \right) +D\left( g \right)$

Initial

$Pi$

$0$

at time

$t$ ,

$Pi-p$

$p$

$\therefore$ total pressure at

$t$ ,

${ P }_{ t }={ P }_{ i }-p+p+p+p={ P }_{ i }+2p$

Since the reaction is

${ 1 }^{ st }$ order reaction.

$\therefore$ rate constant,

$K=\dfrac { 2.303 }{ t } log\left( \dfrac { { P }_{ i } }{ { P }_{ i }-p } \right)$

$\Rightarrow K=\dfrac { 2.303 }{ t } log\dfrac { { P }_{ i } }{ \left( { P }_{ i }-\dfrac { { P }_{ t }-{ P }_{ i } }{ 2 } \right) }$

$\Rightarrow K=\dfrac { 2.303 }{ t } log\left( \dfrac { 2{ P }_{ i } }{ { 3P }_{ i }-{ P }_{ t } } \right)$

$\therefore$ Correct answer is C.

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

0%
Increase by n times
0%
Decrease by factor of n
0%
Not change
0%
None of these

Explanation

Solution:-(C) Not change

According to Arrhenius equation-

Rate constant

$\left( K \right) = A {e}^{-{{E}_{a}}/{RT}}$

whereas,

$T =$ Temperature at which reaction is carried out.

$A =$ Arrhenius Parameter

${E}_{a} =$ Activation Energy of a reaction.

$R =$ Universal Gas Constant. .

For a particular reaction, rate constant depends on the temperature only and thus is independent of concentration of reactant.

Hence there will be no change in the value of rate constant w.r.t. the concentration of reactant.

The time elapsed of a certain between 33% and 67% completion of a first order reaction is 30 minutes. What is the time needed for 25% completion?

0%
$150.5$ minutes
0%
$12.5$ minutes
0%
$180.5$ minutes
0%
$165.5$ minutes

Explanation

$30 = \dfrac{2.303}{k} log \dfrac{100}{33} - \dfrac{2.303}{k} log \dfrac{100}{67}$

$k = 0.0236\space {min}^{-1}$

$0.0236 = \dfrac{2.303}{t} log \dfrac{100}{75}$

$t = 12.5\space minutes$

For which order reaction a straight line is obtained along with $x - axis$ by plotting a graph between half-life $\left( {{t_{1/2}}} \right)$ and initial concentration $'a'$ .

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

Explanation

We know that

$t_{\frac{1}{2}}\propto a^{1-n}$

The graph to be straight line

$t_{\frac{1}{2}}$ should be independent of

$a$ .

That is

$1-n=0$ , hence

$n=1$

It is a first order reaction.

For the decomposition of

$HI$ the following logarithmic plot is shown:

$[R=1.98\ cal/mol-K]$
The activation energy of the reaction is about?

0%
$45600\ cal$
0%
$13500\ cal$
0%
$24600\ cal$
0%
$32300\ cal$

Explanation

$K=Ae^{-Ea/RT}$ Slope=

$y/x=\cfrac {1\times 10^3}{0.1}=10$

$\Rightarrow \log K=-\cfrac {Ea}{2.303 R}\left(\cfrac {1}{T}\right)+\log A$

Slope=

$-\cfrac {Ea}{2.303 R}$

$\therefore E_a= -slope\times 2.303 R$

$=-10\times 2.303 \times 1.98\times 10^3$

$= -45.59 \times 10^3$ cal

How much faster would a reaction proceed at

$25^0$ C than at

$0^0$ C if the activation energy is

$65$ kJ?

0%
$2$ times
0%
$16$ times
0%
$11$ times
0%
$6$ times

Explanation

$log \dfrac{{k}_{2}}{{k}_{1}}$ =

$\dfrac{{E}_{a}}{2.303 \times R} (\dfrac{1}{{T}_{1}} - \dfrac{1}{{T}_{2}})$

$log \dfrac{{k}_{2}}{{k}_{1}} = \dfrac{65000}{2.303 \times 8.314} (\dfrac{1}{273} - \dfrac{1}{298})$

$\dfrac{{k}_{2}}{{k}_{1}}$ =

$11$

So, It is 11 times faster.

A reaction of first - order completed

$90\%$ in

$90$ minutes , hence , it is completed

$50\%$ in approximately :

0%
$50 min$
0%
$54 min$
0%
$27 min$
0%
$62 min$

Explanation

$k = \dfrac{2.303}{t} log \dfrac {100}{100 -x}$

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

$k = 0.0255\space {min}^{-1}$

${t}_{1/2} = \dfrac{2.303}{k} log \dfrac {100}{100 -50}$

${t}_{1/2} = \dfrac{2.303}{0.0255} log \dfrac {100}{100 -50}$

${t}_{1/2} = \dfrac{0.693}{0.0255}$

${t}_{1/2} = 27\space min$

Hence, the correct option is

$C$

Collision frequency of a gas at

$1\ atm$ pressure is

$Z$ . Its value at

$0.5\ atm$ will be:

0%
$0.25Z$
0%
$2Z$
0%
$0.50Z$
0%
$Z$

Explanation

$\rightarrow$ As the pressure increases collision frequency increases.

$\rightarrow$ Vice versa pressure is decreased to half, frequency of collision decreases to half.

Hence option

$C$ is correct.

Collision diameter is least in case of:

0%
${ H }_{ 2 }$
0%
$He$
0%
${ CO }_{ 2 }$
0%
${ N }_{ 2 }$

Explanation

Collision diameter will be least for the low molecular weight.

$d\space \alpha\space \dfrac{1}{M}$

In this case,

$Co_2$ has the highest molecular weight.

So, Option C is correct

The half-life of a zero-order reaction is 30 minutes. What is the concentration of the reactant left after 60 minutes?

0%
$25%$
0%
$50%$
0%
$6.25%$
0%
$0$

Explanation

Its a Zero Order Reaction

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

$k = \dfrac{[{A}_{0}]}{60}$

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

Substituting the value of k, we get

${[{A}_{0}]} = {[{A}_{0}]} - [A]$

$[A] = 0$

A certain zero order reaction has

$k = 0.025 M s^{-1}$ for the disappearance of A. What will be the concentration of A after 15 seconds if the initial concentration is 0.50 M ?

0%
$0.50$ M
0%
$0.375$ M
0%
$0.125$ M
0%
$0.060$ M

Explanation

Its a Zero Order Reaction

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

$0.025 = \dfrac{1}{15}{[{0.5 - A}]}$

${A} = 0.125 M$

The rate constant for a recation is

$10.8 \times 10^{-5} mol L^{-1}S^{-1}$ . The reaction obeys:

0%
First order
0%
Zero order
0%
Second order
0%
All are wrong

Explanation

We can tell order of reaction by its rate constant

$(K)$ units

Reaction	Order	Units
Zero	0	mol $L^{-1} S^{-1}$
First	1	$S^{-1}$
Second	2	mol $^{-1}LS^{-1}$

Here, units are mol

$L^{-1}S^{-1}$ , so zero order.

$60\%$ of a first order reaction was completed in

$60$ minute,

$50\%$ of the same reaction would be completed in approximately?

0%
$46$ minute
0%
$60$ minute
0%
$40$ minute
0%
$50$ minute

Explanation

Its a 1st Order reaction.

$k = \dfrac{2.303}{60} log \dfrac{100}{40}$

$k = 0.015\space {min}^{-1}$

$t = \dfrac{2.303}{0.015} log \dfrac{100}{50}$

$t = 46\space min$

Hence, the correct option is

$A$

$t_{1/2} v/s \dfrac{1}{a^2}$ is straight line graph then determine the order of a reaction:

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

Explanation

In general for an

$n^{th}$ order of a reaction, the half life period is given by,

$t_{1/2} \propto \cfrac {1}{a^{n-1}}$ ,

$a$ = initial concentration;

$n$ = order of reaction

Given,

$t_{1/2}$

$v/s$

$1/a^{2}$ i.e,

$n-1=2,n=3$

$\therefore t_{1/2}$ = (constant)

$\cfrac {1}{a^2}$ is a straight line.

for

$n=3$ i.e, for Third order reaction.

For a first order reaction rate constant is given as

$\log K = 14 - \dfrac{1.2 \times 10^4}{T}$ then what will be value of temperature if its half life period is

$6.93 \times 10^{-3}\ min$ ?

0%
$100\ K$
0%
$1000\ K$
0%
$720\ K$
0%
$327\ K$

Explanation

$k = \dfrac{0.693}{{t}_{1/2}}$

$k = 100\space {min}^{-1}$

Substituting it, we get

$2 = 14 - \dfrac{1.2 \times {10}^{4}}{T}$

$t = 1000K$

CBSE Questions for Class 12 Engineering Chemistry Chemical Kinetics Quiz 7 - MCQExams.com

0:0:3

Practice Class 12 Engineering Chemistry Quiz Questions and Answers

CBSE Questions for Class 12 Engineering Chemistry Chemical Kinetics Quiz 7 - MCQExams.com

0:0:3

Practice Class 12 Engineering Chemistry Quiz Questions and Answers

Support mcqexams.com by disabling your adblocker.