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 =____________

Average kinetic energy of reactants , $$ \dfrac {a}{2k} \ ,\ \quad \dfrac { \Delta [A] }{ \Delta t }\ ,\ \quad \dfrac {dx}{dt} $$

Potential energy , $$\dfrac { 0.693}{k}\ ,\ \quad \dfrac {dx}{dt}\ ,\ \quad \dfrac { \Delta [A] }{\Delta t } $$

Energy of reactants , $$ \dfrac {1}{k}\ ,\ \quad \dfrac { \Delta [A] }{ \Delta t }\ ,\ \quad \dfrac {dx}{dt} $$

Energy of reaction , $$ \dfrac { log k}{t}\ ,\ \quad \dfrac { \Delta [A] }{\Delta t }\ ,\ \quad \dfrac {dx}{dt} $$

MCQ Questions for Class 12 Engineering Chemistry Chemical Kinetics Quiz 16

The half-life of

_{6} C^{14}

$_6 C^{14}$ , if its decay constant is

6.31 \times 10^{- 4}

$6.31 \times 10^{-4}$ is:

0%
1098 yrs
0%
109.8 yrs
0%
10.98 yrs
0%
1.098 yrs

Explanation

t_{1 / 2} = \frac{0.693}{k}

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

= \frac{0.693}{6.31 \times 10^{- 4}}

$= \dfrac{0.693}{6.31 \times 10^{-4}}$

= 0.1098 \times 10^{4} = 1098 y r s

$= 0.1098 \times 10^4 = 1098 yrs$ .

\frac{15}{16} t h

$\dfrac{15}{16} \ th$ of a radioactive sample decays in 40 days half-life of the sample is:

0%
100 days
0%
10 days
0%
1 day
0%
$log_e$ 2 days

Explanation

Quantity of radioactive element decayed

= \frac{15}{16}

$= \dfrac{15}{16}$

Quantity left

= 1 - \frac{15}{16} = \frac{1}{16}

$= 1 - \dfrac{15}{16} = \dfrac{1}{16}$

\frac{1}{16} = 1 \times (\frac{1}{2})^{n}

$\dfrac{1}{16} = 1 \times \Big(\dfrac{1}{2} \Big)^n$ or

(\frac{1}{2})^{4} = (\frac{1}{2})^{n}

$\Big(\dfrac{1}{2} \Big)^4 = \Big(\dfrac{1}{2} \Big)^n$

one half-life =

\frac{40}{4} = 10 d a y s

$\dfrac{40}{4} = 10 \ days$ .

A first-order nuclear reaction is half completed in 45 minutes. How long does it need

99.9 %

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

0%
5 hours
0%
7.5 hours
0%
10 hours
0%
20 hours

Explanation

k = \frac{0.693}{0.75 h r} = \frac{2.303}{t} l o g \frac{a}{a - 0.999 a}

$k = \dfrac{0.693}{0.75 \ hr} = \dfrac{2.303}{t} log \dfrac{a}{a - 0.999a}$

= \frac{2.303}{t} l o g 10^{3} = 7.5 h r s .

$= \dfrac{2.303}{t} log 10^3 = 7.5 hrs.$

Assertion : Order of the reaction can be zero or fractional.
Reason : We cannot determine order from the balanced chemical equation.

0%
Both assertion and reason are correct and the reason is correct explanation of assertion.
0%
Both assertion and reason are correct but reason does not explain assertion.
0%
Assertion is correct but reason is incorrect.
0%
Both assertion and reason are incorrect.
0%
Assertion is incorrect but reason is correct.

Which of the following statement is incorrect ?

0%
Unit of rate of disappearance is $Ms^{-1}$
0%
Unit of rate of reaction is $Ms^{-1}$
0%
Unit of rate constant k depends upon order
0%
Unit of k for first order reaction is $Ms^{-1}$

Explanation

For 1st order reaction:

\frac{d [A]}{d t} = K [A]^{1}

$\dfrac{d[A]}{dt} = K[A]^1$

\frac{M}{s^{- 1}} = K M

$\dfrac{M}{s^{-1}} = KM$

K = s^{- 1}

$K = s^{-1}$

For 2nd order reaction:

- \frac{a [A]}{d t} = K [A]^{2}

$-\dfrac{a[A]}{dt} = K[A]^2$

\frac{M}{s} = K M^{2}

$\dfrac{M}{s} = KM^2$

K = M^{- 1} s^{- 1}

$K = M^{-1} s^{-1}$

Unit of rate of reaction is given by:

\frac{d [A]}{d t} = K [A]

$\dfrac{d[A]}{dt} = K[A]$

This is the rate of reaction

Unit =

M s^{- 1}

$Ms^{-1}$

Rate of disappearance for the reaction:

A \to B

$A \rightarrow B$

\frac{- d [A]}{d t} = M s^{- 1}

$\dfrac{-d[A]}{dt} = Ms^{-1}$

Hence, option D is the answer.

the rate constant for a first order reaction is

2 \times 10^{- 2} m i n^{- 1}

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

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

Explanation

G i v e n : k = 2 \times 10^{- 2} m i n^{- 1}

$Given:k=2\times10^{-2}min^{-1}$

t_{1 / 2} = \frac{0.693}{k} = \frac{0.693}{2 \times 10^{- 2}} = 34.64 m i n

$t_{1/2} = \dfrac { 0.693}{k} = \dfrac { 0.693}{ 2 \times 10^{-2} } = 34.64 min$

Option B is correct.

The activation energy in a chemical reaction is defined as

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

Cyclopropane rearranges to form propene

Δ \to C H_{3} - C H = C H_{2}

$\Delta \rightarrow CH_3 -CH =CH_2$
this follows first order kinetics. the rate constant is

2.714 \times 10^{- 3} s e c^{- 1}

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

0.29 M

$0.29 M$ what will be concentration of cyclopropane after 100 sec ?

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

Explanation

k = \frac{2.303}{t} l o g \frac{a}{(a - x)}

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

(a - x)

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

100 s e c

$100 sec$

2.7 \times 10^{- 3} = \frac{2.303}{100} l o g \frac{0.29}{(a - x)}

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

\Rightarrow \frac{0.27}{2.303} = l o g \frac{0.29}{(a - x)} \Rightarrow 0.117 = l o g \frac{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

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

Fill up the following with suitable terms.

(i)

$(i)$ Activation energy = Threshold energy - _____________

(i i)

$(ii)$ Half-life period of zero order reaction =_____________

(i i i)

$(iii)$ Average rate of reaction = _____________

(i v)

$(iv)$ Instantaneous rate of reaction =____________

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

Which of the following is incorrect statement ?

0%
Stoichiometry of a reaction tells about the order of the elementary reactions
0%
For a zero order reaction, rate and the rate constant are idneitcal.
0%
A zero order reaction is controlled by factors other than concentration of reactants
0%
A zero order reaction is an elementary reaction

Explanation

Elementary reactions that occur are unimolecualar and biomolecular ones,

where unimolecular may include bond dissociation or cis - trans transetion Biomolecular reactions strictly involve two species In comparison, zero order

reactions are complex in nature. (e

g

$\mathrm{g}$ . involving an engyme in

C_{2} H_{5} O H

$\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}$

decomposition)

Thus option

D

$D$ is incorrect statement.

The activation energy of the reaction,

A + B \to C + D + 38 k c a l

$A + B \rightarrow C + D + 38kcal$ is

20

$20$ kcal, What would be the activation energy of the reaction,

C + D \to A + B

$C + D \rightarrow A + B$

0%
$20\,kcal$
0%
$-20 \,kcal$
0%
$18\,kcal$
0%
$58\,kcal$

One or more answers is/are correct.
Select the correct statement(s):

0%
Every substance that appears in the rate law of reaction must be a reactant or product in that reaction
0%
If we know that rate law of a reaction; we can deduce it's mechanism must be elementary
0%
If the reaction has rate $r = k[A][B]^{3/2}$ then reaction may be elementary
0%
A zero order reaction must be a complex reaction

An increase in the concentration of the reactants of reaction leads to change in :

0%
Heat of reaction
0%
Threshold energy
0%
Collision frequency
0%
Activation energy

Explanation

The number of collisions that takes place per second per unit volume of the reaction mixture is known as collision frequency (Z)

As the concentration of reactants increases the number of reactant molecules per unit volume increases which increases the collision frequency.

Z \propto

$Z \propto$ number of reactant molecules per unit volume

Option C is correct.

When the activation energies of the forward and backward reactions are equal, then:

0%
$\Delta E = 0, \Delta S = 0$
0%
$\Delta E = 0, \Delta G = 0$
0%
$\Delta S = 0, \Delta G = 0$
0%
only $\Delta E = 0$

Explanation

$A \rightleftharpoons B \\ \text { The energy of reactant and product becomes } \\ \text { equal when the activation energies of the } \\ \text { forward and back ward reactions are equal. } \\ \text { Thus } \triangle E=E_{A}-E_{B}=0 \text { . } \\ \text { The correct option is } D \text { . }$

CBSE Questions for Class 12 Engineering Chemistry Chemical Kinetics Quiz 16 - 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 16 - MCQExams.com

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Practice Class 12 Engineering Chemistry Quiz Questions and Answers

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