Chemical Kinetics Test

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Chemical Kinetics Test - 21

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Weekly Quiz Competition

Question 1
1 / -0

The most common molecular collisions are in between:

A
2 molecules

B
3 molecules

C
4 molecules

D
5 molecules

Solution
Question 2
1 / -0

The half life of first order reaction $$PCl_{5} \rightarrow PCl_{3}+Cl_{2}$$ is 10 minutes at a certain temperature. The time in which the concentration of $$PCl_{5}$$ would be reduced to 10% of the initial concentration is:

A
26 min

B
33 min

C
71 min

D
90 min

Solution

The relationship between the rate constant and half life period is $$k=\displaystyle \frac{0.693}{t}=\displaystyle \frac{0.693}{10}=0.0693$$.
The integrated rate law expression is $$k=\displaystyle \frac{2.303}{t}\log\frac{a}{a-x}.$$
Substitute values in the above expression.
$$0.0693=\displaystyle \frac{2.303}{t}\log\frac{100}{10}.$$
$$t=33 min$$.
Question 3
1 / -0

For the gas phase decomposition $$A\rightarrow2B$$, the rate constant is $$6.93\times 10^{-3}$$ $$min^{-1}$$ at 300K. The percentage of A remaining at the end of 300 minutes is :

A
75

B
50

C
25

D
12.5

Solution

$$ln\left [ \frac{Ao}{A} \right ]=kt$$
$$=6.93\times 10^{-3}\times 300$$
$$\frac{Ao}{A}=8$$
$$\frac{A}{Ao}=.1250$$
$$\frac{A}{Ao}\times 100$$ =12.5%
Question 4
1 / -0

The rate of a certain reaction at different times are as follows:
Time rate ($$mole$$ $$lit^{-1}\sec^{-1}$$)
I) 0 a) $$2.8 \times 10^{-2}$$
II) 10 b) $$2.78\times 10^{-2}$$
III) 20 c) $$2.81\times 10^{-2}$$
IV) 30 d) $$2.79\times 10^{-2}$$
The correct matching is:

A
zero order

B
first order

C
second order

D
third order

Solution

At different times, the rate of the reaction is same.
Hence, the rate of the reaction is independent of time.
This is the characteristics of a zero order reaction.
Question 5
1 / -0

From the graph pick out the correct one:

A
$$\Delta E$$ for forward reaction is $$B-A$$

B
$$\Delta E$$ for the forward reaction $$C-A$$

C
$$\Delta E$$ reverse is greater than forward

D
$$\Delta E$$ for forward reaction is $$A + B$$

Solution

The change in energy of a reaction is equal to the energy of products minus the energy of reactants.
$$\Delta E=E_{C}-E_{A}$$.
Question 6
1 / -0

For a zero-order reaction $$A$$ $$\rightarrow $$ product, the rate constant is $$10^{-2}\ mol\ L^{-1}\ s^{-1}$$ . Starting with 10 moles of $$A$$ in a 1 L vessel, how many moles of $$A$$ would be left unreacted after 10 minutes?

A
5 moles

B
6 moles

C
4 moles

D
10 moles

Solution

$$k=10^{-2}$$
$$kt = R_o - R$$
$$10^{-2}\times 10\times 60=\dfrac{10}{1}-R$$
$$R$$ $$= 6\ mol L^{-1}$$
6 mole react unreacted mole $$= 10 -6$$
$$= 4 moles$$
Question 7
1 / -0

Which is the graphical representation for the zeroth order reaction?

A

B

C

D

Solution

rate = k
$$\dfrac{-dx}{dt}=k$$
Where k is rate constant. Thus, the rate of reaction is independent of concentration of reactant.
So $$\dfrac{dx}{dt}$$ is constant.
Hence graph will be constant straight line parallel to x axis.
Question 8
1 / -0

For the reaction $$2N_2O_5 \rightarrow 4NO_2+ O_2$$ rate at a particular time and rate constant are $$1.02\times 10^{-4}\ M s^{-1}$$ and $$3.4\times10^{-5}\ Ms^{-1}$$ respectively. Then the concentration of $$N_2O_5$$ at that time will be:

A
$$1.732$$

B
$$3$$

C
$$1.02\times10^{4}$$

D
$$3.4\times 10^5$$

Solution

Unit of rate = $$M{ s }^{ -1 }$$
Let the rate be of the form
$$r=K{ \left[ { N }_{ 2 }{ O }_{ 5 } \right] }^{ x }$$
Given unit of rate constant = $${ S }^{ -1 }$$
For dimensional consistenty x=1
and hence, the given reaction is a first order reaction.
$$\Rightarrow $$ rate $$=K\left[ { N }_{ 2 }{ O }_{ 5 } \right] $$
$$\Rightarrow \quad 1.02\times { 10 }^{ -4 }=\left[ { N }_{ 2 }{ O }_{ 5 } \right] 3.4\times { 10 }^{ -5 }$$
$$\Rightarrow \quad \left[ { N }_{ 2 }{ O }_{ 5 } \right] =3$$
Question 9
1 / -0

The temperature dependence of rate constant (k) of a chemical reaction is written in terms of Arrhenius equation, $$k = A.e^{E^{\ast }/RT}. Activation energy $$(E^{\ast })$$ of the reaction can be calculated by plotting

A
K vs T

B
K vs $$\frac{1}{log T}$$

C
log K vs $$\frac{1}{T}$$

D
log k vs $$\frac{1}{log T}$$
Question 10
1 / -0

If 60% of a first order reaction was completed in 60 minutes, 50% of the same reaction would be completed in approximately: (log 4 = 0.60, log 5 = 0.69)

A
40 minutes

B
50 minutes

C
45 minutes

D
60 minutes

Solution

Rate of a general $${ 1 }^{ st }$$ order reaction.
$$\dfrac { -d\left[ A \right] }{ dt } =k\left[ A \right] $$
$$\Rightarrow \quad \int _{ { \left[ A \right] }_{ 0 } }^{ { \left[ A \right] }_{ t } }{ \dfrac { d\left[ A \right] }{ \left[ A \right] } } =-\int _{ 0 }^{ t }{ kdt } $$
$$ln\dfrac { { \left[ A \right] }_{ t } }{ { \left[ A \right] }_{ 0 } } =-kt$$
$$\Rightarrow \quad \boxed { 2.303log\frac { { \left[ A \right] }_{ t } }{ { \left[ A \right] }_{ 0 } } =-kt } \quad \longrightarrow \quad \left( 1 \right) $$
Given : 60% of reaction completes in 60 minutes.
Applying (1).
$$2.303log\dfrac { 0.4{ \left[ A \right] }_{ 0 } }{ \left[ { A }_{ 0 } \right] } =-60k$$
$$2.303log0.4\quad =\quad -60k$$
$$\Rightarrow \quad $$$$k=\dfrac { -2.303log0.4 }{ 60 } $$
Using the value of k
$$2.303log\dfrac { 0.5{ \left[ A \right] }_{ 0 } }{ { \left[ A \right] }_{ 0 } } =-xk$$
$$2.303log0.5=-x\left( \dfrac { -2.303log0.4 }{ 60 } \right) $$
$$\Rightarrow \quad \dfrac { 60log0.5 }{ log0.4 } =x$$
$$\Rightarrow \quad x=\dfrac { 60(log5-1) }{ (log4-1) } $$
$$\Rightarrow \quad x=\dfrac { 60(0.69-1) }{ 0.60-1 } $$
$$\Rightarrow \quad x=\dfrac { 60\times 0.31 }{ 0.4 } $$
$$\therefore \quad \boxed { x\quad \simeq \quad 45min. } $$

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Re-Attempt Weekly Quiz Competition

Time	rate ($$mole$$ $$lit^{-1}\sec^{-1}$$)
I) 0	a) $$2.8 \times 10^{-2}$$
II) 10	b) $$2.78\times 10^{-2}$$
III) 20	c) $$2.81\times 10^{-2}$$
IV) 30	d) $$2.79\times 10^{-2}$$

Chemical Kinetics Test - 21

Result

Chemical Kinetics Test - 21

Score

-

Rank

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SHARING IS CARING

Question 1

2 molecules

3 molecules

4 molecules

5 molecules

Solution

Question 2

26 min

33 min

71 min

90 min

Solution

Question 3

75

50

25

12.5

Solution

Question 4

zero order

first order

second order

third order

Solution

Question 5

$$\Delta E$$ for forward reaction is $$B-A$$

$$\Delta E$$ for the forward reaction $$C-A$$

$$\Delta E$$ reverse is greater than forward

$$\Delta E$$ for forward reaction is $$A + B$$

Solution

Question 6

5 moles

6 moles

4 moles

10 moles

Solution

Question 7

Solution

Question 8

$$1.732$$

$$3$$

$$1.02\times10^{4}$$

$$3.4\times 10^5$$

Solution

Question 9

K vs T

K vs $$\frac{1}{log T}$$

log K vs $$\frac{1}{T}$$

log k vs $$\frac{1}{log T}$$

Question 10

40 minutes

50 minutes

45 minutes

60 minutes

Solution

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