Chemical Kinetics Test

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Chemical Kinetics Test - 65

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Weekly Quiz Competition

Question 1
1 / -0

What is the formula to find the value of $$t_{1/2}$$ for a zero order reaction?

A
$$\dfrac {k}{[R]_{0}}$$

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

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

D
$$\dfrac {0.693}{k}$$

Solution

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 $$\dfrac {[R]_{0}}{2k}$$.
Question 2
1 / -0

In the presence of an acid, the initial concentration of cane sugar was reduced from $$ 0.20 $$ to $$ 0.10 $$ molar in $$ 5 $$ hours and from $$0.2 $$ to $$ 0.05 $$ molar in $$10 $$ hours. The reaction is of:

A
Zero order

B
First order

C
Second order

D
Third order
Question 3
1 / -0

Directions For Questions

Population growth of humans and bacteria follow first-order growth kinetics. Suppose, $$100$$ bacteria are placed in a flask containing nutrients for the bacteria so that they can multiply. A study at $${ 35 }^{ \circ}C$$ gave the following results:

$$\begin{matrix} Time\left( minutes \right) & 0 & 15 & 30 & 45 & 60 \\ Number\ of\ bacteria & 100 & 200 & 400 & 800 & 1600 \end{matrix}$$

...view full instructions

The rate constant for the first-order growth of bacteria can be calculated using:

A
$$k=\dfrac { 2.303 }{ t } \log _{ 10 }{ \left( \dfrac { a }{ a-x } \right) } $$

B
$$k=-\dfrac { 2.303 }{ t } \log_{10} { \left( \dfrac { a }{ a+x } \right) } $$

C
$$k=\dfrac { 0.693 }{ t } $$

D
$$k=\dfrac { x }{ t } $$

Solution
Question 4
1 / -0

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):

A
$$-149$$ kJ

B
$$+149$$ kJ

C
$$-5$$ kJ

D
$$+5$$ kJ

Solution

Relation used;

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

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

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

Question 5

1 / -0

Match the List-I and List-II:

List-I	List-II
(A) Rate constant has the same unit as the rate of reaction.	(i) One
(B) Reactions having apparent molecularity more than three reaction	(ii) Zero order
(C) Reactions having molecularity two but order of reaction is one.	(iii) Complex reaction
(D) For a reaction, $$A \rightarrow B$$, the rate of reaction doubles as the ocular reaction concentration of A is doubled.	(iv) Pseudo unimolecular reaction

(A) - ii (B) - iv (C) - iii (D) - i

(A) - ii (B) - iii (C) - iv (D) - i

(A) - iii (B) - ii (C) - iv (D) - i

(A) - ii (B) - iv (C) - i (D) - iii

Solution

(A) Rate constant has the same unit as the rate of reaction for zero order reactions.
(B) Reactions having apparent molecularity more than three reaction for pseudo uni-molecular reactions.
(C) Reactions having molecularity two but order of reaction is one are Complex reaction.
(D) For a reaction, $$A \rightarrow B$$, the rate of reaction doubles as the ocular reaction concentration of A is doubled in first order reactions.
So the correct options are (A) - ii (B) - iv (C) - iii (D) - i

Hence option A is correct.

Question 6
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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:

A
The intrinsic energy of the reaction $$2NO+O_2\rightleftharpoons 2NO_2$$ is less

B
$$CO$$ is smaller in size than that of $$NO$$

C
$$CO$$ is poisonous

D
The activation energy for the reaction $$2NO+O_2\rightleftharpoons 2NO_2$$ is less

Solution

Evidently, less is the activation energy, faster is the reaction or greater is the activation energy, slower is the reaction.
Question 7
1 / -0

A certain reaction rate increases $$1000$$ folds in the presence of a cataylst at $$27^{\circ}C$$. The activation energy of the original pathway is $$98\ kJ/mol$$. What is the activation energy of the new pathway?

A
$$80.77\ kJ$$

B
$$56.38\ kJ$$

C
$$24.67\ kJ$$

D
$$90.34\ kJ$$

Solution

Given that rate increase by 1000 tines

$$\Rightarrow \dfrac{K_2}{K_1}=1000$$

$$\log \dfrac {K_{2}}{K_{1}} = \log 1000=3$$

$$3= \dfrac {E_{1} - E_{2}}{2.303RT}$$

$$3= \dfrac {98000 - E_{2}}{2.303\times 8.314 \times 300}$$

$$E_{2}(catalyst) = 80.77\ kJ$$.
Question 8
1 / -0

A first-order reaction is $$40$$% complete after $$8$$ min. How long will it take before it is $$90$$% complete?

A
40.20 min

B
36.36 min

C
24.52 min

D
None of these

Solution

For first order reaction we have:

$$K= \cfrac {1}{t}$$ $$ln \cfrac {a}{a-x}$$ $$(i)$$

where, $$K$$= rate constant of reaction
$$t$$= time
$$a$$= initial concentration of the reactant
$$x$$= amount of reactant reacted in time $$t$$

Since, the reactant has reacted $$40$$% in $$8$$ min, the remaining reactant is $$60$$%
$$\Rightarrow$$ At $$t=8$$ min , $$(a-x)=0.6a$$

Now, we have to find time when the reaction is $$90$$% complete i.e. $$(a-x)=0.1a$$

So Case I: $$t=8$$ min; $$(a-x)=0.6a$$
Case II: $$t=?$$ ; $$(a-x)=0.1a$$

Since the reaction is of first order $$K$$ is same in both cases so using the formula $$(i)$$ :-
$$\cfrac {1}{2}\times ln \left(\cfrac {1}{0.6}\right)=\cfrac {1}{t}\times ln \left(\cfrac {1}{0.1}\right)$$
$$\Rightarrow$$ $$t=8\times \cfrac {ln(1/0.1)}{ln(1/0.6)}=8\times \cfrac {2.302}{0.511}=36.36$$ min

Answer: (B) $$36.36$$ min
Question 9
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Which among the following is wrong for $$1^{st}$$ order reaction?

A
$$t_{99.9\%}$$ $$= \dfrac {5}{2}t_{99\%}$$

B
$$t_{99.9\%}$$$$ = 10t_{1/2}$$

C
$$t_{99.9\%}$$$$ = 3t_{90\%}$$

D
$$t_{96.875\%}$$$$ = 5t_{1/2}$$
Question 10
1 / -0

Which of the following are the correct representations of a zero order reaction, where A represents the reactant?

A
a, b, c

B
a, b, d

C
b, c, d

D
a, c, d

Solution

The graphs a, b, d are the correct representations of a zero-order reaction, where A represents the reactant.

For a zero order reaction, rate $$ \displaystyle =k[A]^0=k=$$ constant. Hence, the graph of rate vs $$[A]$$ is a straight line parallel to [A] axis.

Also, $$ \displaystyle [A]_t = -kt +[A]_0 $$

The plot of $$ \displaystyle [A]_t$$ vs time (t) is a straight line with a negative slope.

Slope $$ \displaystyle = -k$$

The half life period $$ \displaystyle t_{1/2} = \dfrac {[A]_0}{2k}$$

The graph of $$ \displaystyle t_{1/2}$$ with $$ \displaystyle [A]_0$$ is a straight line with positive slope. The slope is $$ \displaystyle \dfrac {1}{2k}$$

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Re-Attempt Weekly Quiz Competition

Chemical Kinetics Test - 65

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Chemical Kinetics Test - 65

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SHARING IS CARING

Question 1

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

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

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

$$\dfrac {0.693}{k}$$

Solution

Question 2

Zero order

First order

Second order

Third order

Question 3

$$k=\dfrac { 2.303 }{ t } \log _{ 10 }{ \left( \dfrac { a }{ a-x } \right) } $$

$$k=-\dfrac { 2.303 }{ t } \log_{10} { \left( \dfrac { a }{ a+x } \right) } $$

$$k=\dfrac { 0.693 }{ t } $$

$$k=\dfrac { x }{ t } $$

Solution

Question 4

$$-149$$ kJ

$$+149$$ kJ

$$-5$$ kJ

$$+5$$ kJ

Solution

Question 5

(A) - ii (B) - iv (C) - iii (D) - i

(A) - ii (B) - iii (C) - iv (D) - i

(A) - iii (B) - ii (C) - iv (D) - i

(A) - ii (B) - iv (C) - i (D) - iii

Solution

Question 6

The intrinsic energy of the reaction $$2NO+O_2\rightleftharpoons 2NO_2$$ is less

$$CO$$ is smaller in size than that of $$NO$$

$$CO$$ is poisonous

The activation energy for the reaction $$2NO+O_2\rightleftharpoons 2NO_2$$ is less

Solution

Question 7

$$80.77\ kJ$$

$$56.38\ kJ$$

$$24.67\ kJ$$

$$90.34\ kJ$$

Solution

Question 8

40.20 min

36.36 min

24.52 min

None of these

Solution

Question 9

$$t_{99.9\%}$$ $$= \dfrac {5}{2}t_{99\%}$$

$$t_{99.9\%}$$$$ = 10t_{1/2}$$

$$t_{99.9\%}$$$$ = 3t_{90\%}$$

$$t_{96.875\%}$$$$ = 5t_{1/2}$$

Question 10

a, b, c

a, b, d

b, c, d

a, c, d

Solution

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