Chemical Kinetics Test

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Chemical Kinetics Test - 32

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Weekly Quiz Competition

Question 1
1 / -0

In a given reaction under standard conditions in a closed container, which type of reaction would show no real increase in the rate of the reaction when the concentration of each reactant is doubled?

A
A zero order

B
A first order

C
A second order

D
A third order

Solution

For a zero-order reaction, the rate of reaction is constant and is equal to the rate constant. i.e. Rate=K where k is rate constant.
It implies, a zero-order reaction is independent of the concentration of reactants.
Therefore, there will be no change in the rate of a zero order reaction when the concentration of each order reaction when the concentration of each reactant is doubled.
Question 2
1 / -0

The unit of zero order rate constant is:

A
$$litre$$ $$mol^{-1} sec^{-1}$$

B
$$mol$$ $$litre^{-1} sec^{-1}$$

C
$$sec^{-1}$$

D
$$litre^{2} sec^{-1}$$

Solution

The unit of zero order rate constant is $$\displaystyle mol \ litre^{-1} sec^{-1}$$.

For zero order reaction rate $$\displaystyle = k[A]^0=k$$.

The unit of the rate constant is equal to the unit of rate which is concentration per unit time.
Question 3
1 / -0

In the following image, which reverse reaction has the highest activation energy?

A
E to D

B
G to E

C
C to A

D
This cannot be determined from the given information.

Solution

In reverse reaction, reaction starts from G so it can be seen clearly that most of activation energy is needed from product E to D
Question 4
1 / -0

A first order reaction is 60% complete in 20 minutes. How long will the reaction take to be 84% complete?

A
54 mins

B
68 mins

C
40 mins

D
76 mins

Solution

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

The reaction is 60% complete in 20 minutes.

$$k = \dfrac {2.303}{20}log \dfrac {100}{100-60}$$.....(1)

The reaction is 84% complete

$$k = \dfrac {2.303}{t}log \dfrac {100}{100-84}$$.....(2)

But (1) = (2),

$$ \dfrac {2.303}{20}log \dfrac {100}{100-60}=\dfrac {2.303}{t}log \dfrac {100}{100-84}$$

$$ \dfrac {1}{20}log \dfrac {100}{100-60}=\dfrac {1}{t}log \dfrac {100}{100-84}$$

$$ t=\dfrac {20}{log \dfrac {100}{100-60}}log \dfrac {100}{100-84}$$

$$ t=\dfrac {20}{ 0.3979 } \times 0.79588 $$

$$ t=40 $$ minutes.
The reaction will take 40 minutes to be 84% complete.
Question 5
1 / -0

The activation energy of a chemical reaction can be determined by :

A
evaluating rate constants at two different temperatures

B
changing the concentration of reactants

C
evaluating the concentration of reactants at two different temperatures

D
evaluating rate constant at standard temperature

Solution

Explanation:
Activation energy is the minimum amount of energy that must be provided for compounds to result in a chemical reaction.

The formula to be used to find the activation energy is :

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

where, $${k}_{1}$$ and $${k}_{2}$$ are the rate constants for the reaction at two different temperatures $${T}_{1}$$ and $${T}_{2}$$ .

Thus, the activation energy can be calculated by evaluating the value of rate constants at two different temperatures.

Hence, The correct answer is option $$(A)$$.
Question 6
1 / -0

A plot of $$\dfrac { 1 }{ T } $$ vs. $$logk$$ for a reaction gives the slope $$(-1\times { 10 }^{ 4 }\ K)$$. The energy of activation for the reaction is :
(Given $$R=8.314\ J{ K }^{ -1 }{ mol }^{ -1 }$$)

A
$$1.202\ kJ{ mol }^{ -1 }$$

B
$$83.14\ kJ{ mol }^{ -1 }$$

C
$$8314\ J { mol }^{ -1 }$$

D
$$191.47 \ kJ mol^{-1}$$

Solution

$$log\ k=log\ A-\dfrac{E_a}{2.303RT}$$

Slope $$=\dfrac{-E_a}{2.303R}=-1\times 10^{4}$$

$$\Rightarrow E_a=2.303\times 8.314\times 10^4=191.47\ kJmol^{-1}$$
Question 7
1 / -0

The correct statement regarding the following energy diagrams is

A
Reaction M is faster and less exothermic than Reaction N

B
Reaction M is slower and less exothermic than Reaction N

C
Reaction M is faster and more exothermic than Reaction N

D
Reaction M is slower and more exothermic than Reaction N
Question 8
1 / -0

A given sample of milk turns sour at room temperature $$(27 ^oC)$$ in 5 hours. In a refrigerator at $$-3 ^oC$$, it can be stored 10 times longer. The energy of activation for the souring of milk is :

A
$$2.303 \times 10 R \ \ kJ \cdot mol^{-1}$$

B
$$2.303 \times 5 R \ \ kJ \cdot mol^{-1}$$

C
$$2.303 \times 3 R \ \ kJ \cdot mol^{-1}$$

D
$$2.303 \times 2.7 R \ \ kJ \cdot mol^{-1}$$

Solution

The energy of activation is given by the expression

$$\displaystyle E_a = \dfrac {2.303RTT'}{T'-T}log\dfrac {k'}{k}$$

$$\displaystyle E_a = \dfrac {2.303 \times R \times 300 \times 270}{300-270}log\dfrac {10k}{k}$$

$$\displaystyle E_a = 2.303 \times 2699 R \ \ J \cdot mol^{-1}$$

$$\displaystyle E_a =2.303 \times 2.7 R \ \ kJ \cdot mol^{-1}$$
Question 9
1 / -0

When the concentration of a reactant of first order reaction is doubled, the rate becomes ________ times, but for ___________ order reaction, the rate remains same.

A
Four; zero

B
Two; zero

C
Zero; four

D
Zero; three

Solution

When the concentration of a reactant of the first-order reaction is doubled, the rate becomes two times, but for zero-order reaction, the rate remains the same.

For first order reaction,
rate $$\displaystyle = k[A]$$
When $$\displaystyle [A]$$ is doubled, the rate is doubled.

For zero order reaction,
rate $$\displaystyle = k[A]^0$$
When $$\displaystyle [A]$$ is doubled, the rate is unchanged.

So, the correct option is $$B$$
Question 10
1 / -0

The excess energy which a molecule must posses to become active is known as:

A
kinetic energy

B
threshold energy

C
potential energy

D
activation energy

Solution

The excess energy which a molecule must possess to become active is known as activation energy.

The molecules with energy less than activation energy will not give fruitful collisions. Such molecules will not form products.

Hence, option D is correct.

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Chemical Kinetics Test - 32

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Chemical Kinetics Test - 32

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Question 1

A zero order

A first order

A second order

A third order

Solution

Question 2

$$litre$$ $$mol^{-1} sec^{-1}$$

$$mol$$ $$litre^{-1} sec^{-1}$$

$$sec^{-1}$$

$$litre^{2} sec^{-1}$$

Solution

Question 3

E to D

G to E

C to A

This cannot be determined from the given information.

Solution

Question 4

54 mins

68 mins

40 mins

76 mins

Solution

Question 5

evaluating rate constants at two different temperatures

changing the concentration of reactants

evaluating the concentration of reactants at two different temperatures

evaluating rate constant at standard temperature

Solution

Question 6

$$1.202\ kJ{ mol }^{ -1 }$$

$$83.14\ kJ{ mol }^{ -1 }$$

$$8314\ J { mol }^{ -1 }$$

$$191.47 \ kJ mol^{-1}$$

Solution

Question 7

Reaction M is faster and less exothermic than Reaction N

Reaction M is slower and less exothermic than Reaction N

Reaction M is faster and more exothermic than Reaction N

Reaction M is slower and more exothermic than Reaction N

Question 8

$$2.303 \times 10 R \ \ kJ \cdot mol^{-1}$$

$$2.303 \times 5 R \ \ kJ \cdot mol^{-1}$$

$$2.303 \times 3 R \ \ kJ \cdot mol^{-1}$$

$$2.303 \times 2.7 R \ \ kJ \cdot mol^{-1}$$

Solution

Question 9

Four; zero

Two; zero

Zero; four

Zero; three

Solution

Question 10

kinetic energy

threshold energy

potential energy

activation energy

Solution

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