Chemical Kinetics Test

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Chemical Kinetics Test - 28

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Weekly Quiz Competition

Question 1
1 / -0

The time required for $$100\%$$ completion of a zero order reaction is:

A
$$ak$$

B
$$\dfrac{a}{2k}$$

C
$$\dfrac{a}{k}$$

D
$$\dfrac{2k}{a}$$

Solution

Hint: The rate of zero reactions is always equal to the rate constant of the reactions.

Correct Option: $$C$$

Explanation for correct option:

Let's write the kinetic equation of zero-order reaction as:
$$x = [A] - [A]_0 =k t$$
Where $$x$$ is the amount of reactant converted into a product,
$$[A] = $$ Final concentration
$$[A]_0 - $$ Initial concentration
$$k$$ is the rate constant and
$$t$$ is time.

If the initial concentration of reactants given is completely converted into a product, then we can say that it is $$100 \%$$ completion reaction.
So, let us consider the initial concentration of reactant as a that is
$$[A]_{\circ}=a$$
$$[A] = [A_0] - kt $$
$$ 0 = a - kt_{100} $$
Time required for $$100\%$$ completion is; $$a=k \times t_{100}$$
Then, $$t_{100}=\dfrac{a}{k}$$
Question 2
1 / -0

A radioactive sample has a half-life period 1500 years. A sealed tube containing 1 g of the sample will contain _________ after 3000 year.

A
1 g of the sample

B
0.5 g of the sample

C
0.25 g of the sample

D
0.025g of the sample

Solution

C. 0.25 g of sample.
Half life 1500 years of 1g sample. After 1500 years it becomes 0.5g, after 3000years it becomes 0.25gm.
Question 3
1 / -0

The half-life period of a radioactive substance is 20 minutes. The time taken for 1 g of the substance to reduce to 0.25g will be:

A
30 minutes

B
40 minutes

C
60 minutes

D
10 minutes

Solution

B. 40min.
If half life time of sample is 20 min for 1g sample, then after 20 min it becomes 0.5g,after 40min it becomes 0.25g.
Question 4
1 / -0

$$99\%$$ at a first order reaction was completed in $$32\:min.$$ when will $$99.9\%$$ of the reaction complete.(in min) :

A
48

B
46

C
50

D
45

Solution

$$k\times32\,=\,ln\begin{pmatrix}\displaystyle\frac{100}{1}\end{pmatrix}\;\;\;\;\;\;...(i)$$
$$\;\;\;\;\;k\times t\,=\,ln\begin{pmatrix}\displaystyle\frac{100}{0.1}\end{pmatrix}\;\;\;\;\;\;...(ii)$$
$$\;\;\;\;\;eq.\;(ii)/(i)$$
$$\;\;\;\;\;\displaystyle\frac{t}{32}\,=\,\displaystyle\frac{3\,ln\,10}{2\,ln\,10}$$
$$\;\;\;\;\;t\,=\,48\:min$$
Question 5
1 / -0

80% of a first order reaction was completed in 70 min. How much time will it take for 90% completion of a reaction?

A
114 min

B
140 min

C
100 min

D
200 min

Solution

Using $$k_{1}=\dfrac{2.303}{t}log\dfrac{a}{a-x}$$, we can find out the value of K. Using K time required for 90% reaction can be calculated.

By putting values in the above equation, we will get

$$K=\dfrac { 2.303 }{ 70 } log\dfrac { 1 }{ 0.2 } = 0.023$$

Now, again using the same equation as above, but for 90% completion, we can calculate the time.

$$t=\dfrac { 2.303 }{ 0.023} log\dfrac { 1 }{ 0.1 }$$

So, we get time $$= 100\ min$$.
Question 6
1 / -0

The rate equation for a reaction, $$\displaystyle A\longrightarrow B$$ is $$\displaystyle r={ k\left[ A \right] }^{ 0 }$$. If the initial concentration of the reactant is $$a$$ $$mol.{ dm }^{ -3 }$$, the half-life period of the reaction is:

A
$$\displaystyle \frac { a }{ 2k } $$

B
$$\displaystyle \frac { k }{ a } $$

C
$$\displaystyle \frac { a }{ k } $$

D
$$\displaystyle \frac { 2a }{ k } $$

Solution

Reaction is $$A\longrightarrow B$$
Rate is given as $$r=k[A]^0=k$$
Order of the reaction is 0.

Initial concentration of A is $$A_0=a$$
Order of the reaction is zero.

Half-life, $$t_{1/2}=\dfrac{[A_0]}{2k}=\dfrac{a}{2k}$$
Question 7
1 / -0

The time taken for $$10$$% completion of a first order reaction is $$20$$ min. Then, for $$19$$% completion, the reaction will take:

A
$$40$$ min

B
$$60$$ min

C
$$30$$ min

D
$$50$$ min

Solution

When the reaction is 10% completed,
$$k=\dfrac { 2.303 }{ 20 } \log { \dfrac { 100 }{ 100-10 } } $$ ...(i)
When the reaction is 19% completed,
$$k=\dfrac { 2.303 }{ t } \log { \dfrac { 100 }{ 100-19 } } $$ ...(ii)
From equation (i) and (ii),
$$\Rightarrow \dfrac { 2.303 }{ 20 } \log { \dfrac { 100 }{ 90 } } =\dfrac { 2.303 }{ t } \log { \dfrac { 100 }{ 81 } } $$
$$\Rightarrow \dfrac { 1 }{ 20 } \times 0.04575=\dfrac { 1 }{ t } \times 0.09151$$
$$\therefore t=40$$ min.
Question 8
1 / -0

For a zero order reaction :

A
$${ t }_{ 1/2 }\propto { R }_{ o }$$

B
$${ t }_{ 1/2 }\propto \dfrac{1}{{ R }_{ o }}$$

C
$${ t }_{ 1/2 }\propto { R }_{ o }^{ 2 }$$

D
$${ t }_{ 1/2 }\propto \dfrac{1}{{ R }_{ o }^{2 }}$$

Solution

For a zero-order reaction ,
$$[R]=-kt+[R_0]$$

When $$t=t_{1/2}$$, $$[R]=[R_0/2]$$
So $$[R_0/2]=-kt_{1/2}+[R_0]$$

$$t_{1/2}=\cfrac{[R_0]}{2k}$$
$$t_{1/2}$$ is proportional to $$[R_0]$$
Hence, the correct option is $$\text{A}$$
Question 9
1 / -0

Speed of a chemical reaction depends on which of the following parameters?

A
Temperature

B
Concentration

C
Pressure

D
All of the above

Solution

The speed of any chemical reaction depends upon the temperature, pressure, concentration, surface area etc. of a substance.
Question 10
1 / -0

The relationship between rate constant and half-life period of zero order reaction is given by:

A
$$\displaystyle t_{1/2} = [A_0]2k$$

B
$$\displaystyle t_{1/2} = \frac{0.693}{k}$$

C
$$\displaystyle t_{1/2} = \frac{[A_0]}{2k}$$

D
$$\displaystyle t_{1/2} = \frac{2[A_0]}{2k}$$

Solution

The relationship between rate constant and half-life period of zero-order reaction is given by:

$$t = \dfrac{[A_0] - [A]}{k}$$
At $$t = t_{1/2}$$, $$[A] = \dfrac 12 [A_0]$$
$$t_{1/2}=\dfrac { [{ A }_{ 0 }] }{ 2k } $$
Hence, option $$C$$ is correct.

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Chemical Kinetics Test - 28

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Chemical Kinetics Test - 28

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Question 1

$$ak$$

$$\dfrac{a}{2k}$$

$$\dfrac{a}{k}$$

$$\dfrac{2k}{a}$$

Solution

Question 2

1 g of the sample

0.5 g of the sample

0.25 g of the sample

0.025g of the sample

Solution

Question 3

30 minutes

40 minutes

60 minutes

10 minutes

Solution

Question 4

48

46

50

45

Solution

Question 5

114 min

140 min

100 min

200 min

Solution

Question 6

$$\displaystyle \frac { a }{ 2k } $$

$$\displaystyle \frac { k }{ a } $$

$$\displaystyle \frac { a }{ k } $$

$$\displaystyle \frac { 2a }{ k } $$

Solution

Question 7

$$40$$ min

$$60$$ min

$$30$$ min

$$50$$ min

Solution

Question 8

$${ t }_{ 1/2 }\propto { R }_{ o }$$

$${ t }_{ 1/2 }\propto \dfrac{1}{{ R }_{ o }}$$

$${ t }_{ 1/2 }\propto { R }_{ o }^{ 2 }$$

$${ t }_{ 1/2 }\propto \dfrac{1}{{ R }_{ o }^{2 }}$$

Solution

Question 9

Temperature

Concentration

Pressure

All of the above

Solution

Question 10

$$\displaystyle t_{1/2} = [A_0]2k$$

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

$$\displaystyle t_{1/2} = \frac{[A_0]}{2k}$$

$$\displaystyle t_{1/2} = \frac{2[A_0]}{2k}$$

Solution

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