Chemical Kinetics Test

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Chemical Kinetics Test - 13

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Weekly Quiz Competition

Question 1
1 / -0

In a first order reaction, the concentration of the reactant, decreases from 0.8 M to 0.4 M in 15 minutes. The time taken for the concentration to change from 0.1 M to 0.025 M is:

A
30 min

B
15 min

C
7.5 min

D
60 min

Solution

As question says concentration becomes half in 15 mins. and using information that is first order reaction we can predict time for 0.1M to 0.025M reaction.
If we take 0.1M of same sub then 1st half life will give 0.05M and second half life gives 0.025
so from above
Time: 30 mins (2 half life).
Question 2
1 / -0

For a first order reaction, (A) $$\rightarrow$$ product, the concentration of A changes from 0.1 M to 0.025 M in 40 minutes. The rate of reaction when the concentration of A is 0.01 M, is:

A
$$ 1.73\displaystyle \times 10^{-5}Mmin^{-1}$$

B
$$ 3.47\displaystyle \times 10^{-4}Mmin^{-1}$$

C
$$ 3.47\displaystyle \times 10^{-5}Mmin^{-1}$$

D
$$ 1.73\displaystyle \times 10^{-4}Mmin^{-1}$$

Solution

For first order reaction,

$$\displaystyle k= \frac{2.303}{40} log\frac{0.1}{0.025} $$

$$= \dfrac{2.303}{40}log\ 4 $$

$$= \dfrac{2.303}{20}log\ 2 $$

$$= \dfrac {2.303}{20} \times 0.301 = 0.0347$$

$$\displaystyle k= 0.0347$$

$$Rate=k[A]$$

$$\displaystyle Rate= 0.0347 \times 10^{-2}=3.47 \times 10^{-4} M min^{-1}$$
Question 3
1 / -0

The time for half-life period of a certain reaction $$\mathrm{A}\rightarrow$$ products is 1 hour. When the initial concentration of the reactant $$A$$ is 2.0 mol L$$^{-1}$$, how much time does it take for its concentration to come from 0.50 to 0.25 mol L$$^{-1}$$ if it is a zero-order reaction?

A
4 h

B
0.5 h

C
0.25 h

D
1 h

Solution

For a zero order reaction $$\displaystyle \mathrm{k}=\frac{\mathrm{x}}{\mathrm{t}} \rightarrow(1)$$
Where $$\mathrm{x}=$$ amount decomposed

For a zero order reaction,
$$\displaystyle \mathrm{k}=\frac{[\mathrm{A}]_{0}}{2\mathrm{t}_{\frac{1}{2}}}$$

Since $$[\mathrm{A}_{0}]=2\mathrm{M}, \mathrm{t}_{1/2}=1$$ hr; $$\mathrm{k}=1$$

From equation (1),
$$\displaystyle \mathrm{t}=\frac{0.25}{1}=0.25$$ hr
Question 4
1 / -0

The rate of a reaction quadruples when the temperature changes from 300 to 310 K. The activation energy of this reaction is :
(Assume activation energy and pre-exponential factor are independent of temperature; $$ln\ 2$$=0.693; R = 8.314 J $$mol^{ -1} K^{-1}$$)

A
26.8 kJ $$mol^{-1}$$

B
414.4 kJ $$mol^{-1}$$

C
107.2 kJ $$mol^{-1}$$

D
53.6 kJ $$mol^{-1}$$

Solution

The rate of the reaction can be written as:

$$ln\left(\dfrac{K_2}{K_1}\right)=\dfrac{E_a}{R}\Bigg[\dfrac{1}{T_1}-\dfrac{1}{T_2}\Bigg]$$

$$ln\left(\dfrac{4}{1}\right)=\dfrac{E_a}{8.314}\Bigg[\dfrac{1}{300}-\dfrac{1}{310}\Bigg]$$

$$\Rightarrow E_a=107.2\ KJ/mol$$
Question 5
1 / -0

Higher order (>3) reactions are rare due to:

A
low probability of simultaneous collision of all the reacting species

B
increase in entropy and activation energy as more molecules are involved

C
shifting of equilibrium towards reactants due to elastic collisions

D
loss of active species on collision

Solution

For higher order (>3) reactions to occur, 3 or more molecules (having energy equal to or greater than activation energy) must simultaneously collide with proper orientation.
The probability for such collisions is very low. Hence these reactions are rare.
Question 6
1 / -0

$$\mathrm{l}\mathrm{n}$$ a first order reaction the concentration of reactant decreases from 800 $$\mathrm{m}\mathrm{o}l/\mathrm{d}\mathrm{m}^{3}$$ to 50 $$\mathrm{m}\mathrm{o}l/\mathrm{d}\mathrm{m}^{3}$$ is $$2\times 10^{4}$$ sec. The rate constant of reaction in $$\sec^{-1}$$ is:

A
$$2\times 10^{4}$$

B
$$3.45\times 10^{-5}$$

C
$$1.386\times 10^{-4}$$

D
$$2\times 10^{-4}$$

Solution

Assuming decay is happening in the above example, we can say it will take 4 half-lives in $$2\times{ 10 }^{ 4 }$$ sec.

So, each half-life will take $$5\times { 10 }^{ 3 }$$ sec.

Now using, $$t_{ 1/2 }=\dfrac { 0.693 }{ k } $$ and putting values we get

$$k=\dfrac { 0.693 }{ 5\times{ 10 }^{ 3 } } $$

$$k=1.386\times{ 10 }^{ -4 }$$

Option C is correct.
Question 7
1 / -0

$$A$$ follows the first-order reaction.

$$(A)$$ $$\longrightarrow$$ product.

The concentration of $$A$$ changes from 0.1 $$\mathrm{M}$$ to 0.025 $$\mathrm{M}$$ in 40 minutes. Find the rate of reaction of $$A$$ when concentration of $$A$$ is 0.01 $$\mathrm{M}$$?

A
$$3.47\displaystyle \times 10^{-4}\mathrm{M}$$ $$\min^{-1}$$

B
$$3.47\displaystyle \times 10^{-5}\mathrm{M}$$ $$\min^{-1}$$

C
$$1.73\displaystyle \times 10^{-4}\mathrm{M} $$ $$\min^{-1}$$

D
$$1.73\displaystyle \times 10^{-3}\mathrm{M}$$ $$\min^{-1}$$

Solution

As given, concentration changes from $$0.01\ M$$ to $$0.025\ M$$ in $$40$$ minutes.

As we know,
$$t_{1/2}$$ is the time when concentration becomes half of the original.

$$\Rightarrow 2\mathrm{t}_{1/2}=40\min$$

$$\mathrm{t}_{1/2}=20\min$$

$$K = \dfrac{0.693}{t_{\frac{1}{2}}}$$

$$k = \dfrac{0.693}{20}$$

For first order reaction,
$$\displaystyle \mathrm{r}=\mathrm{K}[\mathrm{A}]=\frac{0.693}{20}\times 0.01=3.47\times 10^{-4}\ M\ min^{-1}$$
Question 8
1 / -0

Activation energy of a reaction:

A
is independent of temperature

B
increase with temperature

C
gets double in every 10 degree raise in temperature

D
decrease with temperature

Solution

Activation energy is defined as the minimum amount of extra energy required by a reacting molecule to get converted into products. Activation energy can't be altered by temperature. Catalysts are molecules that speed up reactions by decreasing the activation energy by providing an alternate path. The activation energy depends on the nature of the chemical transformation that takes place.
Question 9
1 / -0

The decomposition of phosphine $$(PH_3)$$ on tungsten at low pressure is a first-order reaction. It is because the:

A
rate is independent of the surface coverage

B
rate of decomposition is very slow

C
rate is proportional to the surface coverage

D
rate is inversely proportional to the surface coverage

Solution

The decomposition of $$PH_3$$ on tungsten at low pressure is a first order reaction because surface area coverage is proportional to partial pressure of $$PH_3$$. Hence, rate is proportional to the surface coverage.
Question 10
1 / -0

A first order reaction has a specific reaction rate of $$10^{-2} sec^{-1}$$. How much time will it take for $$20\ g$$ of the reactant to reduce to $$5\ g$$?

A
$$693.0\ sec$$

B
$$238.6 \ sec$$

C
$$138.6\ sec$$

D
$$346.5\ sec$$

Solution

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

$$\displaystyle t = \dfrac {2.303}{10^{-2} \: /s}log\left( \dfrac {20 \: g}{5 \: g}\right) $$

$$\displaystyle t = 138.6$$ sec
Hence, the time required will be 138.6 sec.

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Re-Attempt Weekly Quiz Competition

Chemical Kinetics Test - 13

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Chemical Kinetics Test - 13

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SHARING IS CARING

Question 1

30 min

15 min

7.5 min

60 min

Solution

Question 2

$$ 1.73\displaystyle \times 10^{-5}Mmin^{-1}$$

$$ 3.47\displaystyle \times 10^{-4}Mmin^{-1}$$

$$ 3.47\displaystyle \times 10^{-5}Mmin^{-1}$$

$$ 1.73\displaystyle \times 10^{-4}Mmin^{-1}$$

Solution

Question 3

4 h

0.5 h

0.25 h

1 h

Solution

Question 4

26.8 kJ $$mol^{-1}$$

414.4 kJ $$mol^{-1}$$

107.2 kJ $$mol^{-1}$$

53.6 kJ $$mol^{-1}$$

Solution

Question 5

low probability of simultaneous collision of all the reacting species

increase in entropy and activation energy as more molecules are involved

shifting of equilibrium towards reactants due to elastic collisions

loss of active species on collision

Solution

Question 6

$$2\times 10^{4}$$

$$3.45\times 10^{-5}$$

$$1.386\times 10^{-4}$$

$$2\times 10^{-4}$$

Solution

Question 7

$$3.47\displaystyle \times 10^{-4}\mathrm{M}$$ $$\min^{-1}$$

$$3.47\displaystyle \times 10^{-5}\mathrm{M}$$ $$\min^{-1}$$

$$1.73\displaystyle \times 10^{-4}\mathrm{M} $$ $$\min^{-1}$$

$$1.73\displaystyle \times 10^{-3}\mathrm{M}$$ $$\min^{-1}$$

Solution

Question 8

is independent of temperature

increase with temperature

gets double in every 10 degree raise in temperature

decrease with temperature

Solution

Question 9

rate is independent of the surface coverage

rate of decomposition is very slow

rate is proportional to the surface coverage

rate is inversely proportional to the surface coverage

Solution

Question 10

$$693.0\ sec$$

$$238.6 \ sec$$

$$138.6\ sec$$

$$346.5\ sec$$

Solution

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