Chemical Kinetics Test

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Chemical Kinetics Test - 36

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Weekly Quiz Competition

Question 1
1 / -0

Decomposition of $${ H }_{ 2 }{ O }_{ 2 }$$ is a first order reaction. A $$16$$ volume solution of $${ H }_{ 2 }{ O }_{ 2 }$$ of half life $$30$$ min is present at start. When will the solution become one volume?

A
After $$120$$ min

B
After $$90$$ min

C
After $$60$$ min

D
After $$150$$ min

Solution

Given that $$a_0=16V$$, $$a_t=1V$$, $$t_{1/2}=\frac{In2}{k} $$=30 min
For 1st order reaction, $$t=\frac{1}{k} In\frac{a_0}{a_t} $$
$$t=\frac{t_{1/2}}{In2} .In16$$=$$4t_{1/2}=120$$ min
Question 2
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For a 1st order reaction of the form, $$A\xrightarrow [ ]{ \quad k \quad } B$$, the correct representations are:

A
$$I$$ and $$II$$

B
$$III$$ and $$IV$$

C
$$I$$ and $$IV$$

D
$$II$$ and $$III$$

Solution

We know, $$\ln { \dfrac { \left[ A \right] }{ { \left[ A \right] }_{ 0 } } } =-kt$$ ...(i)
It shows that $$\ln { \dfrac { \left[ A \right] }{ { \left[ A \right] }_{ 0 } } } $$ decreases linearly with increase in time '$$t$$'.
From equation (i) $$\dfrac { \left[ A \right] }{ { \left[ A \right] }_{ 0 } } =\dfrac { 1 }{ { e }^{ kt } } $$ ...(ii)
This equation shows that the ratio $$\dfrac { \left[ A \right] }{ { \left[ A \right] }_{ 0 } } $$ will also decrease with increase in time but exponentially.
Question 3
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The time for half-life period of a certain reaction, $$A\rightarrow $$ Product, is $$1$$ hour, when the initial concentration of the reactant '$$A$$' is $$2$$ $$mol$$ $${ L }^{ -1 }$$. 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
$$0.25\ hr$$

B
$$1\ hr$$

C
$$4\ hrs$$

D
$$0.5\ hr$$

Solution

$$\dfrac { { t }_{ 1 } }{ { t }_{ 2 } } ={ \left( \dfrac { { a }_{ 2 } }{ { a }_{ 1 } } \right) }^{ n-1 }$$
$$\dfrac { 1 }{ { t }_{ 2 } } ={ \left( \dfrac { 0.5 }{ 2 } \right) }^{ 0-1 }$$
$${ t }_{ 2 }=0.25 hr$$
Question 4
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In the reaction, $$A+B\longrightarrow C+D$$, the rate $$\left( \dfrac { dx }{ dt } \right) $$ when plotted against time '$$t$$' gives a straight line parallel to time axis and at some time $$'t'$$, $$\dfrac{dx}{dt}=k$$. The order and rate of reaction will be:

A
$$1, k+1$$

B
$$0, k$$

C
$$\left( 1+k \right) , 1$$

D
$$k, k+1$$

Solution

A straight line parallel to time axis indicate that the rate is constant. This is the case for zero order reaction.
$$\frac{dx}{dt} =k$$
rate of the reaction=k
Question 5
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The reaction, $$A\xrightarrow { \quad k \quad } $$ Product, is zero order while the reaction, $$B\xrightarrow { \quad k\quad } $$ Product, is first order reaction. For what initial concentration of $$A$$ are the half lives of the two reactions equal?

A
$$\left( \ln{ 4 } \right) M$$

B
$$2\ M$$

C
$$(2\log { 2 })\ M$$

D
$$(\ln { 2 })\ M$$

Solution

For zero order reaction,
$$x=kt$$
$$\therefore \dfrac { a }{ 2 } =k\times { t }_{ { 1 }/{ 2 } }$$, i.e., $${ t }_{ { 1 }/{ 2 } }=\dfrac { a }{ 2k } $$ ....(i)
For first order reaction,
$${ t }_{ { 1 }/{ 2 } }=\dfrac { \log _{ e }{ 2 } }{ k } $$ ....(ii)
From equations (i) and (ii), $$\dfrac { a }{ 2k } =\dfrac { \log _{ e }{ 2 } }{ k }$$
$$ a=\log _{ e }{ 4 } M$$
Question 6
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Consider following two competing first order reactions:
$$P\xrightarrow [ ]{ \quad { k }_{ 1 }\quad } A+B;$$ $$Q\xrightarrow [ ]{ \quad { k }_{ 2 }\quad } C+D$$
If $$50$$% of the reaction of $$P$$ was completed when $$96$$% of $$Q$$ was completed, then the ratio $$\left( { { k }_{ 2 } }/{ { k }_{ 1 } } \right) $$ will be:

A
$$4.6$$

B
$$4.06$$

C
$$1.123$$

D
$$2.303$$

Solution

For $$50$$% completion of $$P$$:
$$t=\dfrac { 0.693 }{ { k }_{ 1 } } $$ i.e., $${ k }_{ 1 }=\dfrac { 0.693 }{ t }$$ ....(i)
For $$96$$% completion of $$Q$$:
$$t=\dfrac { 2.303 }{ { k }_{ 2 } } \log { \left( \dfrac { 100 }{ 100-96 } \right) }$$
$$\therefore { k }_{ 2 }=\dfrac { 2.303\times 1.398 }{ t } $$ .....(ii)
Dividing equation (ii) by equation (i)
$$\dfrac { { k }_{ 2 } }{ { k }_{ 1 } } =4.6$$
Question 7
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Which of the following graphs is correct for the following reaction?
$$C{ H }_{ 3 }-C{ H }_{ 2 }-CH=C{ H }_{ 2 }\xrightarrow [ { 300 }^{ o }C ]{ \quad { { H }_{ 2 } }/{ Ni }\quad } C{ H }_{ 3 }-C{ H }_{ 2 }-C{ H }_{ 2 }-C{ H }_{ 3 }$$

A

B

C

D

Solution

Reaction involves adsorption on the surface of $$Ni$$ catalyst; hence it is a zero order reaction.
$$a=2{ t }_{ { 1 }/{ 2 } }k$$ (for zero order)
$$\log { a } =\log { { t }_{ { 1 }/{ 2 } } } +\log { 2k } $$
$$Y=MX+C$$
Here, slope $$M=1$$ i.e., $$\tan { \theta } =1$$
$$\therefore \theta ={ 45 }^{ o }$$
Intercept $$C=\log { 2k } $$
Question 8
1 / -0

For a first-order reaction, $$A\longrightarrow $$ Products, 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
$$1.73\times { 10 }^{ -3 }{ M }/{ min }$$

B
$$3.47\times { 10 }^{ -4 }{ M }/{ min }$$

C
$$3.47\times { 10 }^{ -5 }{ M }/{ min }$$

D
$$1.73\times { 10 }^{ -4 }{ M }/{ min }$$

Solution

$${ t }_{ 1/4 }=\cfrac { 2\ln { (2) } }{ k } \\ k=\cfrac { 2\ln { (2) } }{ { t }_{ 1/4 } } $$
$$=\cfrac { 2\ln { (2) } }{ 40 } mi{ n }^{ -1 }\\ \therefore v=k[A]$$
$$=\cfrac { 2\ln { (2) } }{ { t }_{ 1/4 } } \times 0.01$$
$$=\cfrac { 2\ln { (2) } }{ 40 } \times 0.01$$
$$=3.47\times { 10 }^{ -4 }Mmi{ n }^{ -1 }$$

Question 9

1 / -0

The rate of a reaction at $$10$$ sec intervals are as follows:

Time $$\left( sec \right) $$	Rate $$\left( mol\ { L }^{ -1 }{ sec }^{ -1 } \right) $$
$$0$$	$$4.8\times { 10 }^{ -2 }$$
$$10$$	$$4.79\times { 10 }^{ -2 }$$
$$20$$	$$4.78\times { 10 }^{ -2 }$$
$$30$$	$$4.81\times { 10 }^{ -2 }$$

What will be the order of the reaction?

$$0$$

$$2$$

$$1$$

$$3$$

Solution

As observed from the given data, the rate of the reaction is nearly constant. This is possible for a zero order reaction, where rate of the reaction is equal to rate constant.

Question 10
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The half-life period of a first order process is $$1.6$$min. It will be $$90\%$$ complete in :

A
$$5.3$$ min

B
$$10.6$$ min

C
$$43.3$$ min

D
$$99.7$$ min

Solution

For a first order reaction,
$$k=\displaystyle\frac{2.303}{t}log\frac{a}{a-x}$$

At half-time,
$$x=\displaystyle\frac{a}{2}$$

$$\therefore k=\displaystyle\frac{2.303}{1.6}\times log\frac{a}{\displaystyle a-\frac{a}{2}}$$

$$=\displaystyle\frac{2.303}{1.6}\times log 2$$

$$=\displaystyle\frac{2.303}{1.6}\times 0.3010$$

$$=0.4332$$

For $$90\%$$ completion,
$$x=0.9$$a

$$\therefore k=\displaystyle\frac{2.303}{t}log\frac{a}{a-0.9a}$$

$$0.4332=\displaystyle\frac{2.303}{t}log\frac{a}{0.1a}$$

$$=\displaystyle\frac{2.303}{t}log10$$

$$=\displaystyle\frac{2.303}{t}\times 1$$

$$t=\displaystyle\frac{2.303}{0.4332}$$

$$=5.3$$min.

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Chemical Kinetics Test - 36

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Chemical Kinetics Test - 36

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Question 1

After $$120$$ min

After $$90$$ min

After $$60$$ min

After $$150$$ min

Solution

Question 2

$$I$$ and $$II$$

$$III$$ and $$IV$$

$$I$$ and $$IV$$

$$II$$ and $$III$$

Solution

Question 3

$$0.25\ hr$$

$$1\ hr$$

$$4\ hrs$$

$$0.5\ hr$$

Solution

Question 4

$$1, k+1$$

$$0, k$$

$$\left( 1+k \right) , 1$$

$$k, k+1$$

Solution

Question 5

$$\left( \ln{ 4 } \right) M$$

$$2\ M$$

$$(2\log { 2 })\ M$$

$$(\ln { 2 })\ M$$

Solution

Question 6

$$4.6$$

$$4.06$$

$$1.123$$

$$2.303$$

Solution

Question 7

Solution

Question 8

$$1.73\times { 10 }^{ -3 }{ M }/{ min }$$

$$3.47\times { 10 }^{ -4 }{ M }/{ min }$$

$$3.47\times { 10 }^{ -5 }{ M }/{ min }$$

$$1.73\times { 10 }^{ -4 }{ M }/{ min }$$

Solution

Question 9

$$0$$

$$2$$

$$1$$

$$3$$

Solution

Question 10

$$5.3$$ min

$$10.6$$ min

$$43.3$$ min

$$99.7$$ min

Solution

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