Chemical Kinetics Test

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Chemical Kinetics Test - 55

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Weekly Quiz Competition

Question 1
1 / -0

The rate for a first order reaction is $$0.6932 \times 10^{-2}$$ mol litre$$^{-1}$$ min$$^{-1}$$ and the initial concentration of the reactant is 1 M, $$t_{1/2}$$ is equal to:

A
0.693 x $$10^{-2}$$ minute

B
100 minutes

C
0.693 x $$10^{-3}$$ minute

D
6.932 minutes

Solution

Rate $$= k [A]$$

$$k= \dfrac{0.6932\times 10^{-2}}{1}$$

$$t_{^1/_2}= \dfrac{0.693}{k}$$

$$= \dfrac{0.693}{0.693\times 10^{-2}}$$

$$= 100\ min$$

Hence, the correct option is $$\text{B}$$.
Question 2
1 / -0

For a reaction following first-order kinetics, which of the following statements is/are correct ?

A
The time taken for the completion of 50% of the reaction is $$t_{1/2}$$.

B
A plot of the reciprocal of the concentration of the reactants against time gives a straight line.

C
The degree of dissociation is equal to $$1-e^{-kt}$$.

D
A plot of $$[A]_{0}/[A]$$ versus time gives a straight line.

Solution

The time taken for the completion of $$50$$% of the reaction is called as the half-life period. It is represented as $$t_{1/2}$$.
Thus option A is correct.

A plot of the logarithm of the concentration of the reactants against time gives a straight line.
Thus option B is incorrect.

The integrated rate law for the first order reaction is $$[A]=[A]_0e^{-kt}$$.
Substitute $$[A]=a-x$$ and $$[A]_0=a$$ in the above expression.
$$a-x=ae^{-kt}$$.
Rearrange above expression.
$$x=a\left ( 1-e^{-kt} \right )$$.
Hence option C is incorrect.

A plot of $$log [A]_{0}/[A]$$ versus time gives a straight line.
Hence, option D is incorrect.
Question 3
1 / -0

During the hydrogenation of vegetable oil at $$25^{0}C$$, the pressure of $$H_{2}$$ reduces from 2 atmospheres to 1.2 atmospheres in 50 minutes. The rate of reaction in terms of molarity per second is:

A
$$1.09\times 10^{-6}$$

B
$$1.09\times 10^{-5}$$

C
$$1.09\times 10^{-7}$$

D
$$1.09\times 10^{5}$$

Solution

For hydrogenation reaction the rate of reaction is written as $$ r = -\dfrac{d[H_2]}{dt}$$.

Assuming the hydrogen an ideal gas $$P = CRT$$

Therefore $$ C = \dfrac{P}{RT}$$

Hence $$\Delta C = \dfrac{\Delta P}{RT}$$

$$\Delta P = -0.8$$.

So $$\Delta C = \dfrac{-0.8}{0.0821\times 298}$$

So,
$$ r = -\dfrac{ \dfrac{-0.8}{0.0821\times 298}}{50\times60} = 1.09\times10^{-5}$$

$$r = 1.09\times10^{-5}$$

Option B is correct.
Question 4
1 / -0

The rate constant for two parallel reactions were found to be $$1.0\times 10^{-2}dm^{3}$$ $$mol^{-1}s^{-1}$$ and $$3.0\times 10^{-2}dm^{3}$$ $$mol^{-1}s^{-1}$$. If the corresponding energies of activation of the parallel reactions are 60.0 KJ $$mol^{-1}$$ and 70.0 KJ $$mol^{-1}$$ respectively, then what is the apparent overall energy of activation?

A
130.0 KJ $$mol^{-1}$$

B
65.0 KJ $$mol^{-1}$$

C
67.5 KJ $$mol^{-1}$$

D
100.0 KJ $$mol^{-1}$$

Solution

Overall energy of activation =$$\dfrac{k_{1}E_{1}+k_{2}E_{2}}{k_{1}+k_{2}}$$

$$\implies \dfrac{60 \times 1 \times 10^{-2}+70\times 3\times 10^{-2}}{4\times10^{-2}}$$

$$\implies \dfrac{270}{4}=67.5 \: KJ mol^{-1}$$
Question 5
1 / -0

Consider an endothermic reaction $$X \rightarrow Y$$ with the activation energies $$E_{b}$$ and $$E_{f}$$ for the backward and forward reactions respectively. In general:

A
$$E_{b}> E_{f}$$

B
$$E_{b}< E_{f}$$

C
$$E_{b}=E_{f}$$

D
Any of the above

Solution

Here, $$E_f>E_b$$

Option B is correct.
Question 6
1 / -0

At $$100^{0}C$$ , the gaseous reaction $$A \rightarrow 2B + C$$ is found to be of first order. Starting with pure $$A$$, if at the end of 10 min, the total pressure of the system is 140 mm and after a long time it is 300 nm, the partial pressure of $$A$$ at the end of 10 min is:

A
70 mm

B
160 mm

C
60 mm

D
80 mm

Solution

After a long time, the dissociation of $$A$$ is complete. The dissociation of one mole of $$A$$ gives 3 moles of products.

Hence, the initial pressure of $$A=\dfrac{300}{3}=100\ mm.$$

Let after 10 minutes the change in the pressure of $$A$$ is $$x\ mm$$. The pressure of $$B$$ and $$C$$ would be $$2x$$ and $$x\ mm$$ respectively.

Total pressure $$=\left ( 100-x \right )+2x+x=\left ( 100+2x \right )mm.$$

$$100+2x=140\ mm$$

$$2x=(140-100)\ mm$$

$$x=20\ mm$$

Thus the pressure of $$A$$ after 10 minutes $$=(100-x)mm=(100-20)=80mm$$.
Question 7
1 / -0

A substance $$A$$ decomposes in a solution following first-order kinetics. Flask I contains 1L of a 1M solution of $$A$$ and flask II contains 100 ml of a 0.6 M solution. After 8 hours the concentration of $$A$$ in flask I has become 0.50 M. What will be the time taken for the concentration of $$A$$ in flask II to become 0.3 M?

A
0.4

B
2.4 h

C
8.0 h

D
Can't be calculated since rate constant is not given

Solution

We know, for first order reaction

$$k=\dfrac{2.303}{8}\times log[\dfrac{A_o}{A}]$$

$$k=\dfrac{2.303}{8}\times log[\dfrac{1}{0.5}]$$

$$k= 0.086625\ hr^{-1}$$

$$k$$ will be the same for the second flask.

$$t_2=\dfrac{2.303}{0.086625}\times log\left [ \dfrac{0.6}{0.3} \right ]$$

$$t_2= 8\ hr$$
Question 8
1 / -0

The rate of a first-order reaction is $$1.5\times 10^{-2}\ mol\ L^{-1}\ min^{-1}$$ at $$0.5$$ M concentration of the reactant. The half-life of the reaction is:

A
$$23.1$$ min

B
$$8.73$$ min

C
$$7.53$$ min

D
$$0.383$$ min

Solution

We know that for a first order reaction:-
Rate$$=k[R]$$
$$\therefore 1.5\times { 10 }^{ -2 }{ M }{ min^{-1} }=k\left[ 0.5M \right] $$

$$\Rightarrow k=\cfrac { 1.5\times { 10 }^{ -2 }{ M }/{ min } }{ 0.5M } $$
$$\Rightarrow k=0.03{ min }^{ -1 }$$
We also know that, half life of a first order reaction:-
$$t{ 1 }/{ 2 }=\cfrac { \ln { \left( 2 \right) } }{ k } $$
$$=\cfrac { 0.693 }{ k } $$
$$=\cfrac { 0.693 }{ 0.03{ min }^{ -1 } } $$
$$=23.1$$ min

$$\therefore $$ Correct answer is an option $$A=23.1$$ min is the half-life of the reaction.
Question 9
1 / -0

Directions For Questions

The gas phase decomposition, $$2N_{2}O_{5}\rightarrow 4NO_{2}+O_{2}$$ follows first order rate law. At a given temperature, the rate constant of the reaction is $$23.03\times 10^{-3}s^{-1}$$. The initial pressure of $$N_{2}O_{5}$$ is $$0.09$$ atm.

...view full instructions

The total pressure after 200 seconds, if the initial pressure is $$0.1$$ atm is _______ .

A
$$0.154$$ atm

B
$$0.248$$ atm

C
$$0.174$$ atm

D
$$0.114$$ atm

Solution

After 200 sec, $$log \dfrac{P_{o}}{P}=\dfrac{kt}{2.303}=\dfrac{23.03\times 10^{-3}\times 200}{2.303}=200\times 10^{-2}=2.$$

Hence, $$\dfrac{P_{o}}{P}=100.$$

or $$P=P_{o}\times 0.01.$$

The pressure of dinitrogen pentoxide is $$P_{N_{2}O_{5}}=(P_{o}-2x)=P_{o}\times 0.01$$

Hence, $$x=\dfrac{1}{2}P_{o}(1-0.01)=0.495P_{o}=0.0495.$$

The total pressure is, $$(P_{o}+3x)=P_{o}+3\times0.0495=0.1 +0.1485 = 0.2485\;atm.$$
Question 10
1 / -0

Directions For Questions

The gas phase decomposition, $$2N_{2}O_{5}\rightarrow 4NO_{2}+O_{2}$$ follows first order rate law. At a given temperature, the rate constant of the reaction is $$23.03\times 10^{-3}s^{-1}$$. The initial pressure of $$N_{2}O_{5}$$ is $$0.09$$ atm.

...view full instructions

The time required for the decomposition of $$N_{2}O_{5}$$, so that the total pressure becomes 0.15 atm is ___________.
(Given $$log\ 1.8=0.255$$)

A
25.5 sec

B
35.5 sec

C
45.5 sec

D
55.5 sec

Solution

The reaction for the decomposition of nitrogen pentoxide is as shown below:
$$2N_{2}O_{5}\rightarrow 4NO_{2}+O_{2}$$

Initial pressures of $$N_{2}O_{5}, NO_2$$ and $$O_2$$ are $$P_{o},\;0 $$ and $$0$$ respectively. At equilibrium, the pressures of $$N_{2}O_{5},\;NO_{2}$$ and $$O_2$$ are $$(P_{o}-2x),\; 4x$$ and $$x$$ respectively.

Total pressure is $$(P_{o}-2x)+4x+x=(P_{o}+3x)$$

But, $$P_{o}=0.09\;atm.$$

Hence, $$P_{total}=0.15\;atm=0.09+3x \implies x=0.02\;atm.$$

And $$P_{N_{2}O_{5}}=(P_{o}-2x)=(0.09-0.04)=0.05\;atm.$$

The integrated rate law for the first-order reaction is

$$t =\dfrac{2.303}{k}log\dfrac{a}{(a-x)}=\dfrac{2.303}{23.03\times 10^{-3}}log (\dfrac{0.09}{0.05})=\dfrac{1}{10^{-2}}\times 0.255=25.5\;sec.$$

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Chemical Kinetics Test - 55

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Chemical Kinetics Test - 55

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Question 1

0.693 x $$10^{-2}$$ minute

100 minutes

0.693 x $$10^{-3}$$ minute

6.932 minutes

Solution

Question 2

The time taken for the completion of 50% of the reaction is $$t_{1/2}$$.

A plot of the reciprocal of the concentration of the reactants against time gives a straight line.

The degree of dissociation is equal to $$1-e^{-kt}$$.

A plot of $$[A]_{0}/[A]$$ versus time gives a straight line.

Solution

Question 3

$$1.09\times 10^{-6}$$

$$1.09\times 10^{-5}$$

$$1.09\times 10^{-7}$$

$$1.09\times 10^{5}$$

Solution

Question 4

130.0 KJ $$mol^{-1}$$

65.0 KJ $$mol^{-1}$$

67.5 KJ $$mol^{-1}$$

100.0 KJ $$mol^{-1}$$

Solution

Question 5

$$E_{b}> E_{f}$$

$$E_{b}< E_{f}$$

$$E_{b}=E_{f}$$

Any of the above

Solution

Question 6

70 mm

160 mm

60 mm

80 mm

Solution

Question 7

0.4

2.4 h

8.0 h

Can't be calculated since rate constant is not given

Solution

Question 8

$$23.1$$ min

$$8.73$$ min

$$7.53$$ min

$$0.383$$ min

Solution

Question 9

$$0.154$$ atm

$$0.248$$ atm

$$0.174$$ atm

$$0.114$$ atm

Solution

Question 10

25.5 sec

35.5 sec

45.5 sec

55.5 sec

Solution

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