Chemical Kinetics Test

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Chemical Kinetics Test - 33

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Weekly Quiz Competition

Question 1
1 / -0

The unit of rate constant for zero order reaction is:

A
$$s^{-1}$$

B
$$mol\, L^{-1}s^{-1}$$

C
$$L\, mol^{-1}s^{-1}$$

D
$$L^2mol^{-2}s^{-1}$$

Solution

$$Rate=\dfrac{dC}{dt} = k[R_0]^o = k$$ or $$k=\dfrac{dC}{dt} = \dfrac{conc.}{time}$$

$$=\dfrac{mol\,L^{-1}}{s} = mol\,L^{-1}s^{-1}$$.
Question 2
1 / -0

Which graph shows zero activation energy for reaction?

A

B

C

D

Solution

The graph for zero activation energy is plot A. Because, in plot A, the difference between reactants energy and energy of the transition state is zero.

While in plot C, the energy of reactants is much lower than the transition state energy, which tells that the difference between the two is the activation energy.

Plots B and D are not possible when we plot for energy vs reaction progress.

So, the correct answer is A.
Question 3
1 / -0

For a zero-order reaction with a rate constant k, the slope of the plot of reactant concentration against time is:

A
$$\dfrac{ k} {2.303}$$

B
$$k$$

C
$$-\dfrac {k}{2.303}$$

D
$$-k$$

Solution

For a Zero order reaction,

$$A_t=A_o-kt$$

Slope$$=-k$$

Hence option $$D$$ is correct.
Question 4
1 / -0

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

A
$${ t }_{ \frac { 1 }{ 2 } }={ \left[ A \right] }_{ 0 }2k$$

B
$${ t }_{ \frac { 1 }{ 2 } }=\dfrac { 0.693 }{ k } $$

C
$${ t }_{ \frac { 1 }{ 2 } }=\dfrac { { \left[ A \right] }_{ 0 } }{ 2k } $$

D
$${ t }_{ \frac { 1 }{ 2 } }=\dfrac { 2{ \left[ A \right] }_{ 0 } }{ k } $$

Solution

$${ t }_{ \frac { 1 }{ 2 } }=\frac { { [A] }_{ 0 } }{ 2k } $$
Where $${ t }_{ \frac { 1 }{ 2 } }$$ is half life
$$[A]_0$$ is initial concentration and k is rate constant.
Question 5
1 / -0

The order of reaction for which half-life period is independent of initial concentration is:

A
zero

B
first

C
second

D
third

Solution

For first order reaction, the half-life period is independent of initial concentration.

For first order reaction, the half-life period expression $$\displaystyle (t_{1/2})$$ is given by the expression $$\displaystyle t_{1/2} = \dfrac {0.693}{k}$$, where k is the rate constant.

For nth order reaction, $$\displaystyle t_{1/2} \propto \dfrac {1}{a^{n-1}}$$.
Question 6
1 / -0

The half-life of $$^{14}C$$ is 5570 yr. How many years will it take 90% of a sample to decompose?

A
5,570 yr

B
17,700 yr

C
18,510 yr

D
50,100 yr

Solution

The half-life of $$\displaystyle ^{14}C$$ is $$5570 yr.\ $$

$$\displaystyle t_{1/2} = 5570$$ years

The decay constant $$\displaystyle \lambda = \dfrac {0.693}{t_{1/2}}$$

$$\displaystyle \lambda = \dfrac {0.693}{ 5570 \: yr}$$

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

$$\displaystyle t = \dfrac {2.303}{\dfrac {0.693}{ 5570 \: yr}} log\left( \dfrac {100}{100-90}\right)$$

$$\displaystyle t=18,510 \: yr$$

So, the correct option is $$C$$
Question 7
1 / -0

Collision theory is applicable to

A
First order reactions

B
Zero order reactions

C
Bimolecular reactions

D
Intramolecular reactions

Solution

Collision theory is applicable to Bi-molecular reactions and reactions with molecularity greater than two. It is not applicable to uni-molecular reactions. The basic requirement of collision theory is that the reacting species (atoms, ions or molecules) must come together and collide in order for the reaction to occur. Collisions are possible in bi-molecular reactions.
Question 8
1 / -0

$$3A\longrightarrow B+C$$

It would be a zero-order reaction when:

A
The rate of reaction is proportional to square of concentration of $$A$$

B
The rate of reaction remains the same at any concentration of $$A$$

C
The rate remains unchanged at any concentration of $$B$$ and $$C$$

D
The rate of reaction doubles if concentration of $$B$$ is increased to double

Solution

For reaction, $$3A\longrightarrow B+C$$

If it is zero order reaction, therefore the rate remains same at any concentration of '$$A$$' or $$\dfrac { dx }{ dt } =k {[ A }]^{ 0 } $$.

It means that the rate is independent of the concentration of reactants.

Hence, option B is correct.
Question 9
1 / -0

A certain radioactive isotope decay has $$\alpha$$-emission,
$$ _{ { Z }_{ 1 } }^{ { A }_{ 1 } }{ X }\longrightarrow $$ $$_{ { Z }_{ 1 }-2 }^{ { A }_{ 1 }-4 }{ Y }$$
half life of $$X$$ is $$10$$ days. If $$1$$ $$mol$$ of $$X$$ is taken initially in a sealed container, then what volume of helium will be collected at STP after $$20$$ days?

A
$$22.4 L$$

B
$$11.2 L$$

C
$$16.8 L$$

D
$$33.6 L$$

Solution

$$_{Z_1}^{A_1}X\rightarrow _{Z_1-2}^{A_1-4}Y+_2^4He$$
$$1\quad\quad 0\quad\quad 0$$ .......initially
$$0.25\quad\quad 0.75\quad\quad 0.75$$.... after two half lives(20 days)
After $$20$$ days $$0.75$$ $$mol$$ helium will be formed.
$$\therefore $$ Volume of helium at STP $$=0.75\times 22.4=16.8\ L$$
Question 10
1 / -0

One gram of $$^{ 226 }{ Ra }$$ has an activity of nearly $$1 Ci$$. The half life of $$^{ 226 }{ Ra }$$ is

A
$$1582$$ yrs

B
$$12.5$$ hrs

C
$$140$$ days

D
$$4.5\times { 10 }^{ 9 }$$ yrs

Solution

Use the following relation for calculation of activity:
Activity $$=\dfrac { 0.693 }{ { t }_{ { 1 }/{ 2 } } } \times \dfrac { w }{ At.wt } \times 6.023\times { 10 }^{ 23 }$$
$$3.7\times { 10 }^{ 10 }=\dfrac { 0.693 }{ { t }_{ { 1 }/{ 2 } } } \times \dfrac { 1 }{ 226 } \times 6.023\times { 10 }^{ 23 }$$=$$0.5\times 10^{8}\ sec$$.
In years, it will be $$\dfrac{0.5\times 10^{8}}{86400\times 365}=1582\ years$$