Chemical Kinetics Test

Result

Chemical Kinetics Test - 78

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Question 1
1 / -0

The values of enthalpies of reactants and products are $$x$$ and $$y$$ $$J/mol$$, respectively. If the activation energy for the backward reaction is $$z$$ $$J/mol$$, then the activation energy for forward reaction will be
(in J/mol)

A
$$x-y-z$$

B
$$x-y+z$$

C
$$y-x-z$$

D
$$y-x+z$$

Solution
Question 2
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The initial rate of a zero-order reaction:
$$A(g)\rightarrow 2B(g)$$ is $$0.01M\quad { min }^{ -1 }$$. If the initial concentration of $$A$$ is $$0.1M$$, the concentration of $$B$$ after $$60s$$ is?

A
$$0.09M$$

B
$$0.01M$$

C
$$0.02M$$

D
$$1.2M$$

Solution
Question 3
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Which of the following is incorrect statement?

A
Stoichiometry of a reaction tells about the order of the elementary reactions

B
For a zero-order reaction, rate and the rate constant are identical

C
A zero-order reaction is controlled by factors other than the concentration of reactants

D
A zero-order reaction is always elementary reaction

Solution
Question 4
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A zero-order reaction $$A\rightarrow B$$. At the end of $$1h$$, $$A$$ is $$75$$% reacted. How much of it will be left unreacted at the end of $$2h$$?

A
$$12.5$$%

B
$$6.25$$%

C
$$3.12$$%

D
$$0$$%
Question 5
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At a certain temperature, the reaction between $$NO$$ and $${O}_{2}$$ is fast, while that between $$CO$$ and $${O}_{2}$$ is slow. It may be concluded that

A
$$NO$$ is more reactive than $$CO$$

B
$$CO$$ is smaller in size than $$NO$$

C
activation energy for the reaction:
$$2NO+{ O }_{ 2 }\rightarrow 2{ NO }_{ 2 }$$ is less

D
activation energy for the reaction:
$$2NO+{ O }_{ 2 }\rightarrow 2{ NO }_{ 2 }$$ is high
Question 6
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In general, the rate of a reaction can be increased by all the factors except

A
increasing the temperature

B
increasing the concentration of reactants

C
increasing the activation energy

D
using a catalyst

Solution
Question 7
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For a zero-order reaction,

A
the reaction rate is doubled when the initial concentration is doubled

B
the time for half change is half the time taken for completion of the reaction

C
the time for half change is independent of the initial concentration

D
the time for completion of the reaction is independent of the initial concentration

Solution
Question 8
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Directions For Questions

Three flasks A, B, and C have equal volumes. A is maintained at $$300$$K, B at $$600$$K, and C at $$900$$K. Flask A contains $$H_2$$ gas, flask B contains $$CH_4$$ gas and flask C contains $$O_2$$ gas, and masses of all gases are the same. Assume ideal behavior of gases.

...view full instructions

Flask with greater collision number of the molecules.

A
A

B
B

C
C

D
Same in all

Solution

Flask with greater collision number of molecules

$$Z_A=\dfrac{u_{av}}{\lambda}$$

$$Z_A=\dfrac{(u_{av})_A}{\lambda_A}$$

$$Z_B=\dfrac{(u_{av})_B}{\lambda_B}$$

$$\dfrac{Z_A}{Z_B}=\dfrac{(u_{av})_A}{(u_{av})_B}\times \dfrac{\lambda_B}{\lambda_A}=2\times 2=4$$

Similarly $$Z_B > Z_C$$

$$\therefore Z_A > Z_B > Z_C$$

So $$Z_A$$ is greatest
Option A is the correct answer.
Question 9
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The time taken in $$75$$% completion of zero-order reaction is $$10h$$. In what time, the reaction will be $$90$$% completed?

A
$$12.0h$$

B
$$16.6h$$

C
$$10.0h$$

D
$$20.0h$$

Solution

$$75 %$$ completed reaction means
Left amount $$A=25$$
Initial amount $$A_∘ = 100$$ and the reaction is completed in $$ t=10\ hrs$$

Similarly for $$ 90%,$$
Left amount $$A = 10$$
Initial amount $$ A_∘ = 100$$
$$A = A_∘−kt$$
$$25=100−k×10$$
$$k×10=75$$
$$k=7.5mol \ l^{−1}s^{−1}$$

$$A=A_∘−kt$$
$$10=100−7.5×t$$
$$7.5t=90$$
$$t=12.00 hrs$$

Hence, option is $$(A)$$ is correct answer.
Question 10
1 / -0

The rate expression for a reaction is $$\cfrac { -dC }{ dt } =\cfrac { \alpha C }{ 1+\beta C } $$, where $$\alpha $$ and $$\beta$$ are constants and $$C$$ is the concentration of reactant at time, $$t$$. The half-life for this reaction is

A
$$\cfrac { 1 }{ \alpha } \ln { 2 } +\cfrac { \beta { C }_{ o } }{ 2\alpha } \quad $$

B
$$\cfrac { 1 }{ \beta } \ln { 2 } +\cfrac { \beta { C }_{ o } }{ 2\alpha } $$

C
$$\cfrac { \beta \ln { 2 } }{ \alpha } $$

D
$$\cfrac { \alpha }{ \beta } $$

Solution