Chemical Kinetics Test

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Chemical Kinetics Test - 41

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Weekly Quiz Competition

Question 1
1 / -0

For a first order reaction, the ratio of the time taken for $$7/8^{th}$$ of the reaction to complete to that of half of the reaction to complete is :

A
$$3 : 1$$

B
$$1 : 3$$

C
$$2 : 3$$

D
$$3 : 2$$

Solution

$$k = \dfrac {2.303}{t}\log \dfrac {a}{a - x}$$

For $$7/8$$ of the reaction to complete, $$t = t_{7/8}$$,

$$a - x = a - 7a/8 = a/8$$

$$\therefore t_{7/8} = \dfrac {2.303}{k}\log \dfrac {a}{a/8} = \dfrac {2.303}{k}\log 8$$

For half of the reaction to complete, $$t = t_{1/2}$$,

$$x = a - a/2 = a/2$$

$$\therefore t_{1/2} = \dfrac {2.303}{k}\log \dfrac {a}{a/2} = \dfrac {2.303}{k}\log 2$$

$$\therefore \dfrac {t_{7/8}}{t_{1/2}} = \dfrac {\log 8}{\log 2} = \dfrac {\log 2^{3}}{\log 2} = \dfrac {3\log 2}{\log 2} = 3:1$$.

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

A first order reaction is $$50$$% complete in $$30$$ minutes at $$27^{\circ}C$$ and in $$10$$ minutes at $$47^{\circ}C$$. The reaction rate constant at $$27^{\circ}C$$ and the energy of activation of the reaction are respectively.

A
$$k = 0.0231\ min^{-1}, E_{a} = 43.848\ kJ\ mol^{-1}$$

B
$$k = 0.017\ min^{-1}, E_{a} = 52.54\ kJ\ mol^{-1}$$

C
$$k = 0.0693\ min^{-1}, E_{a} = 43.848\ kJ\ mol^{-1}$$

D
$$k = 0.0231\ min^{-1}, E_{a} = 28.92\ kJ\ mol^{-1}$$

Solution

Since $$t_{1/2} = \dfrac {0.693}{k},$$

$$ \therefore k = \dfrac {0.693}{t_{1/2}}$$

Given: $$t_{1/2} = 30\ min$$ at $$27^{\circ}C$$ and $$t_{1/2} = 10\ min$$ at $$47^{\circ}C$$

$$\therefore k_{27^{\circ}C} = \dfrac {0.693}{30}min^{-1} = 0.0231\ min^{-1}$$

and $$k_{47^{\circ}C} = \dfrac {0.693}{10} = 0.0693\ min^{-1}$$

We know that $$\log \dfrac {k_{47^{\circ}C}}{k_{27^{\circ}C}} = \dfrac {E_{a}}{2.303R}\times \dfrac {(T_{2} - T_{1})}{T_{2}\times T_{1}}$$

$$\therefore E_{a} = \dfrac {2.303R\times T_{1}\times T_{2}}{(T_{2} - T_{1})} \log \dfrac {k_{47^{\circ}C}}{k_{27^{\circ}C}}$$

or $$E_{a} = \dfrac {2.303\times 8.314\times 10^{-3} \times 300\times 320}{(320 - 300)} \log \dfrac {0.0693}{0.0231}$$

$$= 43.848\ kJ\ mol^{-1}$$.

Hence, the correct answer is option $$\text{A}$$.
Question 3
1 / -0

Threshold energy is equal to:

A
Activation energy

B
Activation energy $$-$$ energy of molecules

C
Activation energy $$+$$ energy of molecules

D
None of these

Solution

According to the collision theory, only a small fraction of collisions is effective in bringing about the chemical reaction and the rest of the collisions are ineffective.

For effective collision (to yield product), the colliding molecules must have more than or equal to a certain minimum amount of energy called threshold energy.

Threshold energy $$=$$ Activation energy $$+$$ Average kinetic energy of the molecules.

Hence, the correct answer is option $$\text{C}$$.
Question 4
1 / -0

In the reaction, $$2NO+Cl_2\rightarrow 2NOCl$$, it has been found that doubling the concentration of both the reactants increases the rate by a factor of eight but doubling the chlorine concentration alone only doubles the rate. Which of the following statements is incorrect?

A
The reaction is first order in $$Cl_2$$

B
The reaction is second order in $$NO$$

C
The overall order of reaction is $$2$$

D
The overall order of reaction is $$3$$

Solution

Let the rate of the reaction be given by
rate=$$k[NO]^x[Cl_2]^y$$
By given conditions, 8x rate=$$k[2NO]^x[2Cl_2]^y$$
$$8=26{x+y}$$ or x+y=3
Also, 2x rate=$$k[NO]^x[2Cl_2]^y$$
$$2=2^y$$ or $$y=1$$
x=2
rate=$$k[NO]^2[Cl_2]$$
Overall order=3
Question 5
1 / -0

The rate constant is given by the equation $$k = P.Ze^{-E_{a}/RT}$$. Which factor should register $$a$$ decrease for the reaction to proceed more rapidly:

A
$$T$$

B
$$Z$$

C
$$E$$

D
$$P$$

Solution

$$K=P.ze^{-E_a/RT}$$
$$\Rightarrow log \ k= log (Pze^{-E_a/RT})$$
$$\Rightarrow log \ k= log \ (Pz) + log \ (e^{-E_a/RT})$$
$$\Rightarrow log \ k= log \ (Pz) - \cfrac {E_a}{RT} (log \ e)$$
For $$k$$ to increase, $$\cfrac {-E_a}{RT}$$ shold be small and that can happen only if $$E_a$$ is less or $$T$$ is more. So, reaction proceeds rapidly if and only if $$E_a$$ decreases.
Question 6
1 / -0

Fill up the following with suitable terms.

(i) Activation energy $$=$$ Threshold energy $$-$$ _______.

(ii) Half-life period of zero order reaction $$=$$ ________.

(iii) Average rate of reaction $$=$$ _______.

(iv) Instantaneous rate of reaction $$=$$ ______.

A
Potential energy, $$\dfrac {0.693}{k}, \dfrac {dx}{dt}, \dfrac {\triangle [A]}{\triangle t}$$

B
Energy of reactants, $$\dfrac {1}{k}, \dfrac {\triangle [A]}{\triangle t}, \dfrac {dx}{dt}$$

C
Energy of reaction, $$\dfrac {\log k}{t}, \dfrac {\triangle [A]}{\triangle t}, \dfrac {dx}{dt}$$

D
Average kinetic energy of reactants, $$\dfrac {a}{2k}, \dfrac {\triangle [A]}{\triangle t}, \dfrac {dx}{dt}$$

Solution

a) Activation energy = Threshold energy$$-$$ the Average kinetic energy of molecules(by definition)

b) half-life of a zero order reaction = $${ t }_{ \frac { 1 }{ 2 } } =\frac { a }{ 2k } $$

c) Average rate of reaction = $$\dfrac {\displaystyle \text{ Total change concentration reactants }}{ \displaystyle \text{Total time taken} }$$ = $$\dfrac { \triangle \left[ A \right] }{ \triangle t }$$

d) Instantaneous rate of reaction =$$ \dfrac { \displaystyle \text {Instantaneous change in concentration} }{\displaystyle \text { Instantaneous time} } $$=$$\dfrac { dx }{ dt } $$
Question 7
1 / -0

The rate of a particular reaction doubles when temperature change from $${ 27 }^{ 0 }$$C to $${ 37 }^{ 0 }$$C. Calculate the energy of the activation of such reaction.

A
53582 KJ

B
53.58 KJ

C
29.86 KJ

D
none of these

Solution

We are given that:
When $$T_1 = 27 + 273 = 300 K$$
Let $$K_1 = k$$
When $$T_2 = 37 + 273 = 310 K$$
$$K_2 = 2 k$$
Substituting these values the equation:
$$ log\dfrac {K_2}{ K_1}$$
= $$\dfrac{Ea}{ 2.303 R}\times \dfrac{(T_2 – T_1)}{ T_1 T_2}$$

We will get:
$$log\dfrac {2 k }{ k}$$ $$= \dfrac{E_a}{ 2.303 \times 8.314}\times\dfrac {310 – 300} { 300 \times310}$$

$$log_ 2 = \dfrac{E_a} {2.303 \times 8.314} \times\dfrac {10}{ 300 \times 310}$$

$$E_a = 53598.6 J mol^{-1}$$
$$E_a = 53.6 kJ mol^{-1}$$
Hence, the energy of activation of the reaction is $$53.6 kJ mol^{-1}$$
Question 8
1 / -0

For an elementary bi-molecular reaction of the type $$A + B \rightarrow C$$ activation energy is equal to 20 Kcal. If specific rate of reaction at 500 K is 4.2 $$\times$$ $$10^{-3}M^{-1}sec^{-1}$$, then identify the incorrect option(s). [Given : ln ($$2.1 \times$$ $$10^{-9}$$) = - 20]

A
% of activated molecules at 500 K is $$2.1 \times$$ $$10^{-7}$$%

B
Maximum possible rate of reaction when $$[A]$$ and $$[B]$$ are both equal to 1 is $$2 \times$$ $$10^6 M sec^{-1}$$

C
% of activated molecules at 300 K is $$2 \times 10^{-6}$$%

D
At some temperature rate of reaction can be $$4 \times 10^3 M sec^{-1}$$ if $$[A] =\ 2M,\ [B] = 0.1 M$$
Question 9
1 / -0

The activation energy for the forward reaction X Y is $$60$$ KJ $$mol^{-1}$$ and $$\Delta H$$ is $$-20$$ KJ $$mol^{-1}$$. The activation energy for the backward reaction Y X is:

A
$$80$$ kJ $$mol^{-1}$$

B
$$40$$ kJ $$mol^{-1}$$

C
$$60$$ kJ $$mol^{-1}$$

D
$$20$$ kJ $$mol^{-1}$$

Solution

We know that,

$$\Delta H = E_f - E_b$$

$$-20=60 - E_b$$

$$E_b=(60+20) KJ/mol$$

$$=80KJmol^{-1}$$
Question 10
1 / -0

The half-life of a first-order reaction is $$100$$ seconds. What is the time required in seconds for $$90\%$$ completion of the reaction?

A
$$100$$

B
$$200$$

C
$$333$$

D
$$500$$

Solution

Half life time $$\left( {\dfrac{{{t_1}}}{2}}\right) = $$ 100 sec. (given)
$$K = \dfrac{{0.693}}{{100}}$$
Time required for 90% completion?
Here let initial concentration $$= a$$
Concentration decomposed is 90%
$$ x = \dfrac{{90 \times a}}{{100}}$$
$$ = 0.9a$$
Now,
$$t = \dfrac{{2.303}}{{\dfrac{{0.693}}{{{t_1}/2}}}}\log\dfrac{a}{{a - x}}\, \Rightarrow \,\dfrac{{2.303}}{{100}}\log \dfrac{a}{{a -0.9a}}$$
$$ \Rightarrow \,\,\dfrac{{2.303}}{{0.693}}\log\dfrac{a}{{0.1a}}\,\, \Rightarrow \,\dfrac{{2.303}}{{100}}\log 10$$
$$ \Rightarrow \,\dfrac{{2.303}}{{0.693}}\times 100 \times \log 10$$
$$ \Rightarrow \,339.32\,\sec $$
Approx. $$333$$ seconds

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Chemical Kinetics Test - 41

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Chemical Kinetics Test - 41

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Question 1

$$3 : 1$$

$$1 : 3$$

$$2 : 3$$

$$3 : 2$$

Solution

Question 2

$$k = 0.0231\ min^{-1}, E_{a} = 43.848\ kJ\ mol^{-1}$$

$$k = 0.017\ min^{-1}, E_{a} = 52.54\ kJ\ mol^{-1}$$

$$k = 0.0693\ min^{-1}, E_{a} = 43.848\ kJ\ mol^{-1}$$

$$k = 0.0231\ min^{-1}, E_{a} = 28.92\ kJ\ mol^{-1}$$

Solution

Question 3

Activation energy

Activation energy $$-$$ energy of molecules

Activation energy $$+$$ energy of molecules

None of these

Solution

Question 4

The reaction is first order in $$Cl_2$$

The reaction is second order in $$NO$$

The overall order of reaction is $$2$$

The overall order of reaction is $$3$$

Solution

Question 5

$$T$$

$$Z$$

$$E$$

$$P$$

Solution

Question 6

Potential energy, $$\dfrac {0.693}{k}, \dfrac {dx}{dt}, \dfrac {\triangle [A]}{\triangle t}$$

Energy of reactants, $$\dfrac {1}{k}, \dfrac {\triangle [A]}{\triangle t}, \dfrac {dx}{dt}$$

Energy of reaction, $$\dfrac {\log k}{t}, \dfrac {\triangle [A]}{\triangle t}, \dfrac {dx}{dt}$$

Average kinetic energy of reactants, $$\dfrac {a}{2k}, \dfrac {\triangle [A]}{\triangle t}, \dfrac {dx}{dt}$$

Solution

Question 7

53582 KJ

53.58 KJ

29.86 KJ

none of these

Solution

Question 8

% of activated molecules at 500 K is $$2.1 \times$$ $$10^{-7}$$%

Maximum possible rate of reaction when $$[A]$$ and $$[B]$$ are both equal to 1 is $$2 \times$$ $$10^6 M sec^{-1}$$

% of activated molecules at 300 K is $$2 \times 10^{-6}$$%

At some temperature rate of reaction can be $$4 \times 10^3 M sec^{-1}$$ if $$[A] =\ 2M,\ [B] = 0.1 M$$

Question 9

$$80$$ kJ $$mol^{-1}$$

$$40$$ kJ $$mol^{-1}$$

$$60$$ kJ $$mol^{-1}$$

$$20$$ kJ $$mol^{-1}$$

Solution

Question 10

$$100$$

$$200$$

$$333$$

$$500$$

Solution

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