Chemical Kinetics Test

Result

Chemical Kinetics Test - 14

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

The rate constant of the reaction $$A\rightarrow B$$ is $$0.6\times 10^{-3} \ mole \ per \ litre \ per \ second$$. If the concentration of $$A$$ is $$5 \ M$$, then concentration of $$B$$ after $$20 \ minutes$$ is:

A
$$0.36 M$$

B
$$0.72 M$$

C
$$1.08 M$$

D
$$3.60 M$$

Solution

The rate constant has unit mole per second which indicates zero order reaction.
x = kt
The concentration of $$B$$ after 20 minutes is $$\displaystyle 20 \times 60 \times 0.6\times 10^{-3} = 0.72 \: M$$
Question 2
1 / -0

The thermal decomposition of $$HCOOH$$ is a first-order reaction with a rate constant of $$2.4\times {10}^{-3}{s}^{-1}$$ at certain temperature. How long will it take for three-fourths of the initial quantity of $$HCOOH$$ to decompose?

A
$$578\ sec$$

B
$$225\ sec$$

C
$$436\ sec$$

D
$$57.8\ sec$$

Solution

Half-life for a first order reaction is $$\dfrac {0.693}{k}$$.

$${t}_{1/2}=\cfrac{0.693}{k}\cfrac{0.693}{2.4\times {10}^{-3}{S}^{-1}}=288.75\ sec$$

$${t}_{3/4}=2\times {t}_{1/2}=577.50s=578\ sec$$

Hence, option A is correct.
Question 3
1 / -0

In a homogeneous reaction $$A\longrightarrow B+C+D$$ the initial pressure was $${P}_{0}$$ and after time $$t$$ it was $$P$$. Expression for rate constant $$k$$ in terms of $${P}_{0},$$ $$P$$ and $$t$$ will be:

A
$$k=\cfrac { 2.303 }{ t } \log { \cfrac { 2{ P }_{ 0 } }{ { 3P }_{ 0 }-P } } $$

B
$$k=\cfrac { 2.303 }{ t } \log { \cfrac { 2{ P }_{ 0 } }{ { P }_{ 0 }-P } } $$

C
$$k=\cfrac { 2.303 }{ t } \log { \cfrac { { 3P }_{ 0 }-P }{ 2{ P }_{ 0 } } } $$

D
$$k=\cfrac { 2.303 }{ t } \log { \cfrac { 2{ P }_{ 0 } }{ { 3P }_{ 0 }-2P } } $$

Solution

$$A\longrightarrow B + C + D$$
Initial $$a$$ $$0$$ $$0$$ $$0$$
After time $$t$$ $$a-x$$ $$x$$ $$x$$ $$x$$
It is given time $$a={P}_{0}...................(i)$$
$$a-x+x+x+x=P.....(ii)$$
From $$(i)$$
$${P}_{0}+2x=P$$ or $$x=\cfrac{P-{P}_{0}}{2}$$
From rate equation
$$k=\cfrac{2.303}{t}\log{\cfrac{a}{a-x}}$$
$$k=\cfrac { 2.303 }{ t } \log { \cfrac { 2{ P }_{ 0 } }{ { 3P }_{ 0 }-P } } $$
Question 4
1 / -0

What is the activation energy (KJ/mol) for a reaction if its rate constant doubles when the temperature is raised from $$300$$K to $$400$$K? $$(R=8.314$$ J $$mol^{-1}K^{-1}$$)

A
$$68.8$$

B
$$6.88$$

C
$$34.4$$

D
$$3.44$$

Solution

$$log\dfrac{K_2}{K_1}=\dfrac{E_a}{2.3R}\left(\dfrac{1}{T_1}-\dfrac{1}{T_2}\right)$$
$$log\dfrac{2}{1}=\dfrac{E_a}{2.33\times 8.31}\left(\dfrac{1}{300}-\dfrac{1}{400}\right)$$
$$=0.3\times 2.3\times 8.31\times 3\times 400=E_a$$
$$E_a=6.88$$kJ.
Question 5
1 / -0

For a reaction $$X\longrightarrow\ Y$$, the graph of the product concentration (x) versus time (t) came out to be straight line passing through the origin. Hence the graph of $$\cfrac{-d[X]}{dt}$$ and time would be:

A
straight line with a negative slope and an intercept on y-axis

B
straight line with positive slope and an intercept on y-axis

C
a straight line parallel to x-axis

D
a hyperbola

Solution

If the product concentration is $$x$$
For a zero-order reaction $$\cfrac{x}{t}=k$$
Thus, the graph would be a straight line passing through the origin. So the given information is for zero order reaction. for zero order reaction, the rate of reaction is constant. Thus, the plot of rate vs time ie., $$-\cfrac{d[X]}{dt}$$ vs time will be a straight line parallel to the x-axis
Question 6
1 / -0

For a first order gas phase reaction:

$$A_{(g)} \rightarrow 2B_{(g)} + C_{(g)}$$.

$$P_{0}$$ be initial pressure of $$A$$ and $$P$$, the total pressure at time '$$t'$$. Integrated rate equation is:

A
$$\dfrac {2.303}{t}\log \left (\dfrac {P_{0}}{P_{0} - P_{t}}\right )$$

B
$$\dfrac {2.303}{t}\log \left (\dfrac {2P_{0}}{3P_{0} - P_{t}}\right )$$

C
$$\dfrac {2.303}{t}\log \left (\dfrac {P_{0}}{2P_{0} - P_{t}}\right )$$

D
$$\dfrac {2.303}{t}\log \left (\dfrac {2P_{0}}{2P_{0} - P_{t}}\right )$$

Solution

The given reaction is :

$$A_{(g)} \rightarrow 2B_{(g)} + C_{(g)}$$

$$A$$ $$B$$ $$C$$
Initial $$P_0$$ $$0$$ $$0$$
Final $$P_0-P$$
$$2P$$ $$P$$
Total pressure at time $$(t) = P_{0} - P + 2P + P = P_{t}$$

$$\Rightarrow P_{t} = P_{0} + 2P$$

$$P_{t} - P_{0} = 2P \Rightarrow P = \dfrac {P_{t} - P_{0}}{2}$$

$$k = \dfrac {2.303}{t} \log \left [\dfrac {P_{0}}{P_{0} - P}\right ] = \dfrac {2.303}{t} \log \left [\dfrac {P_{0}}{P_{0} -\left (\dfrac {P_{t} - P_{0}}{2}\right )}\right ]$$

$$= \dfrac {2.303}{t}\log \left (\dfrac {2P_{0}}{2P_{0} - P_{t} + P_{0}}\right ) = \dfrac {2.303}{t} \log \left (\dfrac {2P_{0}}{3P_{0} - P_{t}}\right )$$

Hence, option $$B$$ is correct.
Question 7
1 / -0

Consider the two equations at a particular temperature ;
$$2N_2O_5\rightarrow 4NO_2+O_2$$

$$N_2O_5\rightarrow 2NO_2+\frac{1}{2}O_2$$

If $$E_1$$ and $$E_2$$ represents the activation energy for the first and second reaction respectively then :

A
$$E_1 > E_2$$

B
$$E_1 < E_2$$

C
$$E_1 = 2E_2$$

D
$$E_1 = E_2$$

Solution

Activation Energy ($$E_a$$) is defined as the least possible amount of energy (minimum) which is required to start a reaction or the amount of energy available in a chemical system for a reaction to take place.

For a chemical reaction, activation energy remains the same, not dependent on a number of molecules undergoing the reaction.

It is changed by the presence of catalyst. In the given reactions, only the number of molecules has changed. Thus the activation energy will remain the same.

So, the correct answer is option D.
Question 8
1 / -0

In a first order reaction with time the concentration of the reactant decreases:

A
Linearly

B
Exponentially

C
No change

D
None of these

Solution

For first order reaction $$\left[ A \right] =\left[ { A }_{ 0 } \right] { e }^{ -kt }\quad $$

$$\therefore$$ The concentration of reactants will exponentially decrease with time.
Question 9
1 / -0

On increasing the temperature by $$10^{0}$$C:

A
number of collisions get doubled

B
value of rate constant does not change

C
energy of activation increases

D
number of fruitful collisions gets doubled

Solution

Increasing the temperature of the substance, increases the fraction of molecules which collide with energies greater than $$Ea$$. For every $$10^0 C$$ rise in temperature, the fraction of molecules having energy equal to or greater than Ea gets doubled leading to doubling the rate of reaction.
Hence, the number of fruitful collisions gets doubled.
Question 10
1 / -0

The activation energy of a reaction is dependent on:

A
temperature

B
pressure

C
concentration

D
nature of reactants

Solution

Activation energy is the minimum amount of energy needed for a reaction to take place.

It depends only on the nature of reactants. It is independent of temperature and concentration.

Hence, option D is correct.

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

	$$A$$	$$B$$	$$C$$
Initial	$$P_0$$	$$0$$	$$0$$
Final	$$P_0-P$$	$$2P$$	$$P$$

Chemical Kinetics Test - 14

Result

Chemical Kinetics Test - 14

Score

-

Rank

-

SHARING IS CARING

Question 1

$$0.36 M$$

$$0.72 M$$

$$1.08 M$$

$$3.60 M$$

Solution

Question 2

$$578\ sec$$

$$225\ sec$$

$$436\ sec$$

$$57.8\ sec$$

Solution

Question 3

$$k=\cfrac { 2.303 }{ t } \log { \cfrac { 2{ P }_{ 0 } }{ { 3P }_{ 0 }-P } } $$

$$k=\cfrac { 2.303 }{ t } \log { \cfrac { 2{ P }_{ 0 } }{ { P }_{ 0 }-P } } $$

$$k=\cfrac { 2.303 }{ t } \log { \cfrac { { 3P }_{ 0 }-P }{ 2{ P }_{ 0 } } } $$

$$k=\cfrac { 2.303 }{ t } \log { \cfrac { 2{ P }_{ 0 } }{ { 3P }_{ 0 }-2P } } $$

Solution

Question 4

$$68.8$$

$$6.88$$

$$34.4$$

$$3.44$$

Solution

Question 5

straight line with a negative slope and an intercept on y-axis

straight line with positive slope and an intercept on y-axis

a straight line parallel to x-axis

a hyperbola

Solution

Question 6

$$\dfrac {2.303}{t}\log \left (\dfrac {P_{0}}{P_{0} - P_{t}}\right )$$

$$\dfrac {2.303}{t}\log \left (\dfrac {2P_{0}}{3P_{0} - P_{t}}\right )$$

$$\dfrac {2.303}{t}\log \left (\dfrac {P_{0}}{2P_{0} - P_{t}}\right )$$

$$\dfrac {2.303}{t}\log \left (\dfrac {2P_{0}}{2P_{0} - P_{t}}\right )$$

Solution

Question 7

$$E_1 > E_2$$

$$E_1 < E_2$$

$$E_1 = 2E_2$$

$$E_1 = E_2$$

Solution

Question 8

Linearly

Exponentially

No change

None of these

Solution

Question 9

number of collisions get doubled

value of rate constant does not change

energy of activation increases

number of fruitful collisions gets doubled

Solution

Question 10

temperature

pressure

concentration

nature of reactants

Solution

User

Question Analysis

My Perfomance

Score

-

Rank

-

Results Announcement on: coming Monday

Stay Tuned: We’ll update you on your result via email or SMS.