Chemical Kinetics Test

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Chemical Kinetics Test - 63

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Question 1
1 / -0

A vessel contains dimethyl ether at a pressure of 0.4 atm. Dimethyl ether decomposes as $$CH_{3}OCH_{3}(g)\rightarrow CH_{4}(g)+CO(g)+H_{2}(g)$$. The rate constant of decomposition is $$4.78\times10^{-3}\:min^{-1}$$. Calculate the ratio to initial rate of diffusion of rate of diffusion after 4.5 hour of initiation of decomposition. Assume the composition of gas present and composition of gas diffusing to be same.

A
$$0.26 : 1$$

B
$$0.13 : 1$$

C
$$1: 0.26$$

D
$$1 : 0.13$$

Solution

$$CH_{3}OCH_{3}\rightarrow CH_{4}+CO+H_{2}$$
$$t = 00.4$$
$$t = 4.5 hr.$$ 0.110.290.290.29
$$\displaystyle kt=\ln \:\frac{P_{0}}{P}\Rightarrow 4.78\times 10^{-3}\times 4.5\times 60=\ln\frac{0.4}{P}$$
$$P=0.11\:atm$$
at$$t=0\:P_{0}=0.4$$
$$M_0=46$$
at $$t=4.5\:hr.\:P=0.11+0.29\times 3=0.98\:atm$$
$$\displaystyle M=\frac{0.11\times 46+0.29(16+28+2)}{0.98}=18.77$$
$$\displaystyle \frac{r_{0}}{r}=\frac{P_{0}}{P}\sqrt{\frac{M}{M_{0}}}=\frac{0.4}{0.98}\sqrt{\frac{18.77}{46}}=0.26$$
Question 2
1 / -0

In this case we have
$$A\rightarrow B+C$$
Time t $$\infty $$
Total $$press^{r}$$ $$P_2$$ $$P_3$$
Find k?

A
$$\displaystyle k=\frac{1}{t}ln\frac{P_{3}}{2(P_{3}-P_{2})}$$

B
$$\displaystyle k=\frac{1}{t}ln\frac{P_{3}}{2(P_{3}+P_{2})}$$

C
$$\displaystyle k=\frac{1}{t}ln\frac{P_{2}}{2(P_{3}-P_{2})}$$

D
$$\displaystyle k=\frac{1}{t}ln\frac{P_{2}}{2(P_{3}+P_{2})}$$

Solution

Let P be the initial pressure of A and x be the change in the pressure of A to reach equilibrium.

When time $$\displaystyle t = \infty P = \dfrac {P_3}{2}$$

When time is t, the total pressure $$\displaystyle P-x+x+x=P+x = P_2 \\ x = P_2 - P=P_2 - \dfrac {P_3}{2} \\ P-x = \dfrac {P_3}{2}-[P_2 - \dfrac {P_3}{2} ]= P_3 - P_2$$

$$\displaystyle k = \dfrac {1}{t}ln \dfrac {P}{P-x}$$

$$\displaystyle k = \dfrac {1}{t} ln \dfrac {\dfrac {P_3}{2}}{(P_3 - P_2)}$$

$$\displaystyle \displaystyle k=\frac{1}{t}ln\frac{P_{3}}{2(P_{3}-P_{2})}$$
Question 3
1 / -0

In a first order reaction, the initial concentrated of the reactant was $$M/10.$$ After $$8$$ minutes $$20$$ seconds the concentrated becomes $$\,M/100.\,$$. What is the rate constant?

A
$$\;5\times10^{-3}\,sec^{-1}$$

B
$$\;2.303\times10^{-5}\,sec^{-1}$$

C
$$\;2.303\times10^{-4}\,sec^{-1}$$

D
$$\;4.606\times10^{-3}\,sec^{-1}$$

Solution

$$K\,=\,\displaystyle\frac{2.303}{t}log\left[\displaystyle\frac{a_0}{a} \right]$$

$$K\,=\,\displaystyle\frac{2.303}{500}log\displaystyle\frac{\displaystyle\frac{1}{10}}{\displaystyle\frac{1}{100}}$$

$$=\,\displaystyle\frac{2.303}{500}log\,10\,=\,4.606\times10^{-3}\,sec^{-1}$$
Question 4
1 / -0

After how many second will the concentration of the reactant in a first order reaction be halved if the rate constant is $$1.155 \times {10}^{-3} {s}^{-1}$$?

A
600

B
100

C
60

D
10

Solution

Rate constant $$k=1.155\times { 10 }^{ -3 }{ s }^{ -1 }$$
$$k=\dfrac { 2.303 }{ t } \log { \dfrac { a }{ \left( a-x \right) } } \because a=a,\left( a-x \right) =\dfrac { a }{ 2 } $$
$${ t }_{ { 1 }/{ 2 } }=\dfrac { 2.303 }{ k } \log { \dfrac { a }{ { a }/{ 2 } } } $$
$$=\dfrac { 2.303 }{ 1.155\times { 10 }^{ -3 } } \log { 2 } $$
$$=\dfrac { 2.303 }{ 1.155\times { 10 }^{ -3 } } \times 0.3010$$
$$=\dfrac { 0.693\times { 10 }^{ 3 } }{ 1.155 } $$
or $${ t }_{ { 1 }/{ 2 } }=\dfrac { 0.693 }{ k } =\dfrac { 0.693 }{ 1.155\times { 10 }^{ -3 } } $$
$$=600 s$$
Question 5
1 / -0

Why does the probability curve become narrower when gas particles are more massive?

A
The gas particles can travel at higher speeds

B
The gas particles are less likely to travel at slower speeds

C
The gas particles are more likely to travel at higher speeds

D
The gas particles can not travel at higher speeds

E
There are more collisions between gas particles

Solution

(D) The gas particles can not travel at higher speed.
Due to increase in mass of the particles the gas particles can not travel at higher speeds, hence, the distribution of the kinetic energy or speed is less, hence, the curve becomes narrower.
Question 6
1 / -0

_________ increases effective collisions without increasing average energy.

A
An increase in the reactant concentration

B
An increase in the temperature

C
A decrease in pressure

D
Catalysts

E
$$\displaystyle pH$$

Solution

An increase in the reactant concentration increase effective collisions without increasing average energy.
This is due to the fact that molecules comes more and more closer, hence, they tends to collide more easily.

Question 7

1 / -0

Consider the three statements about reaction energy diagrams and the relative magnitudes of the activation energy, $$E_{a}$$, and the enthalpy of reaction, $$\Delta H$$.
One or more of the statements is true.
Identify the correct statement or combination of statements from the four choices below.

	Statement
I	For an endothermic reaction, the magnitude of $$E_{a}$$ is always greater than $$\triangle H$$.
II	For an exothermic reaction, the magnitude of $$E_{a}$$ is always greater than $$\triangle H$$.
III	For an exothermic reaction, adding a catalyst will decrease the magnitude of $$\triangle H$$.

Statement I only is true.

Statement II only is true.

Both statements I and II are true.

Both statements I and III are true.

Solution

For endothermic reaction, $$ \Delta H={ H }_{ P }-{ H }_{ R }=positive.$$

For exothermic reaction, $$ \Delta H=negative$$ From the above figures, it is confirmed that

(i) For an endothermic reaction, $$E_a$$ is always greater than $$\Delta H.$$

(ii) For an exothermic reaction, $$E_a$$ may or may not be greater than $$\Delta H.$$ While catalysts decrease the activation energy only and does not affect the $$\Delta H $$ of the reaction.

Therefore, option $$(A)$$ is the correct answer.

Question 8
1 / -0

Among the following which will decrease the rate of the reaction?
i. Using highly concentrated reactants
ii. Decreasing the temperature by $$25\ K$$
iii. Stirring the reactants

A
i only

B
ii only

C
i and iii only

D
ii and iii only

E
i, ii, and iii

Solution

$$\bullet $$ Using highly concentrated reactants will increase the rate of the reaction as rate is directly proportional to the
concentration of reactants.
$$\bullet $$ Decreasing the temperature by 25 K will decrease the rate constant and hence the rate of reaction will decrease.
$$\bullet $$ Stirring the reactants increases the rate of interaction between the reactants and hence the rate of reaction increases.
$$\therefore $$ The correct answer is (ii) only.
Question 9
1 / -0

What is the activation energy of the reverse reaction by this diagram?

A
$$E$$

B
$$D$$

C
$$C$$

D
$$B$$

Solution

The activation energy of the reverse reaction by this diagram is given by $$E$$.
Activation energy is the amount of energy required to raise reactants to transition state.
Question 10
1 / -0

Which of the following choice is correct regarding to increase the rate of a reaction?

A
Decreasing the temperature

B
Increasing the volume of the reaction vessel

C
Reducing the activation energy

D
Decreasing the concentration of the reactant in the reaction vessel

Solution

The relation between specific rate (k), temperature (T) and activation energy (E_a) can be given by Arrhenius equation as follows,
$$k = A e^{-E_a/RT}$$
$$k = p.Z.e^{-Ea/RT}$$
Where,

k = specific rate (or) rate constant

A = pZ = Frequency factor

p = probability factor or steric factor or orientation factor

Z = no. of binary collisions per unit time

E_a = Activation energy

R = Universal Gas constant

T = Absolute Temperature

It is clear that the rate of a reaction is proportional to the temperature but inversely proportional to the activation energy.

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Chemical Kinetics Test - 63

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Chemical Kinetics Test - 63

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SHARING IS CARING

Question 1

$$0.26 : 1$$

$$0.13 : 1$$

$$1: 0.26$$

$$1 : 0.13$$

Solution

Question 2

$$\displaystyle k=\frac{1}{t}ln\frac{P_{3}}{2(P_{3}-P_{2})}$$

$$\displaystyle k=\frac{1}{t}ln\frac{P_{3}}{2(P_{3}+P_{2})}$$

$$\displaystyle k=\frac{1}{t}ln\frac{P_{2}}{2(P_{3}-P_{2})}$$

$$\displaystyle k=\frac{1}{t}ln\frac{P_{2}}{2(P_{3}+P_{2})}$$

Solution

Question 3

$$\;5\times10^{-3}\,sec^{-1}$$

$$\;2.303\times10^{-5}\,sec^{-1}$$

$$\;2.303\times10^{-4}\,sec^{-1}$$

$$\;4.606\times10^{-3}\,sec^{-1}$$

Solution

Question 4

600

100

60

10

Solution

Question 5

The gas particles can travel at higher speeds

The gas particles are less likely to travel at slower speeds

The gas particles are more likely to travel at higher speeds

The gas particles can not travel at higher speeds

There are more collisions between gas particles

Solution

Question 6

An increase in the reactant concentration

An increase in the temperature

A decrease in pressure

Catalysts

$$\displaystyle pH$$

Solution

Question 7

Statement I only is true.

Statement II only is true.

Both statements I and II are true.

Both statements I and III are true.

Solution

Question 8

i only

ii only

i and iii only

ii and iii only

i, ii, and iii

Solution

Question 9

$$E$$

$$D$$

$$C$$

$$B$$

Solution

Question 10

Decreasing the temperature

Increasing the volume of the reaction vessel

Reducing the activation energy

Decreasing the concentration of the reactant in the reaction vessel

Solution

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