Chemical Kinetics Test

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Chemical Kinetics Test - 26

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Question 1
1 / -0

What specific name can be given to the following sequence of steps: $$Hg + hv \rightarrow\, Hg^*$$
$$Hg^*\, +\, H_2\, \rightarrow\, H_2^*\, +\, Hg$$

A
Fluorescence

B
Phosphorescence

C
Photosensitization

D
Chemilumionescence

Solution

$$Hg + hv \rightarrow\, Hg^*$$
$$Hg^*\, +\, H_2\, \rightarrow\, H_2^*\, +\, Hg$$
It is fluorescence as $$Hg$$ emitted the light which it absorbs in first step. (Fluorescence is the emission of light by a substance that has absorbed light or other electromagnetic radiation. It is a form of luminescence. In most cases, the emitted light has a longer wavelength and therefore, lower energy than the absorbed radiation)
Question 2
1 / -0

In a multistep reaction such as $$A + B$$ $$\rightarrow\, Q\, \rightarrow\, C.$$ The potential energy diagrame is shown below. What is $$E_a$$ for the reaction $$Q\, \rightarrow\, C\, ?$$

A
$$3 kcal$$ $$mol^{-1}$$

B
$$5 kcal$$ $$mol^{-1}$$

C
$$8 kcal$$ $$mol^{-1}$$

D
$$11 kcal$$ $$mol^{-1}$$

Solution

For $$Q\, \rightarrow\, C$$
$$E_a$$ $$= 23 - 20$$
$$= 3 kcal$$ $$mol^{-1}$$
Question 3
1 / -0

If the activation energies of the forward and backward reactions of a reversible reaction are $$E_a(f)$$ and $$E_a(b)$$, respectively. The $$\Delta E$$ of the reaction is ...........

A
$$E_a(F)-E_a(b)$$

B
$$E_a(F)+E_a(b)$$

C
$$E_a(F)=E_a(b)$$

D
$$-E_a(F)+E_a(b)$$

Solution

As we know,
$$\Delta E = E_a(F)-E_a(b)$$
Question 4
1 / -0

The rate of reaction increase by the increase of temperature because:

A
collision frequency is increased.

B
energy of products decreases.

C
the fraction of molecules possessing energy $$\geq\, E_T$$ (threshold energy) increases.

D
mechanism of a reaction is changed.

Solution

Every chemical reaction whether exothermic or endothermic has an energy barrier that has to be overcome before reactants can be transformed into products. If the reactant molecules have sufficient energy, they can reach the peak of the energy barrier after the collision and then they can go to the right side of the slope and consequently, change into products.

If the activation energy for a reaction is low, the fraction of effective collisions will be large and the reaction will be fast. On the other hand, if the activation energy is high, then the fraction of effective collisions will be small and the reaction will be slow.

When the temperature is increased, the number of active molecules increases, i.e., the number of effective collisions will increase and the rate of reaction will increase.
Question 5
1 / -0

The activation energy of reactant molecules in a reaction depends upon:

A
Temperature

B
Nature of the reactants

C
Collision per unit time

D
Concentration of reactants

Solution

activation energy:
The minimum amount of energy that molecules must have in order for a reaction to occur upon collision.

Nature of the Reactants:
Substances differ markedly in the rates at which they undergo chemical change. The differences in reactivity between reactions may be attributed to the different structures of the materials involved; for example, whether the substances are in solution or in the solid state matters. Another factor has to do with the relative bond strengths within the molecules of the reactants. For example, a reaction between molecules with atoms that are bonded by strong covalent bonds will take place at a slower rate than would a reaction between molecules with atoms that are bonded by weak covalent bonds. This is due to the fact that it takes more energy to break the bonds of the strongly bonded molecules.
Question 6
1 / -0

What can you say about the existence of A if the potential energy diagram for the reaction
$$A\, \rightarrow\, B$$ looks like :

A
A will exist

B
A will not exist

C
B will not exist

D
A and B are in equilibeium

Solution

From figure it can be seen that, A is at highest energy point so it will have tendency to loose energy and it will not exists.
Question 7
1 / -0

In a certain reaction. $$10$$% of the reactant decomposes in one hour, $$20$$% in two hours, $$30$$% in three hours, and so on. The dimension of the velocity constant (rate constant) are:

A
$$hr^{-1}$$

B
$$Mol\, L^{-1}\, hr^{-1}$$

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

D
$$Mol\, s^{-1}$$

Solution

here, $$t_{1/2}\, \propto$$ a. Hence, zero order reaction. so, dimension of k = mol $$L^{-1}\, hr^{-1}$$
For zero order: t = $$\displaystyle \frac{x}{k}\, or\, k\, =\, \frac{x}{t}$$
If t = $$t_{10\%}\, =\, 10\, min,\, x\, =\, 10,$$
Then, k = $$\displaystyle \frac{10}{10}\, =\, 1\, mol\, L^{-1}\, hr^{-1}$$
If t = $$t_{20\%}\, =\, 20\, min,\, x\, =\, 20,$$
Then, k = $$\displaystyle \frac{20}{20}\, =\, 1\, mol\, L^{-1}\, hr^{-1}.$$
Similarly, for $$30$$% and so on.
Thus, reaction is of zero order.
Question 8
1 / -0

For a hypothetical reaction: A + B $$\rightarrow$$ Products, the rate law is r = k[A]$$[B]^0$$. The order of reaction is:

A
0

B
1

C
2

D
3

Solution

r = $$[A][B]^0$$
OR = 1 + 0 = 1
Question 9
1 / -0

The rate constant of a zero order reaction is $$0.2\:mol\:dm^{-3}h^{-1}$$. If the concentration of the reactant after 30 minutes is $$0.05\:mol\:dm^{-3}$$. Then its initial concentration would be:

A
$$0.15\:mol\:dm^{-3}$$

B
$$1.05\:mol\:dm^{-3}$$

C
$$0.25\:mol\:dm^{-3}$$

D
$$4,00\:mol\:dm^{-3}$$

Solution

For a zero order reaction,
$$t = \frac {[A_0] -[A] }{ k } $$

$$t = 30 min = \frac{1}{2} hr$$

$$k = 0.2 mol dm^{-3} h^{-1}$$

$$[A] = 0.05 mol dm^{-3}$$

On putting these values in the equation above, we get $$A_o = 0.15 mol dm^{-3}$$
Question 10
1 / -0

90% of a first order reaction was completed in 100 min. How much time it will take for 80% completion of a reaction :

A
90 min

B
80 min

C
70 min

D
60 min

Solution

We can use $$k_{1}=\frac{2.303}{t}log\frac{a}{a-x}$$ formula first to calculate the K (reaction constant) for the reaction.
Now,putting given conditions in above equation we get
$$k_{ 1 }=\frac { 2.303 }{ 100 } log\frac { 1 }{ 0.1 } $$
this gives K=0.02303
now using same formula used above but for 80% yield we can calculate time required
$$0.02303=\frac { 2.303 }{ x } log\frac { 1 }{ 0.2 } $$
from this we get time =70 min

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Chemical Kinetics Test - 26

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Chemical Kinetics Test - 26

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Question 1

Fluorescence

Phosphorescence

Photosensitization

Chemilumionescence

Solution

Question 2

$$3 kcal$$ $$mol^{-1}$$

$$5 kcal$$ $$mol^{-1}$$

$$8 kcal$$ $$mol^{-1}$$

$$11 kcal$$ $$mol^{-1}$$

Solution

Question 3

$$E_a(F)-E_a(b)$$

$$E_a(F)+E_a(b)$$

$$E_a(F)=E_a(b)$$

$$-E_a(F)+E_a(b)$$

Solution

Question 4

collision frequency is increased.

energy of products decreases.

the fraction of molecules possessing energy $$\geq\, E_T$$ (threshold energy) increases.

mechanism of a reaction is changed.

Solution

Question 5

Temperature

Nature of the reactants

Collision per unit time

Concentration of reactants

Solution

Question 6

A will exist

A will not exist

B will not exist

A and B are in equilibeium

Solution

Question 7

$$hr^{-1}$$

$$Mol\, L^{-1}\, hr^{-1}$$

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

$$Mol\, s^{-1}$$

Solution

Question 8

0

1

2

3

Solution

Question 9

$$0.15\:mol\:dm^{-3}$$

$$1.05\:mol\:dm^{-3}$$

$$0.25\:mol\:dm^{-3}$$

$$4,00\:mol\:dm^{-3}$$

Solution

Question 10

90 min

80 min

70 min

60 min

Solution

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