Chemical Kinetics Test

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Chemical Kinetics Test - 25

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Question 1
1 / -0

Effective collisions are those in which molecules must:

A
have energy equal to or greater than the threshold energy

B
have proper orientation

C
acquire the energy of activation

D
all of the above

Solution

Option (D) is correct. These are the characteristics of effective collisions.
Effective collisions are collision between two reactants with the appropriate orientation & with sufficient energy to overcome the activation energy barrier.
The number of effective collisions increases exponentially with an increase in temperature.
Question 2
1 / -0

The rate constant of a reaction is 0.0693 $$min^{-1}$$. Starting with 10 mol, the rate of the reaction after 10 min is:

A
0.0693 mol $$min^{-1}$$

B
0.0693 $$\times$$ 2 mol $$min^{-1}$$

C
0.0693 $$\times$$ 5 mol $$min^{-1}$$

D
0.0693 $$\times\, 5^2$$ mol $$min^{-1}$$

Solution

As we know,
for a first order reaction,
$$\displaystyle t_{1/2}\, =\, \frac{0.693}{k}\, =\, \frac{0.693}{0.0693}\, =\, 10\, min$$
Reactant after 10 min = 5 mol
Rate $$\displaystyle \left ( \frac{dx}{dt}\right )\, =\, k[A]\, =\, 0.0693\, \times\, 5\, mol\, min^{-1}$$
Question 3
1 / -0

The activation energy for a simple chemical reaction $$A\, \rightarrow\, B\, is\, E_a$$ in the forward reaction. The activation energy of the reverse reaction:

A
Is negative of $$E_a$$

B
Is always less than $$E_a$$

C
Can be less than or more than $$E_a$$

D
Is always double of $$E_a$$

Solution

As we know,
For exothermic reaction,
$$\Delta H^{\ominus}$$ = (PE of product) - (PE of reactant) < 0.
$$(E_a)_f\, <\, (E_a)_b.$$
For endothermic reaction:
$$(E_a)_f\, >\, (E_a)_b.$$
Question 4
1 / -0

The decomposition of $$Cl_2O_7$$ at $$400$$ K in the gas phase to $$Cl_2$$ and $$O_2$$ is of $$1^{st}$$ order. After $$55$$ sec at 400 K, the pressure of $$Cl_2O_7$$ falls from $$0.062$$ to $$0.044$$ atm.

The rate constant is :

A
$$6.24\times 10^{-3} sec^{-1}$$

B
$$5.20\times 10^{-3} sec^{-1}$$

C
$$7.34\times 10^{-3} sec^{-1}$$

D
none of the above

Solution

The given reaction is:

$$Cl_2O_7\rightarrow Cl_2+\dfrac{7}{2}O_2$$

Mole at $$t=0$$ sec $$a$$ 0 0
Mole at $$t=55$$ sec $$a-x$$ $$x$$ $$7x/2$$
Since, mole at any time $$\propto$$ pressure at that time (at constant V and T)

$$\therefore a\propto 0.062$$

and $$(a-x)\propto 0.044$$ at $$t=55 sec$$

$$\because K=\dfrac {2.303}{t}log \dfrac {a}{(a-x)}=\dfrac {2.303}{55}log \dfrac {0.062}{0.044}$$

$$\quad \quad =6.24\times 10^{-3} sec^{-1}$$

Hence, the correct option is $$A$$
Question 5
1 / -0

The reaction $$A\, \rightarrow\, B$$ follows first-order kinetics. The time taken for 0.8 mol of A to produce 0.6 mol of B is 1 hr. What is the time taken for the conversion of 0.9 mol of A to produce 0.675 mol of B?

A
1 hr

B
0.5 hr

C
0.25 hr

D
2 hr

Solution

Time for the conversion of 0.8 mol of A to 0.6 mol of B :
$$\displaystyle B\, =\, \frac{0.6}{0.8}\, =\, 0.75\, =\, t_{75\%}$$

Similarly, time for the conversion of 0.9 mol of A to 0.675 mol of B = $$\displaystyle \frac{0.675}{0.9}\, =\, 0.75\, =\, t_{75\%}$$

Hence, $$t_{75\%}$$ in both case $$ = 1 hr$$.
Question 6
1 / -0

The activation energy for a simple chemical reaction, $$A\rightarrow{B}$$ is $$E_a$$ in forward direction. The activation energy for the reverse reaction:

A
is negative of $$E_a$$

B
is always less than $$E_a$$

C
can be less than or more than $$E_a$$

D
is always double of $$E_a$$

Solution

As we know,
$$\Delta H=E_{a_{(f)}}-E_{a_{(b)}}$$; if $$\Delta H=-ve, E_{a_{(b)}} > E_{a_{(f)}}$$
if $$\Delta H=+ve, E_{a_{(b)}} < E_{a_{(f)}}$$
Question 7
1 / -0

80% of a first order reaction was completed in 70 min. How much it will take for 90% completion of a reaction ?

A
114 min

B
140 min

C
100 min

D
200 min

Solution

Hint: A first-order reaction can be defined as a chemical reaction in which the reaction rate is linearly dependent on the concentration of only one reactant.

Formula used:
For a first-order reaction,

$$t=\dfrac{2.303}{k} \log \left[\frac{R_{0}}{R}\right]$$

Here, $$R_{0}=$$ initial concentration of reactant $$R=$$ present concentration of reactant

STEP-1
Using $$k_{1}=\dfrac{2.303}{\mathrm{t}} \log \dfrac{\mathrm{a}}{\mathrm{a}-\mathrm{x}}$$,

we can find out the value of $$\mathrm{K}$$. Using $$\mathrm{K}$$ time required for $$90 \%$$ reaction can be calculated.
By putting values in the above equation, we will get

$$\mathrm{K}=\dfrac{2.303}{70} \log \dfrac{1}{0.2}=0.023$$

STEP-2
Now, again using the same equation as above, but for $$90 \%$$ completion, we can calculate the time.

$$\mathrm{t}=\dfrac{2.303}{0.023} \log \dfrac{1}{0.1}$$

So, we get time $$=100 \mathrm{~min}$$.

Final Answer:
Hence, the Correct answer is option $$C$$
Question 8
1 / -0

For a given reaction of the first order, it takes 15 minutes for the concentration to drop from 0.8 M to 0.4 M. The time required for the concentration to drop from 0.1 M to 0.025 M will be:

A
60 minutes

B
15 minutes

C
7.5 minutes

D
30 minutes

Solution

The reactant concentration drop from 0.8 to 0.4 M, i.e., 50% takes place in 15 minutes.

$$K=\dfrac {2.303}{15} log \dfrac {0.8}{0.4}=\dfrac {0.693}{15}=0.0462\ min^{-1}$$

Also,

$$t=\dfrac {2.303}{K}log \dfrac {0.1}{0.025}$$

$$t=\dfrac {2.303}{0.0462} log \dfrac {0.1}{0.025}=30\ min$$
Question 9
1 / -0

The rate of a chemical reaction generally increases rapidly even for small temperature increase because of a rapid increase in:

A
Collision frequency

B
Fraction of molecules with energies in excess of the activation energy

C
Activation energy

D
Average kinetic energy of molecules

Solution

Every chemical reaction, whether exothermic or endothermic has an energy barrier that has to be overcome before reactants are transformed into products.

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 increases.
Question 10
1 / -0

In a reaction, 5 g ethyl acetate is hydrolyzed per litre in the presence of dil. HCl in 300 min.If the reaction is of the first order and the initial concentration of ethyl acetate is 22 g $$L^{-1}$$, the rate constant of the reaction is:

A
$$k \,=\, 8.6\,\times\, 10^{-4}\, min^{-1}$$

B
$$k \,=\, 1.4\,\times\, 10^{-4}\, min^{-1}$$

C
$$k \,=\, 6.9\,\times\, 10^{-4}\, min^{-1}$$

D
$$None \:of \:these $$

Solution

a = 22 g $$L^{-1}$$
x = 5 g $$L^{-1}$$, t = 300 min
k = $$\displaystyle \frac{2.3}{300\,min} \, log\, \frac{22\, g\, L^{-1}}{(22\,-\,5)g\,L^{-1}}\,=\, 8.6\,\times\, 10^{-4}\, min^{-1}$$

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Mole at $$t=0$$ sec	$$a$$	0	0
Mole at $$t=55$$ sec	$$a-x$$	$$x$$	$$7x/2$$

Chemical Kinetics Test - 25

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Chemical Kinetics Test - 25

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Question 1

have energy equal to or greater than the threshold energy

have proper orientation

acquire the energy of activation

all of the above

Solution

Question 2

0.0693 mol $$min^{-1}$$

0.0693 $$\times$$ 2 mol $$min^{-1}$$

0.0693 $$\times$$ 5 mol $$min^{-1}$$

0.0693 $$\times\, 5^2$$ mol $$min^{-1}$$

Solution

Question 3

Is negative of $$E_a$$

Is always less than $$E_a$$

Can be less than or more than $$E_a$$

Is always double of $$E_a$$

Solution

Question 4

$$6.24\times 10^{-3} sec^{-1}$$

$$5.20\times 10^{-3} sec^{-1}$$

$$7.34\times 10^{-3} sec^{-1}$$

none of the above

Solution

Question 5

1 hr

0.5 hr

0.25 hr

2 hr

Solution

Question 6

is negative of $$E_a$$

is always less than $$E_a$$

can be less than or more than $$E_a$$

is always double of $$E_a$$

Solution

Question 7

114 min

140 min

100 min

200 min

Solution

Question 8

60 minutes

15 minutes

7.5 minutes

30 minutes

Solution

Question 9

Collision frequency

Fraction of molecules with energies in excess of the activation energy

Activation energy

Average kinetic energy of molecules

Solution

Question 10

$$k \,=\, 8.6\,\times\, 10^{-4}\, min^{-1}$$

$$k \,=\, 1.4\,\times\, 10^{-4}\, min^{-1}$$

$$k \,=\, 6.9\,\times\, 10^{-4}\, min^{-1}$$

$$None \:of \:these $$

Solution

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