Chemical Kinetics Test

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Chemical Kinetics Test - 23

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Weekly Quiz Competition

Question 1
1 / -0

The initial rate of zero-order reaction of the gaseous reaction $$A(g)\rightarrow2B(g)$$ is $$10^{-2}M min^{-1}$$. If the initial concentration of A is 0.1 M, then the concentration of B after 60 sec would be:

A
0.09 M

B
0.01 M

C
0.02 M

D
0.002 M

Solution

$$\left[A \right ]_{t}$$ = $$\left [A \right ]_{0}$$– $$kt$$

$$\left[A \right ]_{t}$$ = $$(0.1) - (\dfrac{ (10) ^ {-2}\times60}{60})$$

$$\left[A \right ]_{t}$$ = $$0.09$$ M

$$\left [B \right ]_{t}$$ = $$ {2}\times(.0.045) $$

$$\left [B \right ]_{t}$$ = $$0.09$$ M

hence the correct answer is -
(a) $$ 0.09 $$ M
Question 2
1 / -0

In the following reaction, $$ A\rightarrow B$$, rate constant is $$1.2\times10^{-2} Ms^{-1}$$. What is the concentration of $$B$$ after $$10$$ min, if we start with $$10\ M$$ of $$A$$?

A
7.2 M

B
4.5 M

C
6.7 M

D
7.0 M

Solution

Unit of k indicates that it's a zero-order reaction.

According to zero-order reaction.
$$Rate \ of \ reaction = k$$
$$Rate \ of \ formation \ of \ B = \frac{Concenteration\ of\ B}{time(\ in \ seconds)} = k$$
$$Concentration\ of\ B\ after\ 10 mins=kt = 1.2\times10^{-2}\times10\times60 = 7.2M$$
Question 3
1 / -0
To initiate a reaction, minimum energy which is required to break bonds is called
1. bond energy
2. activation energy
3. breaking energy
4. ionization energy
A
1

B
2

C
3

D
4

Solution

Activation energy, in chemistry, the minimum amount of energy that is required to activate atoms or molecules to a condition in which they can undergo chemical transformation.
Question 4
1 / -0

Radioactive disintegration follows _____ order kinetic.

A
zero

B
pseudo first

C
first

D
second

Solution

All radioactive decay follows first order kinetics and this is where the similarity ends.
Question 5
1 / -0

For the first-order reaction $$A\rightarrow product$$s, the half-life is $$200$$s. The rate constant of the reaction is:

A
$$3.46\times 10^{-2} s^{-1}$$

B
$$3.46\times 10^{-3} s^{-1}$$

C
$$3.46\times 10^{-4} s^{-1}$$

D
$$3.75\times 10^{-5} s^{-1}$$

Solution

$$k=0.693/200=0.00346s^{-1}$$
Question 6
1 / -0

For a first order reaction $$A\rightarrow B$$ the rate constant is $$x\ min^{−1}$$ . If the initial concentration of A is 0.01M , the concentration of A after one hour is given by the expression:

A
$$0.01e^{-x}$$

B
$$1\times10^{-2}(1-e^{-60x})$$

C
$$(1\times10^{-2}) e^{-60x}$$

D
None of these
Question 7
1 / -0

The decomposition of an equal mole of two reactants P and Q obey 1st order kinetics with half-life 54 and 18 minutes respectively. If both are allowed to run for 54 minutes. The ratio of mole of Q and P left after decomposition will be:

A
$$3$$

B
$$\dfrac{1}{4}$$

C
$$4$$

D
$$16$$

Solution

Let initially $$X$$ moles of P and $$X$$ moles of Q are present.

For P, $$18$$ minutes corresponds to $$1$$ half-life and for Q, $$54$$ minutes corresponds to $$3$$ half-life periods.

After $$1$$ half life period, the number of moles of P left $$\displaystyle = \frac {X}{2}$$

After $$3$$ half life periods, number of moles of Q left $$\displaystyle = \frac {X}{2^3}=\frac {X}{8}$$

The ratio of mole of Q and P left after decomposition will be $$\displaystyle =\frac {\dfrac {X}{8}}{\dfrac {X}{2}}=\dfrac {1}{4}$$

Hence, option B is correct.
Question 8
1 / -0

If 'a' is the initial concentration of a substance which reacts according to zero order kinetics and K is rate constant, the time for the reaction to go to completion is:

A
a/K

B
2/Ka

C
K/a

D
2K/a

Solution

For zero order reaction, $$K=\frac {x}{t}$$
if $$x=a$$ (complete reactant to react);
$$t=\frac {a}{K}$$.
Question 9
1 / -0

Activation energy of a reaction is:

A
the energy released during the reaction.

B
the energy evolved when activated complex is formed.

C
additional amount of energy needed by the reactants to overcome the potential barrier of reaction.

D
the energy needed to form one mole of the product.

Solution

Hint: Activation energy is the barrier energy required to initiate the reaction and reach transition state.

Correct Answer: Option $$C$$.

Explanation:

We know that activation energy is the minimum amount of energy required by the reactants to undergo a particular chemical reaction.

Final Answer : Hence, the option $$C$$ is the correct answer.
Question 10
1 / -0

Calculate the partial pressure of reactants and products, when azomethane decomposed at an initial pressure of 200 mm for 30 minutes according to
$$(CH_3)_2N_2\rightarrow C_2H_6+N_2$$
The rate constant is $$2.5\times 10^{-4} sec^{-1}$$.

A
$$P'(CH_3)_2N_2=12.755 mm, P'C_2H_6=72.45 mm, P'N_2=724.5 mm$$

B
$$P'(CH_3)_2N_2=127.55 mm, P'C_2H_6=72.45 mm, P'N_2=72.45 mm$$

C
$$P'(CH_3)_2N_2=1275.5 mm, P'C_2H_6=72.45 mm, P'N_2=724.5 mm$$

D
None of these

Solution

30 min corresponds to 1800 seconds.
$$k=\displaystyle \frac {2.303}{t} log \frac {P}{P-x}$$
$$2.5 \times 10^{-4}=\displaystyle \frac {2.303}{1800}log \frac {200}{P-x}$$
$$\displaystyle 1.568 = log \frac {200}{P-x}$$
$$\displaystyle P-x =127.55 \: mm$$
Hence, the partial pressure of azomethane is 127.55 mm Hg.
The partial pressures of ethane and nitrogen are $$\displaystyle200 \: mm-127.55 \: mm=72.45 \: mm$$ respectively.

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Chemical Kinetics Test - 23

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Chemical Kinetics Test - 23

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Question 1

0.09 M

0.01 M

0.02 M

0.002 M

Solution

Question 2

7.2 M

4.5 M

6.7 M

7.0 M

Solution

Question 3

1

2

3

4

Solution

Question 4

zero

pseudo first

first

second

Solution

Question 5

$$3.46\times 10^{-2} s^{-1}$$

$$3.46\times 10^{-3} s^{-1}$$

$$3.46\times 10^{-4} s^{-1}$$

$$3.75\times 10^{-5} s^{-1}$$

Solution

Question 6

$$0.01e^{-x}$$

$$1\times10^{-2}(1-e^{-60x})$$

$$(1\times10^{-2}) e^{-60x}$$

None of these

Question 7

$$3$$

$$\dfrac{1}{4}$$

$$4$$

$$16$$

Solution

Question 8

a/K

2/Ka

K/a

2K/a

Solution

Question 9

the energy released during the reaction.

the energy evolved when activated complex is formed.

additional amount of energy needed by the reactants to overcome the potential barrier of reaction.

the energy needed to form one mole of the product.

Solution

Question 10

$$P'(CH_3)_2N_2=12.755 mm, P'C_2H_6=72.45 mm, P'N_2=724.5 mm$$

$$P'(CH_3)_2N_2=127.55 mm, P'C_2H_6=72.45 mm, P'N_2=72.45 mm$$

$$P'(CH_3)_2N_2=1275.5 mm, P'C_2H_6=72.45 mm, P'N_2=724.5 mm$$

None of these

Solution

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