Chemical Kinetics Test

Result

Chemical Kinetics Test - 22

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 half life of a radioactive isotope is $$3$$ hours. What is the value of its disintegration constant?

A
$$0.3 hour^{-1}$$

B
$$0.693 hour^{-1}$$

C
$$0.231 hour^{-1}$$

D
$$0.231 min^{-1}$$

Solution

Half life, $$ \lambda = 3 \ hrs$$

$$R = \dfrac{ln 2}{\lambda}= \dfrac{0.693}{3} $$ [radioactive decay is always of order 1]

$$ = 0.231 \ hr^{-1}$$
Question 2
1 / -0

Two radioactive isotopes P and Q have half-lives of 10 minutes and 15 minutes, respectively. Freshly prepared samples of each isotope initially contain the same number of atoms as one another, After 30 minutes, the ratio of $$\displaystyle \frac{{the\ number\ of\ atoms\ of\ P}}{{the\ number\ of\ atoms\ of\ Q}}$$ will be:

A
0.5

B
2.0

C
1.0

D
0.25

Solution
Question 3
1 / -0

For an endothermic reaction, energy of activation is E$$_{a}$$ and enthalpy of reaction is H (both of these in kJ/mol). Minimum value of E$$_{a}$$ will be:

A
less than H

B
equal to H

C
more than H

D
equal to zero

Solution

Activation energy is the minimum quantity of energy which the reacting species must possess in order to undergo a specified reaction.
(Image)
$${ E }_{ a }$$-Activation energy
Here, energy of products is higher than that of reactants.
This graph is for an endothermic reaction.
$$\triangle H$$ for an endothermic reaction is positive
See figure above, when the reaction reaches point P, the reactants possess enough energy for the reaction to take place.
In further reaction, some energy is use up for conversion of reactants to products.
This used up energy must be less than $${ E }_{ a }$$, for an endothermic reaction
Since,
$${ E }_{ a }$$- H = (used up energy)
$$\therefore { E }_{ a }>H$$
Hence, the activation energy must be more than H
Question 4
1 / -0

For a first-order reaction; A $$\rightarrow $$ B, the reaction rate at a reactant concentration of 0.01 M is found to be $$2.0 \times 10^{-5}$$ mole L$$^{-1}$$s$$^{-1}$$. The half-life period of the reaction is:

A
$$300s$$

B
$$30s$$

C
$$220s$$

D
$$347s$$

Solution

$$A\quad \rightarrow \quad B$$

Rate$$\quad \rightarrow \quad \dfrac { -d\left[ A \right] }{ dt } =K\left[ A \right] $$

when $$\left[ A \right] $$ = 0.01M, given that rate = $$2\times { 10 }^{ -5 }M{ s }^{ -1 }$$

$$K=2\times { 10 }^{ -3 }$$

Using integrated rate law expression,

the half life $$\left( { t }_{ 1/2 } \right) $$ is given by $$\dfrac { 0.693 }{ K } $$

$${ t }_{ 1/2 }=\dfrac { 0.693 }{ K } $$

$${ t }_{ 1/2 }=\dfrac { 0.693 }{ 2\times { 10 }^{ -3 } } $$

$$\Rightarrow \quad { t }_{ 1/2 }=346.5s$$

$${ t }_{ 1/2 }\quad \approx \quad 347s$$
Question 5
1 / -0

In a first-order reaction $$A\rightarrow B$$, if k is rate constant and initial concentration of the reactant A is 0.5 M then the half-life is:

A
$$\dfrac{ln2}{k}$$

B
$$\dfrac{0.693}{0.5k}$$

C
$$\dfrac{log2}{k}$$

D
$$\dfrac{log2}{k\sqrt{0.5}}$$

Solution

For a first-order reaction,
$$A\rightarrow B$$

Rate = $$k\left[ A \right] $$

Initial concentration$$=0.5M$$

We know that the rate constant of a first order reaction has only time unit. It has no concentration unit, which means that numerical value of k for a first order reaction is independent of concentration unit.
$$\therefore t_{\frac{1}{2}}=\cfrac { \ln { 2 } }{ k } \approx \dfrac { 0.693 }{ k } $$ or $$\cfrac { 2.303 }{ k\log { 2 } } $$

Hence, option A is correct.
Question 6
1 / -0

Statement 1:
In a zero order reaction, if the concentration of the reactant is doubled, the half-life period is also doubled.
Statement 2:
For a zero-order reaction, the rate of the reaction is independent of initial concentration.

A
Statement 1 is True, statement 2 is True, statement 2 is a correct explanation of statement 1.

B
Statement 1 is True, statement 2 is True, statement 2 is not a correct explanation of statement 1.

C
Statement 1 is true, Statement 2 is False.

D
Statement 1 is False, Statement 2 is True.

Solution

For a zero order reaction, the expression for the half life period is $$t_{\frac{1}{2}}=\frac{[A_{o}]}{2k}.$$
Thus half life period is directly proportional to the initial concentration. When the concentration of the reactant is doubled, the half-life period is also doubled. Hence, the statement 1 is true.
The rate of a zero order reaction is independent of the concentration of the reaction. The reaction proceeds at constant rate throughout. Hence, statement 2 is True.
Question 7
1 / -0

For the reaction:

$$N_2 + 3H_2\rightarrow 2NH_3$$,

If $$\cfrac{d[NH_3]}{dt}= 2 \times 10^{-4} mol\ L^{-1}s^{-1}$$, the value of $$\cfrac{-d[H_2]}{dt}$$ would be:

A
$$1 \times 10^{-4} mol \; L^{-1}s^{-1}$$

B
$$3 \times 10^{-4} mol\; L^{-1}s^{-1}$$

C
$$4 \times 10^{-4} mol\; L^{-1}s^{-1}$$

D
$$6 \times 10^{-4} mol\; L^{-1}s^{-1}$$

Solution

$${ N }_{ 2 }+{ 3H }_{ 2 }\longrightarrow { 2NH }_{ 3 }$$

Applying rate law.

$$\dfrac { -1 }{ 3 } \dfrac { d\left[ { H }_{ 2 } \right] }{ dt } =\dfrac { 1 }{ 2 } \dfrac { d\left[ { NH }_{ 3 } \right] }{ dt } $$

Now, given that $$\dfrac { d\left[ { NH }_{ 3 } \right] }{ dt } =2\times { 10 }^{ -4 }mol\ { L }^{ -1 }{ S }^{ -1 }$$

Using above relation,

$$\dfrac { -d\left[ { H }_{ 2 } \right] }{ dt } =\dfrac { 3 }{ 2 } \dfrac { d\left[ { NH }_{ 3 } \right] }{ dt } $$

$$\dfrac { -d\left[ { H }_{ 2 } \right] }{ dt } =\dfrac { 3 }{ 2 } \times 2\times { 10 }^{ -4 }\ mol\ { L }^{ -1 }{ S }^{ -1 }$$

$$\therefore $$ $$\dfrac { -d\left[ { H }_{ 2 } \right] }{ dt } =3\times { 10 }^{ -4 }\ mol\ { L }^{ -1 }{ S }^{ -1 }$$

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

What is the hybridisation in $$AsF_4^-$$ ion :

A
sp

B
$$sp^2$$

C
$$sp^3$$

D
$$sp^3 d$$

Solution

See-saw shape
Since there is 4 bp and 1 lp.

$$\therefore sp^3 d$$ hybridisation'
Question 9
1 / -0

Decomposition of $$H_2O$$ is a first order reaction. A 16 volume solution of $$H_2O_2$$ of half-life period 30 minutes is present at the start. When will the solution become one volume?

A
After 120 minutes

B
After 90 minutes

C
After 60 minutes

D
After 140 minutes

Solution

$$16 \xrightarrow[t\cfrac{1}{2}]{30\quad min}8 \xrightarrow[t\cfrac{1}{2}]{30\quad min}4 \xrightarrow[t\cfrac{1}{2}]{30\quad min}2 \xrightarrow[t\cfrac{1}{2}]{30\quad min}1$$
After $$120\quad min$$ solution become $$1$$ vol.
Question 10
1 / -0

The activation energy of the forward reaction and backward reaction decreases by $$30 kcal$$ then which of the following statement is correct?

A
$$\Delta H$$ will remain constant.

B
$$\Delta H$$ will decrease.

C
$$\Delta H$$ will increase.

D
The reaction will be exothermic.

Solution

$$\Delta H=E_{af}-E_{ab}$$.
So if Activation energy of the forward reaction and backward reaction decreases by 30kcal then
$$\Delta H=[E_{af}+30]-[E_{ab}+30]$$
$$\Delta H=E_{af}-E_{ab}$$.

So $$\Delta H$$will remain same.

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Chemical Kinetics Test - 22

Result

Chemical Kinetics Test - 22

Score

-

Rank

-

SHARING IS CARING

Question 1

$$0.3 hour^{-1}$$

$$0.693 hour^{-1}$$

$$0.231 hour^{-1}$$

$$0.231 min^{-1}$$

Solution

Question 2

0.5

2.0

1.0

0.25

Solution

Question 3

less than H

equal to H

more than H

equal to zero

Solution

Question 4

$$300s$$

$$30s$$

$$220s$$

$$347s$$

Solution

Question 5

$$\dfrac{ln2}{k}$$

$$\dfrac{0.693}{0.5k}$$

$$\dfrac{log2}{k}$$

$$\dfrac{log2}{k\sqrt{0.5}}$$

Solution

Question 6

Statement 1 is True, statement 2 is True, statement 2 is a correct explanation of statement 1.

Statement 1 is True, statement 2 is True, statement 2 is not a correct explanation of statement 1.

Statement 1 is true, Statement 2 is False.

Statement 1 is False, Statement 2 is True.

Solution

Question 7

$$1 \times 10^{-4} mol \; L^{-1}s^{-1}$$

$$3 \times 10^{-4} mol\; L^{-1}s^{-1}$$

$$4 \times 10^{-4} mol\; L^{-1}s^{-1}$$

$$6 \times 10^{-4} mol\; L^{-1}s^{-1}$$

Solution

Question 8

sp

$$sp^2$$

$$sp^3$$

$$sp^3 d$$

Solution

Question 9

After 120 minutes

After 90 minutes

After 60 minutes

After 140 minutes

Solution

Question 10

$$\Delta H$$ will remain constant.

$$\Delta H$$ will decrease.

$$\Delta H$$ will increase.

The reaction will be exothermic.

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.