Chemical Kinetics Test

Result

Chemical Kinetics Test - 44

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 rate of chemical reaction is directly proportional to the equilibrium constant.
In which of the following process reaction will be completed first?

A
$$K=10$$

B
$$K=1$$

C
$$K={ 10 }^{ 3 }$$

D
$$K={ 10 }^{ -2 }$$

Solution

The rate of chemical reaction is directly proportional to the equilibrium constant.
$$\therefore $$ For $$K={ 10 }^{ 3 }$$, the reaction will complete fastest.
Question 2
1 / -0

In a reaction involving one single reactant, the fraction of the reactant consumed may be defined as $$f=\left(1-\dfrac {C}{C_{0}}\right)$$ where $$C_{0}$$ and $$C$$ are the concentrations of the reactant at the after time, t. For a first order reaction:

A
$$\dfrac {df}{dt}=k(1-f)$$

B
$$-\dfrac {df}{dt}=kf$$

C
$$-\dfrac {df}{dt}=k(1-f)$$

D
$$\dfrac {df}{dt}=kf$$

Solution

Integrated rate equation for 1st order reaction
$$K=\dfrac{2.303}{t}\log\dfrac{[A]_1}{[A]_2}$$
Where $${[A]_1}$$ is the initial concentration and $${[A]_2}$$ is the concentration of A at time t
Given initial concentration is $$C_0$$ and concentration at time t is C
Integrating option A we get
$$\ln(1-f)=-kt$$
given $$f=(1-\dfrac{C}{C_0})$$
$$ln\dfrac{[C_0]}{[C]}=kt$$
$$K=\dfrac{2.303}{t}\log\dfrac{[C_0]}{[C]}$$
Question 3
1 / -0

In a first order reaction the amount of reactant decayed in three half lives (let $$a$$ be is initial amount) would be:

A
$$7a/8$$

B
$$a/8$$

C
$$a/6$$

D
$$5a/6$$

Solution

The amount of reactant left after n half lives for a 1st order reaction is $$({\dfrac{1}{2}})^{n}$$ times the Initial concentration.
Here n =3 . So Amount of Reactant left is $$\dfrac{a}{8}$$
So Amount of reactant decayed is $$\dfrac{7a}{8}$$
Question 4
1 / -0

If a first order reaction is completed to the extent of $$75$$% and $$50$$% in time interval, $${t}_{1}$$ and $${t}_{2}$$, what is the ratio $${t}_{1}:{t}_{2}$$?

A
$$\ln {2}$$

B
$$\cfrac{\ln{(3/4)}}{\ln{2}}$$

C
$$2$$

D
$$1/2$$

Solution

For a first order reaction
$$ t=\cfrac { 2.303 }{ k } \log { \cfrac { { [A] }_{ o } }{ { [A] }_{ t } } } $$
Assuming initial concentration as 100;
$$ { t }_{ 1 }(75\%)=\cfrac { 1 }{ k } \ln { \left[ \cfrac { 100 }{ 25 } \right] } =\cfrac { \ln { (4) } }{ k } \\ { t }_{ 2 }(50\%)=\cfrac { 1 }{ k } \ln { \left[ \cfrac { 100 }{ 25 } \right] } =\cfrac { \ln { (2) } }{ k } \\ \cfrac { { t }_{ 1 } }{ { t }_{ 2 } } =\cfrac { \ln { (4) } }{ \ln { (2) } } =2$$
Hence answer answer is option [C]
Question 5
1 / -0

75% of a first order reaction was completed in 32 minutes. When was 50% of the reaction completed?

A
4 min

B
8 min

C
24 min

D
16 min

Solution

$$K \times t= ln \left( \dfrac{[n_0]}{100\times [n_0]-75 [n_0]}\times 100\right)$$

$$\Rightarrow K \times 32 = ln \left( \dfrac{[n_0]\times100 }{25[n_0]}\right)$$

$$\Rightarrow \ K \times 32=2 \ln 2$$

$$K=\dfrac{\ln2}{16}$$

$$t_{1/2}=\dfrac{\ln2}{K}$$

$$\Rightarrow t_{1/2} =16\ min$$
Question 6
1 / -0

For the consecutive unimolecular-type first order reaction $$A \xrightarrow{k_1} R \xrightarrow{k_2} S$$, the concentration of component $$R, C_R$$ at any time $$t$$ is given by $$: C_R = C_{AO} K_1 \left[\dfrac{e^{k_1t}}{(k_2 - k_1)} + \dfrac{e^{-k_2t}}{(k_1 - k_2)}\right]$$ if $$C_A = C_{AO}, CR = C_{RO} = 0$$ at $$t = 0$$ the time at which the maximum concentration of $$R$$ occurs is:

A
$$t_{max} = \dfrac{k_2 - k_1}{ln(k_2/k_1)}$$

B
$$t_{max} = \dfrac{ln(k_2/k_1)}{k_2 - k_1}$$

C
$$t_{max} = \dfrac{e^{k_2/k_1}}{k_2 - k_1}$$

D
$$t_{max} = \dfrac{e^{k_2 - k_1}}{k_2 - k_1}$$

Solution
Question 7
1 / -0

The time for half-life period of a certain reacting $$A\rightarrow $$ product is an hour. How much time does it take for its concentration to come from 0.50 to 0.25 mol $${ L }^{ -1 }$$ if it is a zero order reaction?

A
0.25h

B
1h

C
4h

D
0.5h

Solution

concerntration is halued.

$$\therefore $$ time must be $$t_{1/2}$$

$$\therefore t.025n$$
Question 8
1 / -0

The rate constant, the activation energy and the arrhenius parameter of a chemical reaction at $${ 25 }^{ 0 }C$$ are $$3\times { 10 }^{ 14 }{ sec }^{ -1 }:\ 104.4J{ mol }^{ -1 }$$ and $$6.0\times { 10 }^{ 14 }{ sec }^{ -1 }$$ respectively, the value of the rate constant as $$T\rightarrow \infty $$ is:

A
$$2\times { 10 }^{ 8 }{ sec }^{ -1 }$$

B
$$6\times { 10 }^{ 14 }{ sec }^{ -1 }$$

C
$$\infty$$

D
$$3.6\times { 10 }^{ 30 }{ sec }^{ -1 }$$

Solution
Question 9
1 / -0

In the following first-order competing reactions.

$$A+ \text{Reagent}\rightarrow \text{Product},\quad B+ \displaystyle{\text{Reagent}} \rightarrow \text{Product}$$.

The ratio of $${ K }_{ 1 }/{ K }_{ 2 }$$ if only 50% of B will have been reacted, When 94% of A has been reacted is?

A
4.07

B
0.246

C
2.06

D
0.06

Solution

$$A + Reagent \, \xrightarrow{k_1} P$$

$$B + Reagent \, \xrightarrow{k_2} P$$

Let initial care be x and given time be $$t$$ first order

$$\therefore k_1 t = \ln \dfrac{[A_0]}{[A_0]}$$

$$\Rightarrow k_1 \times t = \ln \dfrac{[A_0]}{[A_0] - 0.94[A_0]}$$

$$\Rightarrow k_1 t = \ln \dfrac{[A]_0}{0.06 [A]_0}$$

$$\Rightarrow k_1 t = \ln \dfrac{50}{3}$$ ....$$(I)$$

Similarly

$$k_2 t = \ln \dfrac{[B_0]}{[B_0]}$$

$$\Rightarrow k_2 t = \ln \dfrac{[B_0]}{0.5 [B_0]}$$

$$\Rightarrow k_2 t = \ln 2 $$ .....$$(II)$$

$$(I)/(II) \Rightarrow \dfrac{k_1}{k_2} = \dfrac{\ln(50/3)}{\ln 2} = \dfrac{2.81}{0.693}$$

$$= 4.072$$

Hence, the correct option is $$A$$
Question 10
1 / -0

The rate of the first order reaction is $$0.69\times { 10 }^{ -2 }mol{ L }^{ -1 }{ min }^{ -1 }$$ and the initial concentration is $$0.2mol{ L }^{ -1 }$$. The half life period is:

A
$$1205s$$

B
$$330$$

C
$$600s$$

D
$$1s$$

Solution

Half-life for first order reaction,
$${ t }_{ { 1 }/{ 2 } }=\cfrac { 0.693 }{ K } =\cfrac { 0.693 }{ r } \times \left[ { A }_{ o } \right] \quad \quad \quad \quad \quad \because r=K\left[ { A }_{ o } \right] $$
$$\therefore { t }_{ { 1 }/{ 2 } }=\cfrac { 0.693\times 60\times 0.2 }{ 0.69\times { 10 }^{ -2 } } =1200s(approx)$$

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Chemical Kinetics Test - 44

Result

Chemical Kinetics Test - 44

Score

-

Rank

-

SHARING IS CARING

Question 1

$$K=10$$

$$K=1$$

$$K={ 10 }^{ 3 }$$

$$K={ 10 }^{ -2 }$$

Solution

Question 2

$$\dfrac {df}{dt}=k(1-f)$$

$$-\dfrac {df}{dt}=kf$$

$$-\dfrac {df}{dt}=k(1-f)$$

$$\dfrac {df}{dt}=kf$$

Solution

Question 3

$$7a/8$$

$$a/8$$

$$a/6$$

$$5a/6$$

Solution

Question 4

$$\ln {2}$$

$$\cfrac{\ln{(3/4)}}{\ln{2}}$$

$$2$$

$$1/2$$

Solution

Question 5

4 min

8 min

24 min

16 min

Solution

Question 6

$$t_{max} = \dfrac{k_2 - k_1}{ln(k_2/k_1)}$$

$$t_{max} = \dfrac{ln(k_2/k_1)}{k_2 - k_1}$$

$$t_{max} = \dfrac{e^{k_2/k_1}}{k_2 - k_1}$$

$$t_{max} = \dfrac{e^{k_2 - k_1}}{k_2 - k_1}$$

Solution

Question 7

0.25h

1h

4h

0.5h

Solution

Question 8

$$2\times { 10 }^{ 8 }{ sec }^{ -1 }$$

$$6\times { 10 }^{ 14 }{ sec }^{ -1 }$$

$$\infty$$

$$3.6\times { 10 }^{ 30 }{ sec }^{ -1 }$$

Solution

Question 9

4.07

0.246

2.06

0.06

Solution

Question 10

$$1205s$$

$$330$$

$$600s$$

$$1s$$

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.