Chemical Kinetics Test

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Chemical Kinetics Test - 19

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Weekly Quiz Competition

Question 1
1 / -0

A substance having initial concentration (a) reacts according to zero order kinetics. How much time will the reaction take to reach the completion?

A
$$\dfrac{a}{K}$$

B
$$\dfrac{K}{a}$$

C
$$\dfrac{a}{2K}$$

D
$$\dfrac{2K}{a}$$

Solution

For a zero order reaction, $$[A_{0}] - [A] = Kt$$

When the reaction is complete, $$[A] =0$$.

$$[A_{0}] - [A] =[A_{0}] -0=[A_{0}] = Kt_{completion}$$

$$t_{completion}=\dfrac {[A_{0}] } {K}=\dfrac{a}{K}$$
Question 2
1 / -0

If the initial concentration is reduced to 1/4th in a zero order reaction, then the time taken for half the reaction to complete:

A
remains same

B
becomes 4 times

C
becomes one-fourth

D
doubles

Solution

For a zero order reaction, $$t_{1/2}=\dfrac{[A]_0}{2k}$$

Let $$[A]_0=a$$

$$t_{1/2}=\dfrac{a}{2k}$$ ......(1)

When the intial concentration is reduced to one fourth, $$[A]_0=\dfrac {1} {4} \times a.$$

$$t'_{1/2}=\dfrac{\dfrac {1} {4} \times a}{2k}$$ ......(2)

By dividing equation (2) by (1), we get
$$\dfrac {t'_{1/2}} {t_{1/2}} =\dfrac {\dfrac{\dfrac {1} {4} \times a}{2k}} {\dfrac{a}{2k}} =\dfrac {1} {4}$$

Hence, option C is correct.
Question 3
1 / -0

At 400K, the half-life of a sample of a gaseous compound initially at 56.0 $$kPa$$ is 340s. When the pressure is 28.0 $$kPa$$, the half-life is 170s. The order of the reaction is:

A
0

B
2

C
1

D
1/2

Solution

When the pressure is reduced to half from 56.0 kPa to 28 kPa, the half-life period is reduced to half from 340 s to 170 s.

Thus, the half-life period is proportional to the pressure. This is true for a zero-order reaction.
Question 4
1 / -0

$$N_{2}O_{2} (g)\rightarrow2NO$$ is a first-order reaction interms of the concentration of $$N_{2}O_{2} (g)$$ .Which of the following is valid, $$[N_{2}O_{2} ]$$ being constant?

A
$$[NO]=[N_{2}O_{2}]_{0}e^{-kt}$$

B
$$[NO]=[N_{2}O_{2}]_{0}(1-e^{kt})$$

C
$$[NO]=[N_{2}O_{2}]_{0}(e^{-kt}-1)$$

D
$$[NO]=[N_{2}O_{2}]_{0}(1-e^{-kt})$$

Solution

The reaction is $$N_{2}O_{2} (g)\rightarrow2NO$$
The initial concentration of $$N_{2}O_{2}$$ is $$[N_{2}O_{2}]_0$$.. Assuming initially only $$N_{2}O_{2}$$ is present and no $$NO$$ is formed.
At time t, the concentrtion of $$N_{2}O_{2}$$ and NO are $$[N_{2}O_{2}]$$ and $$[NO]$$ respectively.
The total concentration is equal to $$[N_{2}O_{2}]+[NO]=[N_{2}O_{2}]_0$$ ......(1)
But from integrated rate law equation fo first order reaction , $$[N_{2}O_{2}]=[N_{2}O_{2}]_0e^{-kt}$$......(2)
Substituting equation (2) in equation (1)
$$[N_{2}O_{2}]_0e^{-kt}+[NO]=[N_{2}O_{2}]_0$$
$$[NO]=[N_{2}O_{2}]_0-[N_{2}O_{2}]_0e^{-kt}=[N_{2}O_{2}]_0(1-e^{-kt})$$
Question 5
1 / -0

The time required for the completion of first order reaction is:

A
infinity

B
thrice that of time required for 90% completion

C
3/2 that of time required for 90% completion

D
ten times that of time required for 90% completion

Solution

$$k= \cfrac{1}{t}\times ln\left(\cfrac{a}{a-x}\right)$$

For a First-order reaction, we have

$$(a-x)$$ $$= ae^{-kt}$$

Where $$(a-x)$$ is the concentration of reactant $$A$$ at any time, $$t$$

When the reaction goes to completion

$$(a-x)=0$$ and we get,

$$0= ae^{-kt}$$

$$t=\infty$$

Hence, the correct option is $$\text{A}$$
Question 6
1 / -0

The reaction $$L\rightarrow M$$ is started with $$10.0\ gm$$ of $$L$$. After $$30$$ and $$90$$ minutes, $$5.0\ gm$$ and $$1.25\ gm$$ of $$L$$ respectively are left. The order of the reaction is:

A
0

B
1

C
2

D
3

Solution

After 30 minutes, the concentration is reduced to half.

After the next 60 minutes, the concentration is reduced to one fourth.

Thus, 30 minutes is the half-life period and is independent of the concentration of $$L$$.

Hence, it is a first-order reaction.

Option B is correct.
Question 7
1 / -0

For the reaction A $$\rightarrow $$ B it has been found that the order of the reaction is zero with respect to A. Which of the following expressions correctly describes the reaction?

A
$$K=\dfrac{2.303}{t}log\dfrac{[A_{0}]}{[A]}$$

B
$$[A_{0}] - [A] = Kt$$

C
$$t_{1/2}=\dfrac{0.693}{K}$$

D
$$t_{1/2}\propto \dfrac{1}{[A_{0}]}$$

Solution

For a zero order reaction, $$d[A]=-Kdt.$$

On integrating this equation between limits,

$$[A]=[A]_0$$ at $$ t=0$$

$$[A]=[A]_t $$ at $$ t=t$$, we get $$[A_{0}] - [A] = Kt$$
Question 8
1 / -0

$$SO_{2}Cl_{2} \rightarrow SO_{2}+Cl_{2}$$ is a first order gas reaction with $$k=2.2\times 10^{-5}\sec^{-1}$$ at $$320^o C$$. The percentage of $$SO_{2}Cl_{2}$$ decomposed on heating for 90 minutes is:

A
1.118

B
0.1118

C
18.11

D
11.18

Solution

For a first order reaction,

$$k=\dfrac{ln\left[\dfrac{Ro}{R} \right ]}{t}$$

$$ln\left[\dfrac{Ro}{R} \right ]=2.2\times10^{-5}\times90\times60$$

$$ln\dfrac{Ro}{R}=0.1188$$

$$\dfrac{Ro}{R}=1.126$$

$$R = 0.88 79 Ro$$

Percentage of decomposition = $$R_0-R\ \Rightarrow Ro-0.8879\ Ro$$
$$\Rightarrow 11.18$$ $$Ro$$ %

Hence, option D is correct.
Question 9
1 / -0

Which equation represents the time to complete 90% of first order reaction?

A
$$\dfrac{k}{2.303}log(\frac{4}{3})$$

B
$$\dfrac{2.303}{k}log(\frac{3}{4})$$

C
$$\dfrac{2.303}{k}$$

D
$$\dfrac{2.303}{k}log(3)$$

Solution

$$t = \cfrac{2.303}{k}\ log \left [ \cfrac{Ro}{R} \right ]$$

$$t =\cfrac{2.303}{k}\ log\ \left [ \cfrac{a}{a-0.9a} \right ]$$

$$t = \cfrac{2.303}{k}\ log \left [ \cfrac{1}{0.1} \right ]$$

$$= \cfrac{2.303}{k}$$

Hence, the correct option is $$\text{C}$$
Question 10
1 / -0

At room temperature the reaction between NO and $$O_{2}$$ to give $$NO_{2}$$ is fast while that of between CO and $$O_{2}$$ is slow. It is because:

A
CO is smaller in size than that of NO

B
CO is poisonous

C
the activation energy for the reaction 2NO+$$O_{2}$$ $$\rightarrow $$ 2$$NO_{2}$$ is less than 2CO+$$O_{2}$$ $$\rightarrow $$ 2$$CO_{2}$$

D
the activation energy for the reaction 2CO+$$O_{2}$$ $$\rightarrow $$ 2$$CO_{2}$$ is less than 2NO+$$O_{2}$$ $$\rightarrow $$ 2$$NO_{2}$$

Solution

The reaction between NO and oxygen has lower activation energy than the reaction between CO and oxygen.
Hence, the reaction between NO and oxygen is fast whereas the reaction between CO and oxygen is slow.

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Re-Attempt Weekly Quiz Competition

Chemical Kinetics Test - 19

Result

Chemical Kinetics Test - 19

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SHARING IS CARING

Question 1

$$\dfrac{a}{K}$$

$$\dfrac{K}{a}$$

$$\dfrac{a}{2K}$$

$$\dfrac{2K}{a}$$

Solution

Question 2

remains same

becomes 4 times

becomes one-fourth

doubles

Solution

Question 3

0

2

1

1/2

Solution

Question 4

$$[NO]=[N_{2}O_{2}]_{0}e^{-kt}$$

$$[NO]=[N_{2}O_{2}]_{0}(1-e^{kt})$$

$$[NO]=[N_{2}O_{2}]_{0}(e^{-kt}-1)$$

$$[NO]=[N_{2}O_{2}]_{0}(1-e^{-kt})$$

Solution

Question 5

infinity

thrice that of time required for 90% completion

3/2 that of time required for 90% completion

ten times that of time required for 90% completion

Solution

Question 6

0

1

2

3

Solution

Question 7

$$K=\dfrac{2.303}{t}log\dfrac{[A_{0}]}{[A]}$$

$$[A_{0}] - [A] = Kt$$

$$t_{1/2}=\dfrac{0.693}{K}$$

$$t_{1/2}\propto \dfrac{1}{[A_{0}]}$$

Solution

Question 8

1.118

0.1118

18.11

11.18

Solution

Question 9

$$\dfrac{k}{2.303}log(\frac{4}{3})$$

$$\dfrac{2.303}{k}log(\frac{3}{4})$$

$$\dfrac{2.303}{k}$$

$$\dfrac{2.303}{k}log(3)$$

Solution

Question 10

CO is smaller in size than that of NO

CO is poisonous

the activation energy for the reaction 2NO+$$O_{2}$$ $$\rightarrow $$ 2$$NO_{2}$$ is less than 2CO+$$O_{2}$$ $$\rightarrow $$ 2$$CO_{2}$$

the activation energy for the reaction 2CO+$$O_{2}$$ $$\rightarrow $$ 2$$CO_{2}$$ is less than 2NO+$$O_{2}$$ $$\rightarrow $$ 2$$NO_{2}$$

Solution

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