Chemical Kinetics Test

Result

Chemical Kinetics Test - 47

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 reaction $${ N }_{ 2 }{ O }_{ 5 }\quad \left( In\quad { CCl }_{ 4 } \right) \rightarrow 2{ NO }_{ 2 }+{ 1 }/{ 2 }\quad { O }_{ 2 }\left( g \right) $$ is first order in $${ N }_{ 2 }{ O }_{ 5 }$$ with rate constant $$6.2\times { 10 }^{ -4 }{ S }^{ -1 }$$. What is the value of rate of reaction when $$\left[ { N }_{ 2 }{ O }_{ 5 } \right] =1.25\quad mole\quad { L }^{ -1 }$$

A
$$7.75\times { 10 }^{ -4 }\quad \quad { L }^{ -1 } { S }^{ -1 }$$

B
$$6.35\times { 10 }^{ -3 }\quad \quad { L }^{ -1 } { S }^{ -1 }$$

C
$$5.15\times { 10 }^{ -5 }\quad \quad { L }^{ -1 } { S }^{ -1 }$$

D
$$3.85\times { 10 }^{ -4 }\quad \quad { L }^{ -1 } { S }^{ -1 }$$

Solution

Its a 1st Order Reaction.

$$r = k[A]$$

$$r = 6.2 \times {10}^{-4} \times [1.25]$$

$$r = 7.75 \times {10}^{-4}\space {L}^{-1}{S}^{-1}$$
Question 2
1 / -0

A certain zero order reaction has $$k = 0.025 M s^{-1}$$ for the disappearance of A. What will be the concentration of A after 15 seconds if the initial concentration is 0.50 M ?

A
$$0.50$$ M

B
$$0.375$$ M

C
$$0.125$$ M

D
$$0.060$$ M

Solution

Its a Zero Order Reaction

$$k = \dfrac{1}{t}{[{A}_{0} - A]}$$

$$0.025 = \dfrac{1}{15}{[{0.5 - A}]}$$

$${A} = 0.125 M$$
Question 3
1 / -0

The rate constant for a recation is $$10.8 \times 10^{-5} mol L^{-1}S^{-1}$$. The reaction obeys:

A
First order

B
Zero order

C
Second order

D
All are wrong

Solution

We can tell order of reaction by its rate constant $$(K)$$ units

Reaction Order Units
Zero 0 mol $$L^{-1} S^{-1}$$
First 1 $$S^{-1}$$
Second 2 mol$$^{-1}LS^{-1}$$
Here, units are mol $$L^{-1}S^{-1}$$, so zero order.
Question 4
1 / -0

The reaction $$ A(s) \rightarrow 2 B(g)+C(g) $$is the first order. The pressure after 20 min. and after very long time are 150 mm Hg and 225 mm Hg. The value of rate constant and pressure after 40 min. are:

A
0.05 in 5 $$ min^{-1} $$,200mm

B
0.5 in 2 $$ min^{-1} $$,300mm

C
0.05 in 3 $$ min^{-1} $$,300mm

D
0.05 in 3 $$ min^{-1} $$,200mm

Solution
Question 5
1 / -0

$$t_{1/2} v/s \dfrac{1}{a^2}$$ is straight line graph then determine the order of a reaction:

A
zero order

B
first order

C
second order

D
third order

Solution

In general for an $$n^{th}$$ order of a reaction, the half life period is given by,
$$t_{1/2} \propto \cfrac {1}{a^{n-1}}$$, $$a$$= initial concentration; $$n$$= order of reaction
Given, $$t_{1/2}$$ $$ v/s$$ $$ 1/a^{2}$$ i.e, $$n-1=2,n=3$$
$$\therefore t_{1/2}$$= (constant)$$\cfrac {1}{a^2}$$ is a straight line.
for $$n=3$$ i.e, for Third order reaction.
Question 6
1 / -0

For a first order reaction rate constant is given as $$\log K = 14 - \dfrac{1.2 \times 10^4}{T}$$ then what will be value of temperature if its half life period is $$6.93 \times 10^{-3}\ min$$?

A
$$100\ K$$

B
$$1000\ K$$

C
$$720\ K$$

D
$$327\ K$$

Solution

$$k = \dfrac{0.693}{{t}_{1/2}}$$
$$k = 100\space {min}^{-1}$$
Substituting it, we get
$$2 = 14 - \dfrac{1.2 \times {10}^{4}}{T}$$
$$t = 1000K$$
Question 7
1 / -0

In a first order reaction, the concentration of the reactant decreases from 0.8M to 0.4M in 15 min. The time taken for the concentration to change from 0.1 M to 0.025 M is:

A
60 min

B
15 min

C
7.5 min

D
30 min

Solution

$$k = \dfrac{2.303}{t} log \dfrac{[A]}{[A - x]}$$
$$k = \dfrac{2.303}{15} log \dfrac{0.8}{0.4}$$
$$k = 0.046\space {min}^{-1}$$
The time required
$$0.046 = \dfrac{2.303}{t} log \dfrac{0.1}{0.025}$$
$$t = 30\space min$$
Question 8
1 / -0

K for a zero order reaction is $$2\times { 10 }^{ -2 }\quad { L }^{ -1 }{ Sec }^{ -1 }$$. If the concentration of the reactant after 25 sec is 0.5 M, the initial concentration must have been.

A
0.5 M

B
1.25 M

C
12.5 M

D
1.0 M

Solution

It is a Zero Order Reaction

$$k = \dfrac{[{A}_{0}] -[A]}{t}$$

$$2 \times {10}^{-2} = \dfrac{[{A}_{0}] - [0.5]}{25}$$

$$[{A}_{0}] = 1\space M$$

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

For a first - order reaction, the concentration of reactant:

A
is independent of time

B
varies linearly with time

C
varies exponentially with time

D
None

Solution
Question 10
1 / -0

Acid hydrolysis of ester in first order reaction and rate constant is given by $$k=\dfrac { 2.303 }{ t } log\dfrac { { V }_{ \infty }-{ V }_{ 0 } }{ { V }_{ \infty }-{ V }_{ t } } $$ where $${ V }_{ 0 }$$,$${ V }_{ t }$$ and $$V_{ \infty }$$ are the volume of standard $$NaOH$$ required to neutralise acid present at a given time, if ester is $$50%$$ neutralised then:

A
$${ V }_{ \infty }={ V }_{ t }$$

B
$${ V }_{ \infty }=\left( { V }_{ t }-{ V }_{ 0 } \right)$$

C
$${ V }_{ \infty }=\left( { 2V }_{ t }-{ V }_{ 0 } \right)$$

D
$${ V }_{ \infty }=\left( { 2V }_{ t }+{ V }_{ 0 } \right)$$

Solution

$$\;\;\;\;\;RCOOR'+H_2O\; \;\;\xrightarrow { H^{ + } } \;\;\;RCOOH+R'OH$$
$$t=0$$ $$a$$ $$0$$ $$0$$
$$t$$ $$a-x$$ $$x$$ $$x$$
$$t=\infty$$ $$a-a$$ $$a$$ $$a$$
Let $$V_0=$$ Volume of $$NaOH$$
$$\left( \;at\;\;t=o\right),$$
$$V_t=x+V_0$$
$$V_{\infty}=a+V_0$$
If ester is $$50\%$$ hydrolyzed, $$x=\dfrac{a}{2}$$
$$\therefore V_t=\dfrac{a}{2}+V_0$$
$$a=2V_t-2V_0$$
$$\therefore V_{\infty} =2V_t-2V_0+V_0$$
$$=2V_t-V_0$$
Hence, the answer is $$2V_t-V_0.$$

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Reaction	Order	Units
Zero	0	mol $$L^{-1} S^{-1}$$
First	1	$$S^{-1}$$
Second	2	mol$$^{-1}LS^{-1}$$

Chemical Kinetics Test - 47

Result

Chemical Kinetics Test - 47

Score

-

Rank

-

SHARING IS CARING

Question 1

$$7.75\times { 10 }^{ -4 }\quad \quad { L }^{ -1 } { S }^{ -1 }$$

$$6.35\times { 10 }^{ -3 }\quad \quad { L }^{ -1 } { S }^{ -1 }$$

$$5.15\times { 10 }^{ -5 }\quad \quad { L }^{ -1 } { S }^{ -1 }$$

$$3.85\times { 10 }^{ -4 }\quad \quad { L }^{ -1 } { S }^{ -1 }$$

Solution

Question 2

$$0.50$$ M

$$0.375$$ M

$$0.125$$ M

$$0.060$$ M

Solution

Question 3

First order

Zero order

Second order

All are wrong

Solution

Question 4

0.05 in 5 $$ min^{-1} $$,200mm

0.5 in 2 $$ min^{-1} $$,300mm

0.05 in 3 $$ min^{-1} $$,300mm

0.05 in 3 $$ min^{-1} $$,200mm

Solution

Question 5

zero order

first order

second order

third order

Solution

Question 6

$$100\ K$$

$$1000\ K$$

$$720\ K$$

$$327\ K$$

Solution

Question 7

60 min

15 min

7.5 min

30 min

Solution

Question 8

0.5 M

1.25 M

12.5 M

1.0 M

Solution

Question 9

is independent of time

varies linearly with time

varies exponentially with time

None

Solution

Question 10

$${ V }_{ \infty }={ V }_{ t }$$

$${ V }_{ \infty }=\left( { V }_{ t }-{ V }_{ 0 } \right)$$

$${ V }_{ \infty }=\left( { 2V }_{ t }-{ V }_{ 0 } \right)$$

$${ V }_{ \infty }=\left( { 2V }_{ t }+{ V }_{ 0 } \right)$$

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.