Chemical Kinetics Test

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Chemical Kinetics Test - 38

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Question 1
1 / -0

The unit , $$\\mol L^{-1} s^{-1} $$ is meant for the rate constant of the reaction having the order :

A
0

B
2

C
1

D
3

Solution

General unit for rate constant $$ = \left( \dfrac {mol}{L} \right)^{1-n} s^{-1} $$
(where, $$ n = $$ order of reaction)

Given, unit for rate constant $$ = \left( \dfrac {mol}{L} \right) s^{-1} $$

On comparing,

$$ \left( \dfrac {mol}{L} \right)^{1-n} s^{-1} = \left( \dfrac {mol}{L} \right) s^{-1} $$

$$ 1 - n = 1 $$

$$\Rightarrow n = 0 $$

Hence, the order of the reaction is zero.
Question 2
1 / -0

If for a reaction in which $$A(g)$$ converts to $$B(g)$$ the reaction carried out at const. $$V$$ and $$T$$ results into the following graph.

A
Then the reaction must be $$A(g) \rightarrow 3B(g)$$ and is a first order reaction

B
Then the reaction must be $$A(g) \rightarrow 3B(g)$$ and is a second order reaction

C
Then the reaction must be $$A(g) \rightarrow 3B(g)$$ and is a zero order reaction

D
Then the reaction must be $$A(g) \leftrightarrow 3B(g)$$ and is a first order reaction
Question 3
1 / -0

In a reaction carried out at $$400\ K, 0.0001$$% of the total number of collisions are effective. The energy of activation of the reaction is:

A
zero

B
$$7.37\ kcal/mol$$

C
$$9.212\ kcal/mol$$

D
$$11.05\ k cal/mol$$

Solution

The fraction of molecules having energy equal to or greater than $$E_a$$ is :
$$ x = \dfrac{n}{N} = e^{-E_a/RT} (x = \frac{n}{N} = 0.0001\% = \frac{0.0001}{100} = 10^{-6})$$
$$ \therefore \log x = \dfrac{-E_a}{2.3 \times RT}$$
$$ \implies \ log 10^{-6} = \dfrac{-E_a}{2.3 \times 2 \times 400}$$
$$ \implies E_a = 6 \times 2.3 \times 800$$
$$ \implies E_a = 11.05 \space kcal/mol $$
Question 4
1 / -0

For the zero order reaction $$ A \rightarrow B + C ; $$ initial concentration of $$A$$ is $$ 0.1 M. $$ If $$A = 0.08 M $$ after $$10 $$ minutes, then it's half-life and completion time are respectively :

A
$$10$$ min $$ ; 20 $$ min

B
$$2 \times 10^{-3} $$ min $$ ; 4 \times 10^{-3} $$ min

C
$$25$$ min $$ ; 50 $$ min

D
$$250$$ min $$ ; 500 $$ min

Solution

Its a Zero Order Reaction.

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

$$k = \dfrac{[0.1] - [0.08]}{10}$$ = $$2 \times {10}^{-3}\space M \ {min}^{-1}$$

$${t}_{1/2} = \dfrac{[{A}_{0}]}{2k}$$

$${t}_{1/2} = 25\space min$$

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

$$t = 50\space min$$
Question 5
1 / -0

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

A
30 min

B
40 min

C
35 min

D
25 min

Solution

Order = 1 (given)
$$ \because t_{75} = 2 \times t_{50} $$

$$ \implies 2 \times 15 = 30 $$

$$ k = \frac{0.693 }{t_{50}} = \frac{0.693}{15}$$

$$ a = 0.1 \space M$$

$$ (a-x) = 0.025 \space M$$

For first order reaction $$k = \frac{2.303}{t} \log(\frac{a}{a-x})$$

$$ \implies \frac{2.303 \times \log2}{15} = \frac{2.303}{t} \log(\frac{0.1}{0.025})$$

$$ \implies t= 30 \space minutes $$
Question 6
1 / -0

In which one of the following situations does a conventional electric current flow due north ?

A
Protons in beam are moving due south.

B
A water molecule is moving due north.

C
Electrons in a beam are moving due south.

D
None of these

Solution

When the Electrons in a beam are moving due south, then the conventional electric current flow due north.
So, Option C is correct
Question 7
1 / -0

Which graph represents the zero order reaction?

$$A(g) \rightarrow B(g)$$.

A

B

C

D

Solution

For zero order reaction:
Rate of formation of product i.e.
$$\dfrac {dB}{dt}$$ increases with time $$(t)$$
Hence, $$\dfrac {dB}{dt}\propto t$$.
Question 8
1 / -0

Select incorrect statements :

A
Pre exponential factor for zero order reaction is a unitless quantity

B
If $$t_{1/4}$$ = 30 sec then $$t_{1/2}$$ = 60 sec for first order reaction

C
If $$t_{1/3}$$ = 30 sec then $$t_{2/3}$$ = 90 sec for sec-order reaction

D
If $$t_{1/5}$$ = 30 sec then $$t_{3/5}$$ = 90 sec for zero order reaction.

Solution

Arrhenius Equation is given by:-
$$K=A$$ $$e^{-Ea/RT}$$
where, $$K$$=rate constant of the reaction
$$A$$= pre exponential factor
$$Ea$$= Activation energy
$$R$$= Ideal gas constant
$$T$$= Temperature in Kelvin

Now, $$e^{-Ea/RT}$$ is a dimensionless quantity.
So, unit of $$A$$ will be same at that of $$K$$

For zero order reaction we have,
$$K=\cfrac {1}{t} \left([A]_o-[A]_t\right)$$
so unit of $$K$$ is $$mol L^{-1} time^{-1}$$, so it will be unit of pre exponential factor as well and it will not be an unitless quantity.
Question 9
1 / -0

A first order reaction has $$k = 1.5\times 10^{-6}$$ per second at $$200^0C$$. If the reaction is allowed to run for 10 hrs, what percentage of the initial concentration would have changed into the product?

A
5.213%

B
10.21%

C
20.3%

D
15.2%

Solution

Given,
$$K=1.5\times 10^{-6} s^{-1}$$
We know for a first order reaction,
$$t=\dfrac{2.303}{K}log(\dfrac{N_0}{N})$$
So on substituting given values,
$$\dfrac{N_0}{N}=1.0555$$
$$\implies N=0.9475N_0$$
Now product formed$$=0.9475N_0-N_0=0.0521N_0=$$5.213%
Question 10
1 / -0

In a zero order reaction, 20% of the reaction complete in 10 s. How much time it will take to complete 50% of the reaction ?

A
20 s

B
25 s

C
30 s

D
40 s

Solution

$$20\%\quad completed\quad reaction\quad means\quad \\ left\quad amount\quad A=80\\ initial\quad amount\quad { A }_{ \circ }=100\quad and\quad the\quad reaction\quad is\quad completed\quad in\quad t=10s\\ Similarly\quad for\quad 50\%,\\ left\quad amount\quad A=50\\ initial\quad amount\quad { A }_{ \circ }=100\\ A={ A }_{ \circ }-kt\\ 80=100-k\times 10\\ k\times 10=20\\ k=2mol\quad { l }^{ -1 }{ s }^{ -1 }\quad \\ A={ A }_{ \circ }-kt\\ 50=100-2\times t\\ 2t=50\\ t=25s$$