Chemical Kinetics Test

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Chemical Kinetics Test - 50

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Weekly Quiz Competition

Question 1
1 / -0

For the zero order reaction $$A\rightarrow 2B$$, the rate constant is $$2\times 10^{-6}M min^{-1}$$ the reaction is started with 10 M of A.

A
what will be concentration of A after 2 days

B
what is intial half-line one of reaction?

C
in what time, the reaction will complete?

D
none
Question 2
1 / -0

The time elapsed between $$33\%$$ and $$67\%$$ completion of a first order reaction is $$30$$ minutes. What is the time needed for $$25\%$$ completion?

A
$$15.5\ min$$

B
$$12.5\ min$$

C
$$18.5\ min$$

D
$$16.5\ min$$

Solution

As per the question,

$$30 = -\dfrac{2.303}{k} log \dfrac{[100]}{[100 - 33]}$$ + $$\dfrac{2.303}{k} log \dfrac{[100]}{[100 - 67]}$$

$$\Rightarrow k = 2.36 \times {10}^{-2}\space {min}^{-1}$$

$$t = \dfrac{2.303}{2.36 \times {10}^{-2}} log \dfrac{[100]}{[75]}$$

$$t = 12.5\space min$$
Question 3
1 / -0

A first order reaction follows the expression:

A
$${ C }_{ t }{ e }^{ { k }_{ 1 }t }=C_0$$

B
$$C_t={ C }_{ t }{ e }^{ { k }_{ 1 }t }$$

C
In $$\frac { { C }_{ O } }{ { C }_{ t } } =-k_1t$$

D
In $$\frac { { C }_{ t } }{ { C }_{ o } } =k_1t$$

Solution

1st Order Reaction,

$${k}_{1} = \dfrac{1}{t} ln\dfrac{[{C}_{0}]}{[{C}_{t}]}$$

$${C}_{t}{e}^{{k}_{1}t} = {C}_{0}$$
Question 4
1 / -0

For the first order consecutive reaction $$P\rightarrow Q\rightarrow R$$ under steady state approximation to $$\left[ P \right] ,\left[ Q \right]$$ and $$\left[ R \right] $$ with time are best represented by:

A

B

C

D

Solution
Question 5
1 / -0

Which among the following plots are liner (a-x) is the concentration of the reactant remaining after time, t?
(1):-(a-x) vs t, for a first order reaction
(2):- (a-x) vs t, for a zero order reaction
(3):-(a-x) vs t, for a second order reaction
(4):-1/ (a-x) vs t, for a second order reaction

A
1 and 2

B
1 and 3

C
2 and 3

D
2 and 4

Solution

This is Graph for 2nd Order Reaction.( [A] is [a - x])
The Graph is Linear for Graph between (a-x) vs t, for Zero Order Reaction
So Option 4 is correct
Question 6
1 / -0

A catalyst lower the energy of activation by $$25\%$$. The temperature at which rate of uncatalysed reaction would be equal to that of the catalyzed one at $$27^oC$$ is:

A
$$400^oC$$

B
$$127^oC$$

C
$$300^oC$$

D
$$227^oC$$

Solution

$$\dfrac{{E}_{a(1)}}{{E}_{a(2)}}$$ = $$\dfrac{{T}_{1}}{{T}_{2}}$$

$$\dfrac{4}{3}$$ = $$\dfrac{{T}_{1}}{300}$$

$${T}_{1} = 400 K$$

$${T}_{1} = {127}^{o}C$$
Question 7
1 / -0

Which of the following represents the expression for $$\cfrac{3}{4}$$th life of first order reaction?

A
$$\cfrac{2.303}{k}\log{4/3}$$

B
$$\cfrac{2.303}{k}\log{3/4}$$

C
$$\cfrac{2.303}{k}\log{4}$$

D
$$\cfrac{2.303}{k}\log3$$

Solution

Answer $$C.$$
The expression for rate of first order reaction,
$$k=\cfrac{2.303}{t}\log \cfrac{a}{a-x}$$
or $$t=\cfrac{2.303}{k}\log \cfrac{a}{a-x}$$ where $$x=\cfrac{3}{4}a$$
$$t=\cfrac{2.303}{k}\log \cfrac{a}{a-\cfrac{3}{4}a}$$
$$=\cfrac{2.303}{k}\log \cfrac{a}{\cfrac{1}{4}a}$$
$$=\cfrac{2.303}{k}\log 4$$
Question 8
1 / -0

If a I-order reaction is completed to the extent of $$60\%$$ and $$20\%$$ in time intervals, $$t_1$$ and $$t_2$$ what is the ratio, $$t_1 : t_2$$?

A
6.32

B
5.587

C
4.11

D
8.33

Solution

In 1st Order Reaction,

$$\dfrac {{t}_{1}}{{t}_{2}} = \dfrac {log({\dfrac{100}{100-60}})}{log({\dfrac{100}{100-20}})}$$

$$\dfrac{{t}_{1}}{{t}_{2}} = 4.11$$
Question 9
1 / -0

The reaction $$v_1A + v_2B \rightarrow$$ products is first order with respect to $$A$$ and zero- order with respect to $$B$$.If the reaction is started with $$[A]_0$$ and $$[B]_0$$, the integrated rate expression of this reaction would be:

A
$$ln \dfrac { [A]_ 0 }{ [A]_ 0-x } =k_ 1t$$

B
$$ln\dfrac { [A]_ 0 }{ [A]_ 0-v_ 1x } =k_ 1t$$

C
$$ln\dfrac { [A]_ 0 }{ [A]_ 0-v_ 1x } =v_ 1k_ 1t$$

D
$$ln\dfrac { [A]_0 }{ [A]_ 0-v_ 1x } = -v_ 1k_ 1t$$

Solution

Its 1st Order wrt A and Zero Order wrt B.

So, the integrated rate equation of this reaction is,

$$ln \dfrac{[{A}_{0}]}{[{A}_{0}] - {v}_{1}x}$$ = $${v}_{1}{k}_{1}t$$
Question 10
1 / -0

Fora reaction, $$ A \rightarrow B +C, $$ it was found that at the end of 10 minutes from the start, the total optical rotation of the system was $$ 50^o $$ and when the reaction is complete, it was $$ 100^o C $$ Assuming that only B and C are optically active and dextro rotatory, the rate constant of this first order reaction would be ?

A
$$ 0.069 min^{-1} $$

B
$$ 0.69 min^{-1} $$

C
$$ 6.9 min^{-1} $$

D
$$ 6.9 \times 10^{-2} min^{-1} $$

Solution

Let the rotation of Sucrose be $$r_1^o$$ per mole and the initial moles of Sucrose be a.
$$ \implies r_0 = ar_1^0$$
Let the moles converted be x.
$$\implies r_t = (a-x)r_1^0 $$
$$\implies \frac{a}{a-x} = \frac{r_o}{r_t}$$
$$\implies k = \frac{1}{t} \ln \frac{a}{a-x} = \frac{1}{t} \ln\frac{r_o}{r_t}$$
$$\implies k = \frac{2.303}{10}\log\frac{100}{50}$$
$$\implies k = 0.069 \space min^{-1}$$