Chemical Kinetics Test

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Chemical Kinetics Test - 52

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Weekly Quiz Competition

Question 1
1 / -0

At least how half-lives should elapse for a I order reaction $$A\rightarrow$$ products so that the reaction is at least $$95%$$ completed ? $$(\log 2=0.3)$$

A
$$6$$

B
$$5$$

C
$$4$$

D
$$7$$

Solution
Question 2
1 / -0

Which one of the following plots is correct for a first order reaction:

A

B

C

D

Solution

The plot shown in option C is correct for a first-order reaction, as for first order reaction:
$$log(a-x)=-\dfrac{kt}{2.303}+loga$$
It is similar to straight-line equation $$y=mx+c$$
where intercept is equal to $$loga$$ and slope equal to $$-\dfrac{kt}{2.303}$$

Negative slope shows Option C is correct answer.
Question 3
1 / -0

The rate constant of the first-order reaction may be described by following equation:

log K = $$ 14.34 - \cfrac{1.25 \times 10^4}{T} $$, calculate energy of activation.

A
239.34kJ

B
239.34 kcal

C
239.34J

D
239.34 cal.

Solution

As we know that : $$lnK=ln\ A-\dfrac{E_a}{RT}$$-------(1)
We have given $$logK=14.34-\dfrac{1.25\times 10^4}{T}$$-----(2)

Comparing (1) and (2) we will come to know that $$\dfrac{E_a}{2.303R}=1.25\times 10^4$$

So $$E_a=2.303\times 1.25\times 10^4\times R$$
$$E_a=2.303\times 1.25\times 10^4\times 8.314$$
$$E_a=239.34kJ$$

Option A is correct.
Question 4
1 / -0

The first order reaction has half-life of $$18$$ hrs. What percentage of the reactant will remain after $$24$$hrs $$\left( {\log 2 = 0.3,\log 2.5 = 0.4} \right)$$

A
$$20\% $$

B
$$40\% $$

C
$$45.5\% $$

D
$$68.2\% $$

Solution
Question 5
1 / -0

A reaction proceeds 5 times more at $${ 60 }^{ o }$$C as it does at $${ 30 }^{ o }$$C the energy of activation is $$(\log { 5=0.69990 } )$$

A
12 K cal / mole

B
11 K cal / mole

C
13 K cal / mole

D
16 K cal / mole

Solution

Given:
$$T_2=60+273=333K$$,
$$T_1=30+273=303K$$,
$$R=1.987\times 10^{-3}kcal$$

We know that at a temperature $$T$$

$$\dfrac{r_2}{r_1}=\dfrac{K_2}{K_1}$$
$$\dfrac{r_2}{r_1}=5$$ (at Temperature $$T_2\ and\ T_1$$)

therefore,
$$\dfrac{K_2}{K_1}=5$$

Hence:
$$2.303log_{10}\dfrac{K_2}{K_1}=\dfrac{E_a}{R}\dfrac{T_2-T_1}{T_1\times T_2}$$

$$2.303log_{10}\ 5=\dfrac{E_a}{10^{-3}\times 1.987}\dfrac{333-300}{333\times 300}$$

$$E_a=10.757\ kcalmol^{-1}$$

Option B is correct.
Question 6
1 / -0

In a first order reaction ,if $$25\%$$ of the reaction is completed in $$50$$ minutes $$50\%$$ of the total reaction is completed in _________ minutes (approx):

A
$$120$$

B
$$100$$

C
$$80$$

D
$$150$$

Solution

Given:
First order reaction- $$25$$% complete in $$50\ minutes$$
Then $$50$$% of reaction will complete in ?
For the first order reaction,

$$k=\dfrac{2.303}{t}\dfrac{Log\ a}{a-x}$$

Here $$a=100$$%, $$x=25$$% $$t=50\ minutes$$

so $$k=\dfrac{2.303}{50}\dfrac{log\ 100}{100-25}$$

$$k=\dfrac{2.303}{50}\dfrac{log\ 100}{75}$$

$$k=\dfrac{2.303}{50}\dfrac{log\ 4}{3}$$------------(1)

In Second case $$a=100$$%, $$x=50$$% $$t=? minutes$$

so $$k=\dfrac{2.303}{t}\dfrac{log\ 100}{100-50}$$

so $$k=\dfrac{2.303}{t}\dfrac{log\ 100}{50}$$

so so $$k=\dfrac{2.303}{t}\ log\ 2$$----------------(2)

Compare (1) and (2) we will get:

$$\dfrac{2.303}{50}\dfrac{log\ 4}{3}$$=$$\dfrac{2.303}{t}\ log\ 2$$

on solving we will get $$t=120\ minutes$$ approximately
Option A is correct.
Question 7
1 / -0

What is the activation energy for a reaction if the rate doubles when the temperature is raised from $$20^{o}C$$ to $$35^{o}C$$? $$(R=8.134\ J\ mol^{-1}K^{-1})$$

A
$$15.1\ kJ\ mol^{-1}$$

B
$$342\ kJ\ mol^{-1}$$

C
$$269\ kJ\ mol^{-1}$$

D
$$34.7\ kJ\ mol^{-1}$$

Solution

Given : $$T_1 = 20 + 273 = 293K$$

$$T_2 = 35 + 273 = 308K$$

$$log \dfrac{{k}_{2}}{{k}_{1}} = \dfrac{{E}_{a}}{2.303 \times R} \times (\dfrac{1}{{T}_{1}} - \dfrac{1}{{T}_{2}})$$

Substituting the values, we get

$$log\space 2 = \dfrac{{E}_{a}}{2.303 \times 8.314} \times (\dfrac{1}{293} - \dfrac{1}{308})$$

$${E}_{a}$$ = $$34.7\space kJ\space {mol}^{-1}$$

Therefore, the option is D.
Question 8
1 / -0

K for a zero order reaction is $$2\times10^{-2}\,mol\,L^{-1}sec^{-1}$$. 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
Question 9
1 / -0

For any first order reaction following observation is made. If at temperature (T) half life of the reaction is 6930 second and at temperature ($$T^1$$) half of the reaction is $$0.693 \mu $$ second, then calculate $$\frac{T^1}{T}$$

A
2

B
4

C
6

D
8
Question 10
1 / -0

For the chemical reaction, $$A \to $$ products, the following $$\log \,K = 16.398 - \cfrac{{2800}}{T}$$
Arrhenius factor for the reaction is______________

A
$$2.5 \times {10^{16}}$$

B
$$5 \times {10^{16}}$$

C
$$10^9$$

D
None of these

Solution