Chemical Kinetics Test

Result

Chemical Kinetics Test - 66

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

A reaction takes place in thee steps.The rate constants are $$K_1,K_2$$ and $$k_3$$. The over all rate constant $$k=\cfrac{K_1K_3}{K_2}$$. If (energy of activation) $$E_1, E_2$$ and $$E_3$$ are $$100, 20$$ and $$40 kJ$$.The overall energy of activation is:

A
30

B
40

C
50

D
60
Question 2
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 k cal/mol

C
9.212 k cal/mol

D
11.05 k cal/mol

Solution

We know that, Arrhenius equation for calculation of energy of activation of reaction with rate constant $$K$$ and temeperature $$T$$ is
$$K=A$$ $$e^{-Ea/RT}$$
where, $$Ea$$= Arrhenius activation energy
$$A$$= pre exponential factor (frequency factor)

Now, $$e^{-Ea/K_BT}$$= Fraction of collision having more than activation energy
where, $$K_B=$$ Boltzmann constant
Given, $$T=400$$ $$K$$ and effective collision= $$0.0001$$%

$$\Rightarrow$$ Effective Collision= $$e^{-Ea/K_BT}$$
$$\Rightarrow$$ $$0.0001$$%= $$e^{-Ea/1.3\times 10^{-23}\times 400}$$
$$\Rightarrow$$ $$10^{-6}$$= $$e^{-Ea/1.3\times 10^{-23}\times 400}$$

$$\Rightarrow$$ $$2.303\times \log 10^{-6}$$= $$\cfrac {-Ea}{1.3\times 10^{-23}\times 400}$$
$$\Rightarrow$$ $$2.303\times (-6)$$= $$\cfrac {-Ea}{1.3\times 10^{-23}\times 400}$$

$$\Rightarrow$$ $$Ea$$= $$1.3\times 10^{-23}\times 400\times6\times 2.303=7.19\times 10^{-20}$$ $$J/mol$$
Question 3
1 / -0

What is the activation energy for a reaction if its rate doubles when the temperature is raised from $$20^{\circ}C$$ to $$35^{\circ}C$$?

A
$$521\ kJ\ mol^{-1}$$

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

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

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

Solution

$$K_2=2K_1, T_2=35+273=308$$ $$K, T_1=20+273=293$$ $$K$$

$$\ln \cfrac {K_2}{K_1}=\cfrac {E_a}{R}\left [\cfrac {1}{T_1}-\cfrac {1}{T_2}\right]$$

$$\therefore \ln 2=\cfrac {E_a}{8.314}\left[\cfrac {1}{293}-\cfrac {1}{308}\right] $$

$$\therefore \cfrac {8.314\times \ln 2}{(3.41-3.24)\times 10^{-3}}=E_a$$

$$\therefore E_a\approx 34.7$$ $$kJ/mol$$
Question 4
1 / -0

At a certain temperature, the first order rate constant $$k_{1}$$ is found to be smaller than the second order rate constant $$k_{2}$$. If $$E_{a}(1)$$ of the first order reaction is greater than $$E_{a}(2)$$ of the second order reaction, then as temperature is raised:

A
$$k_{2}$$ will increase faster than $$k_{1}$$

B
$$k_{1}$$ will increase faster than $$k_{2}$$ but will always remain less than $$k_{2}$$

C
$$k_{1}$$ will increase faster than $$k_{2}$$ and become equal to $$k_{2}$$

D
$$k_{1}$$ will increase faster than $$k_{2}$$ and become greater than $$k_{2}$$

Solution
Question 5
1 / -0

For the following reaction at a particular temperature, according to the equations,
$$ 2N_2O_5 \rightarrow 4NO_2 + O_2 $$
$$2NO_2+1/2O_2 \rightarrow N_2O_5$$
the activation energies are $$E_1$$ and $$E_2$$ respectively, then:

A
$$E_1 > E_2$$

B
$$E_1 < E_2$$

C
$$E_1=2E_2$$

D
$$\sqrt{E_1E_2^2}=1$$

Solution
Question 6
1 / -0

A substance $$A$$ undergoes decomposition in solution following the first order kinetics. Flask $$I$$ contains 1 litre of 1 M solution of $$A$$ and flask $$II$$ contains $$100$$ $$ml$$ of 0.6 M solution. After 8 hours, if the concentration of $$A$$ in flask $$I$$ became 0.25 M, what will be the time required in hours for the concentration of $$A$$ in flask $$II$$ to become 0.3 M?

A
0.4

B
2.4

C
4.0

D
Unpredictable as rate constant is not given

Solution

Initial concentration of $$A$$= $$1M$$
Concentration after $$2$$ hours= $$0.25M$$
Since, the reaction follows first order kinetics; so, time taken to have concentration $$0.5M=4$$ hours

This is half life $$t_{1/2}$$ of the reaction.
$$\Rightarrow$$ $$t_{1/2}=4$$ hours

Now in Flask II
Initial concentration of $$A=0.6M$$
Final concentration of $$A=0.3M$$

Since, the concentration is reduced to half of its initial value so time required for it to happen will be $$t_{1/2}$$ i.e. $$4$$ hours.
Question 7
1 / -0

It takes $$1\ h$$ for a first order reaction to go to $$50$$% completion. The total time required for the same reaction to reach $$87.5$$% completion will be:

A
$$1.75\ h$$

B
$$6.00\ h$$

C
$$3.50\ h$$

D
$$3.00\ h$$

Solution

For first order reaction we have:-
$$K=\cfrac {1}{t} ln \left(\cfrac {a}{a-x}\right)$$ $$- (i)$$

where, $$K$$= rate constant
$$t$$=time
$$a$$= initial concentration of the reactant
$$x$$= amount of reactant reacted in time $$t$$

Case I:- $$t=1$$ hour & $$x=0.5a$$
$$(a-x)=0.5a$$

Putting the values in $$(i)$$:-
$$K=\cfrac {1}{1} ln \left(\cfrac {1}{0.5}\right)= ln 2$$ $$- (ii)$$

Case II:- $$t=?$$ & $$x=0.875a$$
$$(a-x)=0.125a$$

Putting the value in $$(ii)$$:-
$$K=\cfrac {1}{t} ln \left(\cfrac {1}{0.125}\right)=\cfrac {1}{t} ln (8)$$ $$- (iii)$$

From, $$(ii)$$ and $$(iii)$$ $$\Rightarrow ln2=\cfrac {1}{t} ln 8$$
$$\Rightarrow t=\cfrac {ln8}{ln2}=3$$ hours $$[\because ln(2^3)=3ln2]$$
Question 8
1 / -0

Calculate the rate constant for decomposition of ammonium nitrate from the following data;
$$NH_4NO_2\rightarrow N_2+2H_2O$$
Time (minutes) 10 25 $$\infty$$
Vol.of $$N_2$$ ml 6.25 13.65 35.05
Given that the reaction follows first order kinectics.

A
$$0.0197 min^{-1}$$

B
$$1.97 min^{-1}$$

C
$$0.197 min^{-1}$$

D
None of these

Solution

The decomposition of ammonium nitrate is given by
$$NH_4NO_2\longrightarrow N_2+2H_2O$$

The kinetics of this reaction is studied by measuring volume of $$N_2$$ formed at different intervals.

Let, $$V_\infty$$ denotes volume of $$H_2$$ formed at $$\infty$$ time.
(time at which reaction is complete)

$$\Rightarrow$$ $$V_\infty \propto$$ amount of $$NH_4NO_2$$ present initially $$(a)$$
$$\Rightarrow$$ $$V_\infty \propto a$$ $$- (i)$$
Let, $$Vt$$ is the volume of $$N_2$$ at any time $$t$$.

$$\Rightarrow$$ $$Vt\propto$$ amount of $$NH_4NO_2$$ reacted $$(x)$$
$$\Rightarrow$$ $$Vt\propto x$$ $$- (ii)$$

Now, $$(a-x)\propto (V_\infty-Vt)$$ $$- (iii)$$

Now, since this reaction follows first order kinetics, so,
$$K=\cfrac {1}{t} ln \left(\cfrac{a}{a-x}\right)$$ $$- (iv)$$

where, $$K$$= rate constant, $$t$$= time, $$a$$= initial concentration of reactant, $$x$$= amount of reactant reacted in time $$t$$

Putting (i) & (iii) in (iv):-
$$K=\cfrac {1}{t} ln \left(\cfrac {V_\infty}{V_\infty-Vt}\right)$$ $$- (v)$$

According to question, $$V_\infty=a=35.05$$

Now, at $$t=10$$ min, $$V_t=6.25$$ ml

Putting the values in $$(v)$$ :-
$$K=\cfrac {1}{10} ln \left(\cfrac {35.05}{35.05-6.25}\right)$$
$$=\cfrac {1}{10} ln \left(\cfrac {35.05}{20.00}\right)$$
$$=\cfrac {1}{10} ln (1.22)=\cfrac {1}{10}\times 0.1900$$
$$=0.01900$$ $$min^{-1}$$
Question 9
1 / -0

The spontaneous decomposition of radio nuclei is a first order rate process.U-238 disintegrates with the emission of $$\alpha$$-particles and has a half-life of $$4.5 \times 10^9$$ years. If at time t=0, 1 mole of U-238 is present, what will be the number of nuclei left after 1 billion years?

A
$$5.16 \times 10^{23}$$

B
$$5.16 \times 10^{20}$$

C
$$5.16 \times 10^{19}$$

D
None of these

Solution

We know that,
Radioactive disintegration follows first order kinetics.
Now, half life period of a reaction is the time taken by reaction to be half completed.
So, amount of reactant left after one half life period= $$\cfrac {[A]_0}{2}$$ ; $$A_0$$ is the amount present at $$t=0$$

Amount present after $$2$$ half lines= $$\cfrac {[A]_0}{2^2}$$
In general, amount of reactant left after $$n$$ half lives=$$\cfrac {[A]_0}{2^n}$$

Now, Half life of $$U-238=4.5\times 10^{9}$$ yrs
Given time, $$t=1$$ billion years
$$=10^{9}$$ years

So, no. of half lives=$$\cfrac {10^9}{4.5\times 10^9}=\cfrac {1}{4.5}$$
Amount of reactant left= $$\cfrac {[A]_0}{2^{4.5}}$$
=$$\cfrac {[A]_0}{22.68}$$

Now, at $$t=0$$, amount of $$U-238$$ present= $$1$$ mole=$$6.022\times10^{23}$$ molecules (nuclei)

$$\therefore$$ No. of nuclei left after $$1$$ billion years= $$\cfrac {6.022\times 10^{23}}{22.68}$$
$$=0.27\times 10^{23}$$
Question 10
1 / -0

For the equilibrium, $$A(g) \rightleftharpoons B(g), \triangle H$$ is $$-40\ kJ/mol$$. If the ratio of the activation energies of the forward $$(E_{f})$$ and reverse $$(E_{b})$$ reactions is $$2/3$$ then:

A
$$E_{f} = 60\ kJ/mol; E_{b} = 100\ kJ/mol$$

B
$$E_{f} = 30\ kJ/mol; E_{b} = 70\ kJ/mol$$

C
$$E_{f} = 80\ kJ/mol; E_{b} = 120\ kJ/mol$$

D
$$E_{f} = 70\ kJ/mol; E_{b} = 30\ kJ/mol$$

Solution

The given reaction is,
$$A(g)\rightleftharpoons B(g)$$
$$\triangle H= -40 KJ/mol$$
& $$\cfrac {E_f}{E_b}=\cfrac {2}{3} \Rightarrow E_f=\cfrac {2}{3} E_b$$
From graph we have,
$$\triangle H=E_b-E_f$$
$$\Rightarrow -40=E_b-\cfrac {2}{3}E_b$$
$$\Rightarrow -40= \cfrac {1}{3}E_b \Rightarrow E_b=-120 KJ/mol$$
& $$E_f=E_b-\triangle H$$
$$=-120- (-40)$$
$$-80$$ $$KJ/mol$$.

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Time (minutes)	10	25	$$\infty$$
Vol.of $$N_2$$ ml	6.25	13.65	35.05

Chemical Kinetics Test - 66

Result

Chemical Kinetics Test - 66

Score

-

Rank

-

SHARING IS CARING

Question 1

30

40

50

60

Question 2

zero

7.37 k cal/mol

9.212 k cal/mol

11.05 k cal/mol

Solution

Question 3

$$521\ kJ\ mol^{-1}$$

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

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

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

Solution

Question 4

$$k_{2}$$ will increase faster than $$k_{1}$$

$$k_{1}$$ will increase faster than $$k_{2}$$ but will always remain less than $$k_{2}$$

$$k_{1}$$ will increase faster than $$k_{2}$$ and become equal to $$k_{2}$$

$$k_{1}$$ will increase faster than $$k_{2}$$ and become greater than $$k_{2}$$

Solution

Question 5

$$E_1 > E_2$$

$$E_1 < E_2$$

$$E_1=2E_2$$

$$\sqrt{E_1E_2^2}=1$$

Solution

Question 6

0.4

2.4

4.0

Unpredictable as rate constant is not given

Solution

Question 7

$$1.75\ h$$

$$6.00\ h$$

$$3.50\ h$$

$$3.00\ h$$

Solution

Question 8

$$0.0197 min^{-1}$$

$$1.97 min^{-1}$$

$$0.197 min^{-1}$$

None of these

Solution

Question 9

$$5.16 \times 10^{23}$$

$$5.16 \times 10^{20}$$

$$5.16 \times 10^{19}$$

None of these

Solution

Question 10

$$E_{f} = 60\ kJ/mol; E_{b} = 100\ kJ/mol$$

$$E_{f} = 30\ kJ/mol; E_{b} = 70\ kJ/mol$$

$$E_{f} = 80\ kJ/mol; E_{b} = 120\ kJ/mol$$

$$E_{f} = 70\ kJ/mol; E_{b} = 30\ kJ/mol$$

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.