Chemical Kinetics Test

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Chemical Kinetics Test - 56

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Weekly Quiz Competition

Question 1
1 / -0

Select the correct statements out of I, II and III for zero order reaction.
I: Quantity of the product formed is directly proportional to time.
II: Larger the initial concentration of the reactant, greater the half-life period.
III: If 50% reaction takes place in 100 minutes, 75% reaction take place in 150 minutes.

A
I only

B
I and II only

C
II and III only

D
I, II and III

Solution

We know for zeroth order Rxn
For (I) $$x=kt$$
For (II) $$t_{1/2} \propto [A_0]$$ [Initial concentration]
For (III) $$\displaystyle A_0- \frac {A_0}{2} = k \times 100 .... (i)$$
$$\displaystyle A_0 - \frac{A_0}{4} = k \times t .... (ii)$$
$$\displaystyle \implies \frac {\frac{A_0}{2}}{(\frac{3A_0}{4})} = \frac{k \times 100}{k \times t}$$

$$\displaystyle t= \frac{100 \times 3}{2} = 150$$ minutes
Question 2
1 / -0

For a reaction $$P \rightarrow Q$$, the half-life of the reaction was 3h, when the initial concentration of P was 0.5M. As the concentration of P was increased to 1.0M, half life changes to 6h. The order of reaction with respect to P is:

A
zero

B
one

C
two

D
three

Solution

We know the relation between half time and concentration i.e.

$$t_{\frac{1} {2}} \propto \displaystyle \frac {1}{a^{(n-1)}}$$ where $$n$$ is the order of reaction

$$t_{\frac{1} {2}} = k \displaystyle \frac {1}{a^{(n-1)}}$$

$$(t_{\frac{1} {2}})_1 \times a_1^{(n-1)}=(t_{\frac{1} {2}})_2 \times a_2^{(n-1)}$$

$$3\times(0.5)^{n-1}=6(1)^{n-1}$$

$$(0.5)^{n-1}=2$$

$$2^{1-n}=2^1$$

So comparing both the sides, we get

$$1-n=1$$

$$n=0$$

Hence, it is a zero-order reaction.
Question 3
1 / -0

For the $$1^{st}$$ order reaction, $$A(g) \rightarrow 2B(g) + C(s),$$ $$t_{1/2}= 24 $$ min. The reaction is carried out by taking a certain mass of 'A' enclosed in a vessel in which it exerts a pressure of 400 mm Hg. The pressure of the reaction mixture after the expiry of 48 min will be:

A
700 mm

B
600 mm

C
500 mm

D
1000 mm

Solution

$$A(g) \longrightarrow 2B(g)+ C(s)$$
$$P_{0}$$ $$0$$
$$(P_{0} - P)$$ $$2P$$

$$ t_{1/2} = 24\ min; \quad k=\dfrac{0.693}{t_{1/2}}$$

$$k = \dfrac{2.303}{t} log \dfrac{P_{0}}{P_{0}-P}$$

$$\dfrac{0.693}{24}= \dfrac{2.303}{48} log \dfrac{400}{400-P}$$

$$4 = \dfrac{400}{400-P}$$

$$400-P = 100$$

$$P = 300$$

$$\therefore$$ The pressureof the reaction mixture after expiry time $$=P_{0} + P = 400+300=700.$$ mm
Question 4
1 / -0

The gaseous decomposition reaction: $$A(g)\rightarrow 2B(g) + C(g)$$, is observed to first order over the excess of liquid water at $$25^oC$$. It is found that after 10 minutes, the total pressure of the system is 188 torr and after a very long time it is 388 torr. Calculate the rate constant of the reaction in $$hr^{-1}$$. The vapour pressure of $$H_2O$$ at $$25^oC$$ is 28 torr. $$[ln 2 = 0.7,\: ln 3 = 1.1, \: ln 10 = 2.3]$$

A
$$0.02$$

B
$$1.2$$

C
$$0.2$$

D
$$0.12$$

Solution

$$A(g)\rightarrow 2B (g) + C(g)$$

Let the initial pressure be $$P_0$$ 0 0
After 10 mins, $$(P_0 - x)$$ $$2x$$ $$ x$$
After long time $$(t\rightarrow \infty )$$ 0 $$2P_{0} $$ $$ P_{0}$$

Now,
$$(P_0 - x) + 2x + x +$$ vapour pressure of $$H_2O = 188$$.

$$P_0 + 2x = 160$$ and $$3P_0 + 28 = 388$$

So, $$P_0=120$$ and $$x =20 \: torr$$.

$$k=\cfrac{1}{t}ln\left ( \cfrac{P_{0}}{P_{0}-x} \right )=0.02\: min^{-1}=1.2 \: hr^{-1}.$$
Question 5
1 / -0

Given $$X \rightarrow$$ product (Taking $$1^{st}$$ order reaction)
conc of $$X$$
(mol/lit.) $$0.01$$ $$0.0025$$
Time (min.) $$0$$ $$40$$
Half life period of this reaction is :

A
$$0\ min$$

B
$$20\ min$$

C
$$40\ min$$

D
$$\sqrt{20}\ min$$

Solution

Initial concentration is 0.01 M. The concentration after 40 min is 0.0025 M.

Ratio of concentrations $$=\frac {0.0025} {0.01}=\frac {1} {4}.$$

Thus in 40 minutes, the concentration is reduced to one fourth. This corresponds to two half-life periods.

Thus the half-life period of the reaction is 20 min.
Question 6
1 / -0

Two substances A $$(t_{\frac 12} =$$ 5 min) and B $$(t_{\frac 12} =$$ 15 min) are taken in such a way that initially $$[A] = 4[B]$$. The time after which both the concentrations will be equal is:

A
5 minutes

B
15 minutes

C
20 minutes

D
concentrations can never be equal

Solution

C$$_{t}$$ = C$$_{0}$$ e$$^{-Kt}$$

According to the question, $$C_{A_t}=C_{B _t}$$

$$C_{A} e^{{-K_A}t}=C_{B} e^{{-k_B}t}$$

$$\dfrac{C_{A}}{C_{B}}=\dfrac{e^{-k_{A}t}}{e^{-K_{A}t}}\Rightarrow
\dfrac{C_{A}}{C_{B}}=e(K_{A}-K_{B})t$$

$$4=e\left [ \dfrac{ln2}{5}- \dfrac{ln2}{15}\right ]\times t$$

$$ln4=\left [ \dfrac{ln2}{5}- \dfrac{ln2}{15}\right ]$$t

$$ln(2)^{2}=\left [ \dfrac{ln2}{5}- \dfrac{ln2}{15}\right ]$$t

$$2ln2=\left [ \dfrac{ln2}{5}- \dfrac{ln2}{15}\right ]$$t

$$2=\left [ \dfrac{1}{5}- \dfrac{1}{15}\right ]$$t

$$2=\dfrac{2}{15}\times t$$

$$t = 15\ minutes$$

Hence, option B is correct.
Question 7
1 / -0

It takes 32 minutes to complete 99% of a first order reaction from start. Calculate the time required (in a minute) to complete 99.9% of the reaction from the start?

A
50

B
48

C
55

D
46

Solution

It takes 32 minutes to complete 99% of a first order reaction.

$$K=\frac{1}{32}ln\frac{a}{0.01a}=\frac{ln 100}{32}$$

$$t=\frac{32}{ln 100}ln\frac{a}{0.01a}$$

$$t=48 min$$
Question 8
1 / -0

The conversion of vinyl allyl ether to pent-4-enol follows first-order kinetics. The following plot is obtained for such a reaction. Determine rate constant for the reaction.

A
$$4.6 \times 10^{-2}s^{-1}$$

B
$$1.2 \times 10^{-2}s^{-1}$$

C
$$2.3 \times 10^{-2}s^{-1}$$

D
$$8.4 \times 10^{-2}s^{-1}$$

Solution

$$ Slope = \frac{k}{-2.303} $$
$$ \implies k = Slope \times (-2.303)$$
$$ \implies k = \frac{-6}{10 \times 60} \times (-2.3) $$
$$ \implies k= 2.3 \times 10^{-2} s^{-1} $$
Question 9
1 / -0

Half-life is independent of the concentration of the reactant. After $$10$$ minutes, the volume of $$N_{2}$$ gas is $$10\ L$$ and after complete reaction, it is $$50\ L$$. Hence, the rate constant is:

A
$$(2.303 /10)\ log\ 5$$ $$min^{-1}$$

B
$$(2.303 /10)\ log\ 1.25\ min^{-1}$$

C
$$(2.303 /10)\ log\ 2\ min^{-1}$$

D
$$(2.303 /10)\ log\ 4\ min^{-1}$$

Solution
Question 10
1 / -0

In the reaction, $$2N_2O_5\rightarrow 4NO_2+O_2$$, initial pressure is $$500\ atm$$ and rate constant $$k$$ is $$3.38\times 10^{-5} sec^{-1}$$. After 10 minutes the final pressure of $$N_2O_5$$ is:

A
490 atm

B
250 atm

C
480 atm

D
420 atm

Solution

$$2N_2O_5\rightarrow 4NO_2+O_2$$,
$$p_0$$
$$p_0-2x$$ $$4x$$ $$x$$

$$k=\dfrac {2.303}{t} log_{10}\dfrac {p_0}{p_1}$$

$$3.38\times 10^{-5}=\dfrac {2.303}{10\times 60} log \dfrac {500}{p_t}$$
$$p_t=490\ atm$$

Hence, option A is correct.

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conc of $$X$$ (mol/lit.)	$$0.01$$	$$0.0025$$
Time (min.)	$$0$$	$$40$$

Chemical Kinetics Test - 56

Result

Chemical Kinetics Test - 56

Score

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SHARING IS CARING

Question 1

I only

I and II only

II and III only

I, II and III

Solution

Question 2

zero

one

two

three

Solution

Question 3

700 mm

600 mm

500 mm

1000 mm

Solution

Question 4

$$0.02$$

$$1.2$$

$$0.2$$

$$0.12$$

Solution

Question 5

$$0\ min$$

$$20\ min$$

$$40\ min$$

$$\sqrt{20}\ min$$

Solution

Question 6

5 minutes

15 minutes

20 minutes

concentrations can never be equal

Solution

Question 7

50

48

55

46

Solution

Question 8

$$4.6 \times 10^{-2}s^{-1}$$

$$1.2 \times 10^{-2}s^{-1}$$

$$2.3 \times 10^{-2}s^{-1}$$

$$8.4 \times 10^{-2}s^{-1}$$

Solution

Question 9

$$(2.303 /10)\ log\ 5$$ $$min^{-1}$$

$$(2.303 /10)\ log\ 1.25\ min^{-1}$$

$$(2.303 /10)\ log\ 2\ min^{-1}$$

$$(2.303 /10)\ log\ 4\ min^{-1}$$

Solution

Question 10

490 atm

250 atm

480 atm

420 atm

Solution

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