Chemical Kinetics Test

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Chemical Kinetics Test - 62

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Weekly Quiz Competition

Question 1
1 / -0

Reaction $$A+B\longrightarrow C+D$$ follow's following rate law rate $$=k={ \left[ A \right] }^{ \frac { 1 }{ 2 } }{ \left[ B \right] }^{ \frac { 1 }{ 2 } }$$. Starting with initial conc. of one mole of A and B each, what is the time taken for amount of A of become 0.25 mole. Given $$k = 2.31\times 10^{-3} sec^{-1}$$.

A
300 sec

B
600 sec

C
900 sec

D
none of these

Solution

$$A+B\longrightarrow C+D$$

$$rate=k\left[ A \right] ^{ 1/2 }\left[ B \right] ^{ 1/2 }$$

$$\displaystyle \frac { dx }{ dt } =k\sqrt { (a-x)(a-x) } $$

$$ \displaystyle \frac { dx }{ dt } =k(a-x)$$

$$ \Rightarrow t= \displaystyle \frac { 2.303 }{ k } \log_ { 10 } \left( \displaystyle \frac { a }{ a-x } \right) $$

$$ t= \displaystyle \frac { 2.303 }{ 2.31\times 10^{ -3 } } \log_ { 10 } \left( \displaystyle \frac { 1 }{ 0.25 } \right) $$

$$t=600\ sec.$$

Hence, the correct option is $$\text{B}$$
Question 2
1 / -0

The gas phase decomposition of dimethyl ether follows first order kinetics.
$$CH_3OCH_3(g)\, \rightarrow\, CH_4(g)\, +\, H_2(g)\, +\, CO(g)$$
The reaction is carried out in a constant volume container at $$500^\circ C$$ and has a half life of 14.5 min. Initially, only dimethyl ether is present at a pressure of 0.40 atm. What is the total pressure of the system after 12 min? Assume ideal gas behaviour.

A
0.74 atm

B
7.4 atm

C
74 atm

D
none

Solution

For first order reaction,
Half life = $$\displaystyle \frac{0.693}{k}$$
or 14.5 = $$\displaystyle \frac{0.693}{k}$$
or k = $$\displaystyle \frac{0.693}{k}\, =\, 4.78\, \times\, 10^{-2}\, min^{-1}$$
Also,
$$\displaystyle k\, =\, \frac{2.303}{t} log \frac{A_0}{A_t}\, =\, \frac{2.303}{t} log \frac{P_0}{P}$$

or $$\displaystyle \frac{0.693}{14.5}\, =\, \frac{2.303}{12} log \frac{0.40}{(0.40\, -\, x)}$$
or x = 0.1746 atm
Total pressure of container after 12 min is
(0.40 - x) + x + x + x = 0.4 + 2x
= 0.4 + 2 $$\times$$ 0.1746
= 0.7492 atm
Question 3
1 / -0

Catalytic decomposition of nitrous oxide by gold at $$900^{0}C$$ at an initial pressure of 200 mm was 50 % in 53 min and 73 % in 100 min, the value of velocity constant is :

A
$$1.308 \, \times\, 10^{-2}\, min^{-1}$$

B
$$1.432 \, \times\, 10^{-2}\, min^{-1}$$

C
$$2.318 \, \times\, 10^{-2}\, min^{-1}$$

D
None of these

Solution

Case I:
$$a\, \propto\, 200 mm$$
$$x\, \propto\, 200\,\times\, \displaystyle \frac{50}{100}$$
or $$x\, \propto\, 100 mm.$$

$$t_{1/2}\,=\, 53 min.$$

$$k_1\,=\, \displaystyle \frac{2.303}{t}\,log\,\frac{a}{a\,-\,x}$$

$$k_1=\displaystyle \frac{2.303}{53} \,log\,\frac{200}{200 \,- \,100}\,=\, 1.308\,\times\,10^{-2}\,min^{-1}$$

Case II :
a $$\propto$$ 200 mm
$$x$$ $$\propto\, 200\,\times\,\displaystyle \frac{73}{100}$$
or, $$x\, \propto\, 146 mm$$

$$ t_{73 \%}$$ = 100 min

$$k_2\,=\, \displaystyle \frac{2.303}{100}\, log\, \left ( \displaystyle \frac{200}{200\, -\, 146} \right )\, =\, 1.309\,\times\, 10^{-2}\, min^{-1}$$

(b) $$k\,=\, \displaystyle \frac{k_1\,+\,k_2}{2}\,=\, \frac{(1.308\,+\, 1.309)\, \times\, 10^{-2}}{2}$$

$$k=1.308 \, \times\, 10^{-2}\, min^{-1}$$
Question 4
1 / -0

In a first order reaction, $$75\%$$ of the reactants disappeared in 1.386 hr. What is the rate constant?

A
$$3.6\, \times\, 10^{-3}\, s^{-1}$$

B
$$2.7\, \times\, 10^{-4}\, s^{-1}$$

C
$$72\, \times\, 10^{-3}\, s^{-1}$$

D
$$1.8\, \times\, 10^{-3}\, s^{-1}$$

Solution

$$t_{75\%}\, =\, 2t_{1/2}$$

$$\displaystyle \therefore\, t_{1/2}\, =\, \frac{1.386\, hr}{2}\, =\, 0.693\, hr$$

$$\displaystyle k\, =\, \frac{0.693}{t_{1/2}}\,= \frac{0.693}{0.693}\, =\, 1\, hr^{-1}\, =\, \frac{1}{3600s}$$

$$=\, 2.7\, \times\, 10^{-4}\, s^{-1}$$
Question 5
1 / -0

In the following reaction, $$A\rightarrow B$$, rate constant is $$1.2\times10^{-2}\:M\:s^{-1}\:$$. What is concentration of B after 10 min, if we start with 10 M of A?

A
7.2 M

B
14.4 M

C
3.6 M

D
None of these

Solution

$$k=1. 2 \times 10^{–2}Ms^{–1}$$

Since unit of rate constant is $$Ms^{–1}$$, it is zero order
reaction
Rate law $$[B] = kt$$

At t = 10 min

$$[B] = 1. 2 \times 10^{–2} \times 10 \times 60\ M$$

$$[B] = 7. 2\ M$$
Question 6
1 / -0

In a certain reaction, 10% of the reactant decomposes in one hour, 20% in two hours, 30% in three hours, and so on. The dimension of the velocity constant (rate constant) is:

A
$$Hr^{-1}$$

B
$$Mol\, L^{-1}\, hr^{-1}$$

C
$$L\, mol^{-1}\, s^{-1}$$

D
$$Mol\, s^{-1}$$

Solution

For zero-order reaction rate is independent of concentration. Hence, it is a zero-order reaction. So, dimension of $$k =$$ $$mol L^{-1}\, hr^{-1}$$

For zero order: t $$=\displaystyle \frac{x}{k}\, or\, k\, =\, \frac{x}{t}$$

If t $$=t_{10\%}\, =\, 60\, min,\, x\, =\, 10,$$

Then, r $$=\displaystyle \frac{-10}{1}\, =\, 10\, mol\, L^{-1}\, hr^{-1}$$

If t $$=t_{20\%}\, =\, 120\, min,\, x\, =\, 20,$$

Then, r $$=\displaystyle \frac{20}{2}\, =\, 10\, mol\, L^{-1}\, hr^{-1}.$$

Similarly, for 30% and so on.

Thus, the reaction is of zero order.
Question 7
1 / -0

$$A\rightleftharpoons B+C$$
Time $$ t$$ $$\infty $$
Total pressure of $$(B + C)$$ $$P_2$$ $$P_3$$
Find equilibrium constant $$k$$.

A
$$\displaystyle k=\frac{1}{t}ln\frac{P_{3}}{2(P_{3}-P_{2})}$$

B
$$\displaystyle k=\frac{1}{t}ln\frac{P_{2}}{2(P_{3}-P_{2})}$$

C
$$\displaystyle k=\frac{1}{t}ln\frac{P_{3}}{2(P_{3}+P_{2})}$$

D
None of these

Solution

$$a=P_{3}$$
$$x=2P_{2}-P_{3}$$
$$x=P_{3}-2(P_{3}-P_{2})$$
$$a-x=2(P_{3}-P_{2})$$
$$\displaystyle k=\frac{1}{t}ln\frac{a}{a-x}$$
$$\displaystyle k=\frac{1}{t}ln\frac{P_{3}}{2(P_{3}-P_{2})}$$
Question 8
1 / -0

80% of a first order reaction was completed in 70 min. How much it will take for 90% completion of a reaction ?

A
114 min

B
100 min

C
140 min

D
70 min

Solution

According to given conditions,
$$\displaystyle t_{80\%}\, \propto \, log\, \frac{100}{100\, -\, 80}\, =\, log\, \frac{10}{2}$$

$$= log\ 5 = 0.69$$ $$\approx$$ $$0.7$$

$$\displaystyle t_{80\%}\, \propto \, log\, \frac{100}{100\, -\, 90}\, =\, log\, 10\, =\, 1$$

$$\displaystyle \frac{t_{80\%}}{t_{90\%}}\, =\, \frac{0.7}{1}\, \Rightarrow\, \frac{70}{t_{90\%}}\, =\, \frac{0.7}{1}$$

$$\displaystyle \therefore t_{90\%}\, =\, \frac{70}{0.7}\, =\, \frac{700}{7}\, =\, 100\, min$$
Question 9
1 / -0

Unit of frequency factor (A) is:

A
mol/L

B
mol/L.s

C
depend upon order of reaction

D
it does not have any unit

Solution
Question 10
1 / -0

The decomposition of a compound P, at temperature T according to the equation
$$2P_{(g)}\rightarrow4Q_{(g)}+R_{(g)}+S_{(l)}$$ is the first order reaction. After 30 minutes from the start of decomposition in a closed vessel, the total pressure developed is found to be 317 mm Hg and after a long period of time the total pressure observed to be 617 mm Hg. Calculate the total pressure of the vessel after 75 minute, if volume of liquid S is supposed to be negligible. Also calculate the time fraction $$t_{7/8}$$?
Given : Vapour pressure of S(l) at temperature T = 32.5 mm Hg.

A
$$P_{t}=37.955\:mm\:Hg,\:t_{7/8}=39.996\:min$$

B
$$P_{t}=379.55\:mm\:Hg,\:t_{7/8}=399.96\:min$$

C
$$P_{t}=179.55\:mm\:Hg,\:t_{7/8}=199.96\:min$$

D
None of these

Solution

$$2P_{(g)}\rightarrow 4Q_{(g)}+R_{(g)}+S_{(l)}$$
$$t=0$$ $$P_0$$
$$t=30\:min.$$ $$P_{0}-P$$2PP/2
$$t=\infty$$- $$2P_0$$ $$P_0/2$$
so $$P_{0}-P+2P+P/2=317-32.5$$
i.e. $$P_{0}+1.5\:P=284.5 $$.........(i)
& $$2.5\:P_{0}=617-32.5=584.5$$
so$$P_{0}=233.8$$
$$P=33.8$$
$$\displaystyle k\times 30=\ln \frac{233.8}{200}\Rightarrow k=0.0052$$
At$$t=75\:min$$
$$\displaystyle 0.0052\times 75=\ln \frac{233.8}{P^{\circ}-P}$$
$$P^{\circ}-P=158.23\Rightarrow P=75.57$$
$$P_{T}=32.5+P_0+1.5P=347.155+32.5$$
$$P_{T}=379.65\:mm\:Hg$$
(ii) $$0.0052\times t=\ln \:8$$
$$t=399.89\:min.$$

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Re-Attempt Weekly Quiz Competition

Chemical Kinetics Test - 62

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Chemical Kinetics Test - 62

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Question 1

300 sec

600 sec

900 sec

none of these

Solution

Question 2

0.74 atm

7.4 atm

74 atm

none

Solution

Question 3

$$1.308 \, \times\, 10^{-2}\, min^{-1}$$

$$1.432 \, \times\, 10^{-2}\, min^{-1}$$

$$2.318 \, \times\, 10^{-2}\, min^{-1}$$

None of these

Solution

Question 4

$$3.6\, \times\, 10^{-3}\, s^{-1}$$

$$2.7\, \times\, 10^{-4}\, s^{-1}$$

$$72\, \times\, 10^{-3}\, s^{-1}$$

$$1.8\, \times\, 10^{-3}\, s^{-1}$$

Solution

Question 5

7.2 M

14.4 M

3.6 M

None of these

Solution

Question 6

$$Hr^{-1}$$

$$Mol\, L^{-1}\, hr^{-1}$$

$$L\, mol^{-1}\, s^{-1}$$

$$Mol\, s^{-1}$$

Solution

Question 7

$$\displaystyle k=\frac{1}{t}ln\frac{P_{3}}{2(P_{3}-P_{2})}$$

$$\displaystyle k=\frac{1}{t}ln\frac{P_{2}}{2(P_{3}-P_{2})}$$

$$\displaystyle k=\frac{1}{t}ln\frac{P_{3}}{2(P_{3}+P_{2})}$$

None of these

Solution

Question 8

114 min

100 min

140 min

70 min

Solution

Question 9

mol/L

mol/L.s

depend upon order of reaction

it does not have any unit

Solution

Question 10

$$P_{t}=37.955\:mm\:Hg,\:t_{7/8}=39.996\:min$$

$$P_{t}=379.55\:mm\:Hg,\:t_{7/8}=399.96\:min$$

$$P_{t}=179.55\:mm\:Hg,\:t_{7/8}=199.96\:min$$

None of these

Solution

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