Chemical Kinetics Test

Result

Chemical Kinetics Test - 60

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

The decomposition of a compound $$P$$, at temperature $$T$$ according to the equation $$\displaystyle2P_{(g)}\rightarrow4Q_{(g)}+R_{(g)}+S_{(l)}$$ 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 $$Hg$$.
The total pressure of the vessel after $$75$$ minute, if the volume of liquid $$S$$ is supposed to be negligible is:
(Given : Vapour pressure of $$S (l)$$ at temperature $$T=32.5 $$mm $$Hg$$)

A
$$P_{t}=379.55$$ mm $$Hg$$

B
$$P_{t}=387.64$$ mm $$Hg$$

C
$$P_{t}=468.23$$ mm $$Hg$$

D
$$P_{t}=489.44$$ mm $$Hg$$

Solution

After long time, total pressure is 617 mm Hg.
$$5P=617-32.5$$
$$2P=255.4$$ .....(1)
After 30 min, $$255+4x=317-32.5$$
$$x=7.375$$......(2)
The expression for the rate of the reaction is $$k=\frac {2.303} {t} log {a}{(a-x)}=\frac {2.303} {30} log ({255}\times {(247.62)})$$......(3)
After 75 min, $$k=\frac {2.303} {t} log {a}{(a-x)}=\frac {2.303} {(75)} log( {255}\times {(255-y)})$$......(4)
From (3) and (4), $$y=23$$
Total pressure $$= 255+4y+32.5=255+4(23)+32.5=379.55\ mm\ Hg$$
Question 2
1 / -0

The decomposition of $$ {\mathrm{N}_{2}}\mathrm{O}_{5} $$ according to the equation $${\mathrm{2N}_{2}}\mathrm{O}_{5}(g)\rightarrow \mathrm{4 NO}_{2}+O_{2}(g) $$ is a first-order reaction. After 30 min from the start of decomposition in a closed vessel the total pressure developed is found to be 284.5 mm Hg. On complete decomposition, the total pressure is 584.5 mm Hg. The rate constant of the reaction is :

A
$$\mathrm{k}_{1} =5.206\times10^{-3}\ \mathrm{min}^{-1}$$

B
$$\mathrm{k}_{1} =3.102\times10^{-3}\ \mathrm{min}^{-1}$$

C
$$\mathrm{k}_{1} =3.126\times10^{-3}\ \mathrm{min}^{-1}$$

D
$$\mathrm{k}_{1} =3.453\times10^{-3}\ \mathrm{min}^{-1}$$

Solution

$$2{ \mathrm{N} }_{ 2 }{ \mathrm{O} }_{ 5 }\longrightarrow \mathrm{4N}{ \mathrm{O} }_{ 2 }+{ \mathrm{O} }_{ 2 }$$

On decomposition of $$2$$ moles of $${ \mathrm{N} }_{ 2 }{ \mathrm{O} }_{ 5 }$$, $$4$$ moles of $$\mathrm{N}{ \mathrm{O} }_{ 2 }$$ and $$1$$ mole of $${ \mathrm{O} }_{ 2 }$$ are produced. Thus, the total pressure after completion corresponds to $$5$$ moles and initial pressure to $$2$$ moles.

Initial pressure of $${ \mathrm{N} }_{ 2 }{ \mathrm{O} }_{ 5 }$$, $${ \mathrm{p }}_{ 0 }=\dfrac { 2 }{ 5 } \times 584.5=233.8$$ mm Hg

After $$30$$ minutes, the total pressure $$=284.5$$ $$mm$$ $$Hg$$

$$\begin{matrix} 2{ N }_{ 2 }{ O }_{ 5 } \\ { p }_{ 0 }-2p \end{matrix}\begin{matrix} \longrightarrow \\ \end{matrix}\begin{matrix} 4N{ O }_{ 2 } \\ 4p \end{matrix}\begin{matrix} + \\ \end{matrix}\begin{matrix} { O }_{ 2 } \\ p \end{matrix}$$

or $${ p }_{ 0 }+3p=284.5$$

or $$3p=284.5-233.8=50.7$$ mm Hg

or $$p=\dfrac { 50.7 }{ 3 } =16.9$$ mm Hg

Pressure of $${ \mathrm{N} }_{ 2 }{ \mathrm{O} }_{ 5 }$$ after $$30$$ minutes $$=233.8-\left( 2\times 16.9 \right) $$

$$=200$$ mm Hg

$$\mathrm{k}=\dfrac { 2.303 }{ 30 } \log _{ 10 }{ \dfrac { 233.8 }{ 200.0 } } $$

$$=5.206\times { 10 }^{ -3 }{ \mathrm{min} }^{ -1 }$$

Hence, the correct option is $$\text{A}$$
Question 3
1 / -0

The reaction is given below, the rate constant for disappearance of A is $$7.48\times10^{-3}sec^{-1}$$. The time required for the total pressure in a system containing A at an initial pressure of 0.1 atm to rise to 0.145 atm is:
$$2A(g)\rightarrow 4B(g)+C(g)$$

A
0.80 min

B
0.567 min

C
0.433 min

D
0.344 min

Solution

Its a first order reaction and for this,
$$kt = 2.303\ log\left( \dfrac{a}{a-x}\right)$$

Here,
$$2A(g)\rightarrow 4B(g)+C(g)$$
P - -
P-2x 4x x

So here, P-2x+4x+x= 0.145
From given condition 2x is 0.03 and $$t = \frac {2.303}{k}\ log\left( \dfrac{p}{p-2x}\right) = 47.63\ seconds = 0.80$$ min.
Question 4
1 / -0

The reaction $$A(aq) \rightarrow B(aq)+C(aq)$$ is monitored by measuring optical rotation of reaction mixture at different time interval. The species A, B and C are optically active with specific rotations $$20^{\circ}, 30^{\circ}$$ and $$-40^{\circ} $$ respectively. Starting with pure A if the value of optical rotation was found to be $$2.5^{\circ}$$ after $$6.93^{\circ}$$ minutes and optical rotation was $$-5^{\circ}$$ after infinite time. Find the rate constant for the first-order conversion into A into B and C:

A
$$0.1 min^{-1}$$

B
$$0.2 min^{-1}$$

C
$$0.3 min^{-1}$$

D
$$0.4 min^{-1}$$

Solution

As after completion of the reaction rotation is $$-5 ^{\circ} $$So according to first order reaction$$k = 2.303log(r_0 - r_{\infty })/(r_t - r_{\infty })/6.93 = 0.1/min$$
Question 5
1 / -0

$$A(aq)\longrightarrow B(aq)+C(aq)$$ is a first order reaction.
Time
$$t$$
$$\infty$$
moles of reagent
$${ n }_{ 1 }$$
$${ n }_{ 2 }$$
Reaction progress is measure with help of titration '$$R$$'. If all $$A,B$$ and $$C$$ reacted with reagent and have '$$n$$' factors [$$n$$ factor; $$eq.wt=\cfrac {mol.wt.}{n}$$] in the ratio of $$1:2:3$$ with the reagent. The $$k$$ in terms of $$t,{ n }_{ 1 }$$ and $${ n }_{ 2 }$$ is :

A
$$k=\cfrac { 1 }{ t } \ln { \left( \cfrac { { n }_{ 2 } }{ { n }_{ 2 }-{ { n }_{ 1 } } } \right) } $$

B
$$k=\cfrac { 1 }{ t } \ln { \left( \cfrac { 2{ n }_{ 2 } }{ { n }_{ 2 }-{ { n }_{ 1 } } } \right) } $$

C
$$k=\cfrac { 1 }{ t } \ln { \left( \cfrac {4 { n }_{ 2 } }{ { n }_{ 2 }-{ { n }_{ 1 } } } \right) } $$

D
$$k=\cfrac { 1 }{ t } \ln { \left( \cfrac {4 { n }_{ 2 } }{5( { n }_{ 2 }-{ { n }_{ 1 }) } } \right) } $$

Solution

The expression for the rate constant for the first order reaction is $$k=\cfrac { 1 }{ t } \ln { \left( \cfrac {[A]_0 }{[A]_t } \right) }$$
But $${ \left( \cfrac {[A]_0 }{[A]_t } \right) }={ \left( \cfrac {4 { n }_{ 2 } }{5( { n }_{ 2 }-{ { n }_{ 1 }) } } \right) }$$
The expression for the rate constant for the first order reaction becomes $$k=\cfrac { 1 }{ t } \ln { \left( \cfrac {4 { n }_{ 2 } }{5( { n }_{ 2 }-{ { n }_{ 1 }) } } \right) }$$
Question 6
1 / -0

Two first order reaction have half-lives in the ratio $$8:1$$. Calculate the ratio of time intervals $${ t }_{ 1 }$$ and $${ t }_{ 2 }$$ are the time period for the $${ \left( \cfrac { 1 }{ 4 } \right) }^{ th }$$ and $${ \left( \cfrac { 3 }{ 4 } \right) }^{ th }$$ completion.

A
$$1:0.301$$

B
$$0.125:0.602$$

C
$$1:0.602$$

D
none of these

Solution

As their half-lives in the ratio $$8:1$$ so their rate constant ratio is $$1:8$$ so ratio of time taken is $$t_1/t_2 = \cfrac { (2.303/k_1)log(1/(1-1/4)) }{(2.303/k_2)log(1/(1-3/4)) } = \cfrac {8 \times log(4/3) }{log4 } = 1 : 0.602 $$
Question 7
1 / -0

$$\displaystyle \:A+B\rightleftharpoons AB+I\xrightarrow{k_{2}}P+A$$
If $$k_1$$ is the rate constant of the reversible step and If $$k_1$$is much smaller than $$k_{2}$$, The most suitable qualitative plot of potential energy (P.E.) versus reaction coordinate for the above reaction.

A

B

C

D

Solution

As k1 is much smaller than k2, Activation energy of the second step will be high. As it is not mentioned that the reaction is exothermic or endothermic hence answer would be Option A and B.
If it would have mentioned endothermic reaction then then the energy of product should be lesser than reactant and in such cases answer would be Option B.
Question 8
1 / -0

Assertion: In a reversible endothermic reaction, $$E_{act}$$ of forward reaction is higher than that of backward reaction.
Reason: The threshold energy of forward reaction is more than that of backward reaction.

A
Both Assertion and Reason are true and Reason is the correct explanation of Assertion

B
Both Assertion and Reason are true but Reason is not the correct explanation of Assertion

C
Assertion is true but Reason is false

D
Assertion is false but Reason is true

E
Both Assertion and Reason are false
Question 9
1 / -0

The gaseous decomposition reaction, $$A(g)\longrightarrow 2B(g)+C(g)$$ is observed to first order the excess of liquid water at $${ 25 }^{ o }C$$. It is found that after $$10$$ minutes the total pressure of system is $$188$$ torr and after very long time it is $$388$$ torr. The rate constant of the reaction (in $$hr^{ -1 }$$) is:
[Given: vapour pressure of $${ H }_{ 2 }O$$ at $${ 25 }^{ o }C$$ 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
none of these

Solution

By putting values in equation $$k=\dfrac{2.303}{t}log\dfrac { P_0 }{ (P_0 - x) } = \dfrac{2.303}{t}log\dfrac { P_\infty/3 }{ (P_\infty /3 - (P_t - P_0)) }$$

We will get answer as $$k=1.2 hr^{-1}$$

$$(A, B, C)$$ let the initial pressure of $$A$$ be $$P_s$$ mm of $$Hg$$ and the pressure of $$A$$ decreases in 10 minute be $$x$$ unit.

$$\begin{matrix} & & \\ &A(g)&\longrightarrow &2B(g)&+&C(g) \\ \text{Initially} & P_0 &&0&&0\\\text{At time 10 min} &P_0-x & &2x& &x \\\text{At}\, \infty \,\text{time} & 0 &&2P_0& &P_0\end{matrix}$$
After long time interval, $$P_{\infty} = 2P_0 + P_0 = 3P_0$$
After $$t$$ time $$P_t = P_0 + 2x$$
For a first order reaction, the rate constant expression would be
$$k=\dfrac{2.303}{t}\log \dfrac{P_0}{P_0-x}$$
Question 10
1 / -0

For a first-order homogeneous gaseous reaction, $$A\longrightarrow 2B+C$$.
If the total pressure after time $$t$$ was $${ P }_{ t }$$ and after long time $$(t\rightarrow \infty )$$ was $${ P }_{ \infty }$$ then $$k$$ in terms of $${ P }_{ t },{ P }_{ \infty }$$ and $$t$$ is :

A
$$k=\cfrac { 2.303 }{ t } \log { \left( \cfrac { 2{ P }_{ \infty } }{ { P }_{ \infty }-{ P }_{ t } } \right) } $$

B
$$k=\cfrac { 2.303 }{ t } \log { \left( \cfrac { 2{ P }_{ \infty } }{2( { P }_{ \infty }-{ P }_{ t }) } \right) } $$

C
$$k=\cfrac { 2.303 }{ t } \log { \left( \cfrac {2 { P }_{ \infty } }{3( { P }_{ \infty }-{ P }_{ t }) } \right) } $$

D
none of these

Solution

                         $$A \rightarrow 2B + C$$
$$t = 0$$           $$P_{i}$$            $$0$$          $$0$$
$$t$$                 $$P_{t} - x$$       $$2x$$        $$x$$
$$t \rightarrow \infty$$       $$0$$       $$2P_{i}$$       $$P_{i}$$

$$P_{\infty} = 3P_{i}$$ or $$P_{i} = \dfrac{P_{\infty}}{3}; P_{i} + 2x = P_{t}$$
$$x = \dfrac{P_t - P_i}{2}$$
As we know $$k = \dfrac{2.303}{t} log \left ( \dfrac{P_i}{P_i - x} \right )$$

so $$k = \dfrac{2.303}{t} log \left ( \dfrac{2P_{\infty}}{3(P_{\infty} - P_t)} \right )$$
Hence optoin (C) is correct.