Chemical Kinetics Test

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Chemical Kinetics Test - 35

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Question 1
1 / -0

$${ T }_{ 50\% }$$ of first order reaction is $$10$$ min. Starting with $$10$$ $$mol$$ $${ L }^{ -1 }$$, rate after $$20$$ min is:

A
$$0.0693$$ $$mol$$ $${ L }^{ -1 }{ min }^{ -1 }$$

B
$$0.0693\times 2.5$$ $$mol$$ $${ L }^{ -1 }{ min }^{ -1 }$$

C
$$0.0693\times 5$$ $$mol$$ $${ L }^{ -1 }{ min }^{ -1 }$$

D
$$0.0693\times 10$$ $$mol$$ $${ L }^{ -1 }{ min }^{ -1 }$$

Solution

$${ t }_{ 1/2 }=10min\\ k=\cfrac { 0.693 }{ { t }_{ 1/2 } } =\cfrac { 0.693 }{ 10 } mi{ n }^{ -1 }$$
Now,
$$ A={ A }_{ o }{ e }^{ -kt }\\ A=10.{ e }^{ \cfrac { 0.693 }{ 10 } .20 }$$
$$ =10{ e }^{ -2\times 0.693 }\longrightarrow$$ Concentration at $$20min.$$
So,
$$ v=kA=\cfrac { 0.693 }{ 10 } .{ 10e }^{ -2\times 0.693 }\\ =0.0693 \times { 10e }^{ -2\times 0.693 } \\ =0.0693\times 2.5mol\ { L }^{ -1 }mi{ n }^{ -1 }$$
Question 2
1 / -0

A substance reacts with initial concentration of $$a$$ $$mol$$ $${ dm }^{ -3 }$$ according to zero order kinetics. The time it takes for the completion of the reaction is: ($$k=$$ rate constant)

A
$$\dfrac { k }{ a } $$

B
$$\dfrac { a }{ 2k } $$

C
$$\dfrac { a }{ k } $$

D
$$\dfrac { 2k }{ a } $$

Solution

Since rate constant of zero order reaction is $$[A]=[A]_o−kt$$
$$\implies t = \frac{[A]_o−[A]}{k} $$
$$ \implies t = \frac{a - 0}{k}$$
$$ \implies t =\frac{a}{k}$$
Question 3
1 / -0

Under the same reaction conditions, initial concentration of $$1.386$$ $$mol$$ $${ dm }^{ -3 }$$ of a substance becomes half in $$40$$ seconds and $$20$$ seconds through first order and zero order kinetics respectivley. Ratio $$\left( \dfrac { { k }_{ 1 } }{ { k }_{ 0 } } \right) $$ of the rate constant for first order $$\left( { k }_{ 1 } \right) $$ and zero order $$\left( { k }_{ 0 } \right) $$ of the reaction is:

A
$$0.5\ { mol }^{ -1 }{ dm }^{ 3 }$$

B
$$1\ mol\ { dm }^{ -3 }$$

C
$$1.5\ mol\ { dm }^{ -3 }$$

D
$$2\ { mol }^{ -1 }{ dm }^{ 3 }$$

Solution

$${ t }_{ { 1 }/{ 2 } }=\dfrac { 0.693 }{ { k }_{ 1 } } $$, $${ t }_{ { 1 }/{ 2 } }=\dfrac { { a }_{ 0 } }{ 2{ k }_{ 0 } } $$
$$40=\dfrac { 0.693 }{ { k }_{ 1 } } $$, $$20=\dfrac { 1.386 }{ 2{ k }_{ 0 } } =\dfrac { 0.693 }{ { k }_{ 0 } } $$
$$\dfrac { 20 }{ 40 } =\dfrac { { 0.693 }/{ { k }_{ 0 } } }{ { 0.693 }/{ { k }_{ 1 } } } =\dfrac { { k }_{ 1 } }{ { k }_{ 0 } } $$
$$\dfrac { { k }_{ 1 } }{ { k }_{ 0 } } =0.5\dfrac { { sec }^{ -1 } }{ mol{ dm }^{ -3 }{ sec }^{ -1 } } =0.5{ mol }^{ -1 }{ dm }^{ 3 }$$
Question 4
1 / -0

The decomposition of $$HI$$ on the surface of gold is:

A
pseudo first order

B
zero order

C
first order

D
second order

Solution
Question 5
1 / -0

A drop of solution (volume $$0.05$$ $$mL$$) contains $$3\times { 10 }^{ -6 }$$ mole of $${ H }^{ + }$$. If the rate constant of disappearance of $${ H }^{ + }$$ is $${ 10 }^{ 7 }$$ $$mol$$ $${ litre }^{ -1 }$$ $${ sec }^{ -1 }$$, how long would it take for $${ H }^{ + }$$ in the drop to disappear?

A
$$6\times { 10 }^{ -8 }\ sec$$

B
$$6\times { 10 }^{ -9 }\ sec$$

C
$$6\times { 10 }^{ -10 }\ sec$$

D
$$6\times { 10 }^{ -12 }\ sec$$

Solution

$$k={ 10 }^{ 7 }{ molL }^{ -1 }{ sec }^{ -1 }\\ [{ H }^{ + }]=\cfrac { 3\times { 10 }^{ -6 } }{ 0.05\times { 10 }^{ -3 } } { molL }^{ -1 }=\cfrac { 3 }{ 5 } \times { 10 }^{ -1 }{ mol }L^{ -1 }\\ \therefore \ln { \left( \cfrac { { A }_{ o } }{ A } \right) } =kt\\ { t }_{ 99\% }=\cfrac { \ln { \left( \cfrac { { A }_{ o } }{ A } \right) } }{ k } \\ =\cfrac { \ln { \left( \cfrac { 3/5\times { 10 }^{ -1 } }{ 3/5\times { 10 }^{ -3 } } \right) } }{ { 10 }^{ 7 } } \\ =9.212\times { 10 }^{ -3 }sec\\ So,\\ v=k[{ H }^{ + }]=0.6\times { 10 }^{ -1 }\times { 10 }^{ 7 }\\ =0.6\times { 10 }^{ 6 }\\ =6\times { 10 }^{ 5 }{ mol }^{ 2 }{ L }^{ -1 }{ s }^{ -1 }\\ \therefore t=6\times { 10 }^{ -9 }sec$$ for $$\cfrac { 3 }{ 5 } \times { 10 }^{ -1 }mol{ L }^{ -1 }$$
Question 6
1 / -0

The rate constant of a reaction is found to be $$3\times { 10 }^{ -3 }$$ $$mol$$ $${ L }^{ -1 }$$ $${ min }^{ -1 }$$. The order of the reaction is:

A
$$0$$

B
$$1$$

C
$$2$$

D
$$1.5$$

Solution

Since the rate constant of a zero-order reaction is Rate = K.
The unit of rate constant will be $$mol . L^{-1} min^{-1}.$$
Question 7
1 / -0

The hydrolysis of ethyl acetate is a reaction of:
$$C{ H }_{ 3 }COO{ C }_{ 2 }{ H }_{ 5 }+{ H }_{ 2 }O\xrightarrow [ ]{ \quad { H }^{ + }\quad } C{ H }_{ 3 }COOH+{ C }_{ 2 }{ H }_{ 5 }OH$$

A
zero order

B
first order

C
second order

D
third order

Solution

$$\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad { H }^{ + }\\ { CH }_{ 3 }{ COOC }_{ 2 }{ H }_{ 5 }+{ H }_{ 2 }O\longrightarrow { CH }_{ 3 }COOH+{ C }_{ 2 }{ H }_{ 5 }OH$$
In the above reaction medium of reaction is water which will be in excess, hence it can be avoided in rate expression, thus rate will be given as
$$Rate=K\left[ { CH }_{ 3 }{ COOC }_{ 2 }{ H }_{ 5 } \right] $$
Hence it is a $${ 1 }^{ st }$$ order reaction.
Question 8
1 / -0

Graph between the concentration of the product '$$x$$' and time '$$t$$' for $$A\rightarrow B$$ is given above.
The graph between $$-\dfrac { d\left[ A \right] }{ dt } $$ and time $$'t'$$ will be of the type:

A

B

C

D

Solution

For a zero order reaction
Rate of reaction=-$$\frac{d[A]}{dt}$$=$$K(a-x)^0$$=K
So the correct answer is (C) because there is no change in rate w.r.t time.
Question 9
1 / -0

For a zero order reaction, $$A\longrightarrow P$$, $${ t }_{ { 1 }/{ 2 } }$$ is:
($$k$$ is the rate constant, $$[A]_0$$ is the initial concentration of $$A$$)

A
$$\dfrac { { \left[ A \right] }_{ 0 } }{ 2k } $$

B
$$\dfrac { \ln { 2 } }{ k } $$

C
$$\dfrac { 1 }{ k{ \left[ A \right] }_{ 0 } } $$

D
$$\dfrac { \ln { 2 } }{ { \left[ A \right] }_{ 0 }k } $$

Solution

For zero order
$$ kt=a\\ { t }_{ 1/2 }=\cfrac { { a }_{ o } }{ 2k } =\cfrac { { \left[ A \right] }_{ o } }{ 2k } $$
Question 10
1 / -0

Following is the graph between $$\log{{ t }_{ { 1 }/{ 2 } }}$$ and $$\log { a } $$ ($$a=$$ initial concentration) for a given reaction at $${ 27 }^{ o }C$$. Hence, order of the reaction is:

A
$$0$$

B
$$1$$

C
$$2$$

D
$$3$$

Solution

$$\cfrac { 1 }{ { a }^{ n-1 } } =\cfrac { 1 }{ { a }_{ 0 }^{ n-1 } } +(n-1)kt$$
for $$n=0$$
$$ \cfrac { 1 }{ { a }^{ -1 } } =\cfrac { 1 }{ { a }_{ 0 }^{ -1 } } +(-1)kt\\ a={ a }_{ o }-kt\\ { kt }_{ 1/2 }=\cfrac { { a }_{ o } }{ 2 } \\ \log { { (t }_{ 1/2 })= } \log { ({ 0 }_{ o }) } -\log { \left( \cfrac { 1 }{ 2k } \right) } $$
So,
$$\log { { (t }_{ 1/2 }) } =\log { ({ 0 }_{ o }) } +\log { (2k) } $$
Which is a straight line as given.