Chemical Kinetics Test

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Chemical Kinetics Test - 48

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Weekly Quiz Competition

Question 1
1 / -0

For an elementary reaction $$X(g)\rightarrow Y(g) +Z(g)$$, the half-life period is $$10$$ min. In what period would the concentration of $$X$$ be reduced to $$10\%$$ of original concentration?

A
$$20$$Min

B
$$33$$Min

C
$$15$$Min

D
$$25$$Min

Solution

It is a 1st Order Reaction.

$${t}_{1/2}$$ = $$\dfrac{0.693}{k}$$

$$10$$ = $$\dfrac{0.693}{k}$$

$$k= 0.0693\space {min}^{-1}$$

$$t = \dfrac{2.303}{0.0693} log \dfrac{100}{10}$$

$$t = 33\space {min}$$

Hence, the correct option is $$B$$
Question 2
1 / -0

$$A + B \longrightarrow C; \Delta H = + 60 KJ/mol$$
$$E_{at}$$ is $$150$$ kj. What is the activation energy for the backward reation?

A
$$210$$kJ

B
$$105$$kJ

C
$$90$$kJ

D
$$145$$kJ

Solution

The Reaction is a Endothermic Reaction. So,

$$[{E}_{a(f)}] - [{E}_{b(f)}] = \Delta H$$

$$[{E}_{b(f)}] = 150 - 60\space kJ$$ = $$90\space kJ$$
Question 3
1 / -0

An endothermic reaction $$A \rightarrow B $$ have an activation energy 15 kcal/mol and the heat of reaction is 5 kcal/mol. The activation energy of the reaction $$B \rightarrow A $$ is:

A
20 kcal/mol

B
15 kcal/mol

C
10 kcal/mol

D
zero

Solution

Its a Endothermic Reaction, So

$$\Delta{H} = [{E}_{a(f)}] - [{E}_{b(f)}]$$

Given, $$[{E}_{a(f)}] = 15kcal/mol$$ and $$\Delta H = 5\space {kcal}/mol$$

$$[{E}_{b(f)}] = 15 - 5 = 10\space kcal/mol$$
Question 4
1 / -0

What is the half - life of a radioactive substance if $$75\%$$ of any given amount of the substance disintegrates in $$60$$ minutes?

A
$$2$$ Hours

B
$$30$$ Minutes

C
$$45$$ Minutes

D
$$20$$ Minutes

Solution

$$k = \dfrac{2.303}{t} log \dfrac{100}{100 - x}$$

Substituting the values,

$$k = \dfrac{2.303}{60} log \dfrac{100}{25}$$

$$k = 0.023\space {min}^{-1}$$

$${t}_{1/2} = \dfrac{0.693}{k}$$

$${t}_{1/2} = 30\space min$$

Hence, the correct option is $$B$$
Question 5
1 / -0

What fraction of a reactant showing first order remains after 40 minute if $$t_{1/2}$$ is 20 minute?

A
1/4

B
1/2

C
1/8

D
1/6

Solution

Its a 1st Order Reaction,
Taking the Initial Concentration as 100 (For Convenience)

$${t}_{1/2} = \dfrac{0.693}{k}$$

$$k = 0.03465\space {min}^{-1}$$
Now, $$k=\dfrac{2.303}{t}log{\dfrac{a}{a-x}}$$
$$0.03465=\dfrac{2.303}{40}log{\dfrac{a}{a-x}}$$
$$\dfrac{a}{a-x}=\dfrac{1}{4}$$
Question 6
1 / -0

Let there be as first-order reaction of the type, $$ A\rightarrow B+C $$. Let us assume that only A is gaseous. We are required to calculate the value of rate constant based on the following data.
Time 0 T $$ \infty $$
Partial pressure of A $$ P_{0} $$ $$ P_{t} $$ -

A
$$ k=\frac{1}{t}In(\frac{P_{0}}{P_{t}})$$

B
$$ k=\frac{1}{t}In(\frac{P_{t}}{P_{0}})$$

C
$$ k=\frac{1}{t}In(\frac{2P_{0}}{P_{t}})$$

D
$$ k=\frac{1}{t}In(\frac{P_{t}}{2P_{0}})$$

Solution

Its a 1st Order Reaction.

At t = 0 , Pressure is $${P}_{0}$$
At t = T , Pressure is $${P}_{t}$$

So,

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

$$k = \dfrac{1}{t} log \dfrac{{P}_{0}}{{P}_{t}}$$
Question 7
1 / -0

Two identical balls A & B having velocity of 0.5 m/s & -0.3 m/s respectively. colloid elastically in 1-0. The velocity of A & B after collision will be?

A
-0.5 m/s, 0.3 m/s

B
-0.3 m/s,0.5 m/s

C
0.5 m/s, 0.3 m/s

D
0.3 m/s,0.5 m/s

Solution
Question 8
1 / -0

$$99\%$$ of a first - order reaction was completed in $$32$$ min When will $$99.9\%$$ of the reaction complete?

A
$$50$$Min

B
$$46$$Min

C
$$49$$Min

D
$$48$$Min

Solution

$$k = \dfrac{2.303}{t} log \dfrac{100}{100 - x}$$

Substituting the values,

$$k = \dfrac{2.303}{32} log \dfrac{100}{1}$$

$$k = 0.144\space {min}^{-1}$$

$$t = \dfrac{2.303}{0.144} log \dfrac{100}{0.1}$$

$$t = 48\space min$$

Hence, the correct option is $$D$$
Question 9
1 / -0

A carbon sample from the frame of picture gives 7 counts of $$C^{14}$$ per minute per gram of carbon. If freshly cut wood gives 15.3 counts of $$C^{14}$$ per minute, calculate the age of frame. ($$t_{1/2} \ of \ C^{14}$$ = 5570 years)

A
6286 years

B
5527 years

C
5570 years

D
4570 years

Solution

For $$1^{st}$$ order radio active disintegration,
$$K=\cfrac{1}{t}ln\cfrac{a_{\circ}}{a_t}$$
here, $$a_{\circ}=15.3$$ counts per minute,
$$a_t=7$$ counts per minute, $$K=\cfrac{ln2}{t_{\frac{1}{2}}}$$
$$\therefore\cfrac{ln2}{5570}=\cfrac{1}{t}ln\cfrac{15.3}{7}$$
$$\therefore t=$$age of frame$$\approx 6286$$ years.
Question 10
1 / -0

$$K_{34^{\circ}\ C} : K_{35^{\circ}\ C} < 1$$, then:

A
rate increases with the rise in temperature

B
rate decreases with rise in temperature

C
rate does not change with rise in temperature

D
none of the above

Solution

According to Arrhenius Equation
$$K \propto e^{-Ea/RT}$$ or $$K \propto \cfrac {1}{e^{Ea/RT}}$$
$$lnK \propto \cfrac {RT}{Ea}$$ Thus $$K \propto T$$
as $$K$$ is directly proportional to Temperature
thus, $$K_{35^oC} > K_{34^oC}$$
$$\Rightarrow \cfrac {K_{34}}{K_{35}} <1 $$
Because Rate increase with rise in temperature.

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Time	0	T	$$ \infty $$
Partial pressure of A	$$ P_{0} $$	$$ P_{t} $$	-

Chemical Kinetics Test - 48

Result

Chemical Kinetics Test - 48

Score

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Question 1

$$20$$Min

$$33$$Min

$$15$$Min

$$25$$Min

Solution

Question 2

$$210$$kJ

$$105$$kJ

$$90$$kJ

$$145$$kJ

Solution

Question 3

20 kcal/mol

15 kcal/mol

10 kcal/mol

zero

Solution

Question 4

$$2$$ Hours

$$30$$ Minutes

$$45$$ Minutes

$$20$$ Minutes

Solution

Question 5

1/4

1/2

1/8

1/6

Solution

Question 6

$$ k=\frac{1}{t}In(\frac{P_{0}}{P_{t}})$$

$$ k=\frac{1}{t}In(\frac{P_{t}}{P_{0}})$$

$$ k=\frac{1}{t}In(\frac{2P_{0}}{P_{t}})$$

$$ k=\frac{1}{t}In(\frac{P_{t}}{2P_{0}})$$

Solution

Question 7

-0.5 m/s, 0.3 m/s

-0.3 m/s,0.5 m/s

0.5 m/s, 0.3 m/s

0.3 m/s,0.5 m/s

Solution

Question 8

$$50$$Min

$$46$$Min

$$49$$Min

$$48$$Min

Solution

Question 9

6286 years

5527 years

5570 years

4570 years

Solution

Question 10

rate increases with the rise in temperature

rate decreases with rise in temperature

rate does not change with rise in temperature

none of the above

Solution

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