Chemical Kinetics Test

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Chemical Kinetics Test - 34

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Question 1
1 / -0

A freshly prepared radio medicine has half life $$2$$ hours. Its activity is $$64$$ times the permissible safe value. The minimum time after which it would be possible to treat the patients with the medicine is:

A
$$3$$ hrs

B
$$9$$ hrs

C
$$24$$ hrs

D
$$12$$ hrs

Solution

$$N={ N }_{ 0 }{ \left( \dfrac { 1 }{ 2 } \right) }^{ n }$$

$$\dfrac { N }{ { N }_{ 0 } } ={ \left( \dfrac { 1 }{ 2 } \right) }^{ n }$$

$$\dfrac { 1 }{ 64 } ={ \left( \dfrac { 1 }{ 2 } \right) }^{ n }$$

$$n=6$$ half lives

$$\therefore $$ time $$=2\times 6=12$$ hrs

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

For an endothermic reaction, where $$\Delta H$$ represents the enthalpy of reaction, the minimum value for the energy of activation will be:

A
less than $$\Delta H$$

B
zero

C
equal to $$\Delta H$$

D
more than $$\Delta H$$

Solution

Plot of energy v/s reaction progress for a typical endothermic reaction :-
The minimum value of the activation energy is the difference between the energy content of reactant and product. $$\Delta H$$ also gives the difference between the reactant and product.
Hence, the minimum value for the activation energy will be equal to $$\triangle H$$.
Question 3
1 / -0

A first-order reaction has a half-life of $$14.5\ hrs$$. What percentage of the reactant will remain after $$24\ hrs$$?

A
$$18.3$$%

B
$$31.8$$%

C
$$45.5$$%

D
$$68.2$$%

Solution

$$k=\dfrac { 2.303 }{ t } \log { \left( \dfrac { a }{ a-x } \right) } $$
$$\dfrac { 0.693 }{ 14.5 } =\dfrac { 2.303 }{ 24 } \log { \dfrac { 100 }{ \left( a-x \right) } } $$
On solving, $$\left( a-x \right) =31.8$$%
Question 4
1 / -0

The rates of a certain reaction at different time intervals are as follows:
$$\begin{matrix} Time(second) & 0 & 10 & 20 \\ Rate(mol\ L^{-1}s^{-1}) & 1.8\times { 10 }^{ -2 } & 1.82\times { 10 }^{ -2 } & 1.79\times { 10 }^{ -2 } \end{matrix}$$
The reaction is of:

A
zero order

B
first order

C
second order

D
third order

Solution

There is no effect of reactant concentration of the rate. this is only possible in a $$0$$ order reaction.
Question 5
1 / -0

Which one of the following is not a first order reaction?

A
$$C{ H }_{ 3 }COOC{ H }_{ 3 }+{ H }_{ 2 }O\xrightarrow [ ]{ \quad { H }^{ + }\quad } C{ H }_{ 3 }COOH+C{ H }_{ 3 }OH$$

B
$$C{ H }_{ 3 }COO{ C }_{ 2 }{ H }_{ 5 }+NaOH\longrightarrow C{ H }_{ 3 }COONa+{ C }_{ 2 }{ H }_{ 5 }OH$$

C
$$2{ H }_{ 2 }{ O }_{ 2 }\longrightarrow 2{ H }_{ 2 }O+{ O }_{ 2 }$$

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

Solution

The hydrolysis of ethyl acetate ($$CH_3COOC_2H_5$$) by sodium hydroxide yields ethanol ($$C_2H_5OH$$) and sodium acetate ($$NaO_2CCH_3$$) by the reaction showed follows second order kinetics.
Question 6
1 / -0

For a zero order reaction which of the following relation is correct, where $$t_{1/2}\ and\ a$$ are half-life and initial concentration respectively.

A
$${ t }_{ { 1 }/{ 2 } }\propto a$$

B
$${ t }_{ { 1 }/{ 2 } }\propto \dfrac { 1 }{ a } $$

C
$${ t }_{ { 1 }/{ 2 } }\propto { a }^{ 2 }$$

D
$${ t }_{ { 1 }/{ 2 } }\propto \dfrac { 1 }{ { a }^{ 2 } } $$

Solution

When zero order kinetic rate law is followed, let $$[A]$$ is the current concentration, $$[A]_0$$ is the initial concentration, and $$k$$ is the reaction constant and $$t$$ is time
Then relation is given by $$[A]=[A]_0−k$$
In order to find the half life we need to isolate t on its own, and divide it by 2. We would end up with a formula as such depict how long it takes for the initial concentration to dwindle by half given by -:
$$ \implies t_{1/2}=\frac{[A]_0}{2k}$$
Question 7
1 / -0

The specific rate constant of a first order reaction depends on:

A
concentration of the reactants

B
concentration of the products

C
time

D
temperature

Solution

$$k={ Ae }^{ \cfrac { -{ E }_{ a } }{ RT } }$$
$$ \therefore$$ $$k$$ depends on temperature.
Question 8
1 / -0

Reaction, $$A+B\longrightarrow C+D+38kcal$$ has activation energy $$20kcal$$. Activation energy for the reaction, $$C+D\longrightarrow A+B\quad $$ is:

A
$$20kcal$$

B
$$-20kcal$$

C
$$18kcal$$

D
$$58kcal$$

Solution

Reaction, $$A+B\longrightarrow C+D+\text {38 kcal }$$ has activation energy $$\text {20 kcal }$$. Activation energy for the reaction, $$C+D\longrightarrow A+B\quad $$ is $$20\text { kcal }+ 38 \text { kcal }=58 \text { kcal }$$.
Note:
The reaction is exothermic as for exothermic reaction (heat liberated during reaction),
$$ \displaystyle E_{a, r} = E_{a, f} + \Delta H_{rxn} $$
$$ \displaystyle E_{a, f} = $$ activation energy for forward reaction
$$ \displaystyle E_{a, r} = $$ activation energy for reverse reaction
$$ \displaystyle \Delta H_{rxn} = $$ enthalpy change for the reaction.
Question 9
1 / -0

At $$500$$ $$K$$, the half-life period of a gaseous reaction at an initial pressure of $$80$$ $$kPa$$ is $$350$$ $$sec$$. When the pressure is $$40$$ $$kPa$$, the half life period is $$175$$ $$sec$$; the order of the reaction is:

A
Zero

B
One

C
Two

D
Three

Solution

$$\dfrac { { \left( { t }_{ { 1 }/{ 2 } } \right) }_{ 1 } }{ { \left( { t }_{ { 1 }/{ 2 } } \right) }_{ 2 } } ={ \left( \dfrac { { p }_{ 2 } }{ { p }_{ 1 } } \right) }^{ n-1 }$$
$$\dfrac { 350 }{ 175 } ={ \left( \dfrac { 40 }{ 80 } \right) }^{ n-1 }$$
$$2={ \left( \dfrac { 1 }{ 2 } \right) }^{ n-1 }$$
$$n-1=-1$$
$$n=0$$ (zero order reaction)
Question 10
1 / -0

Which of the following is a first order reaction?

A
$$2HI\longrightarrow { H }_{ 2 }+{ I }_{ 2 }$$

B
$$2N{ O }_{ 2 }\longrightarrow 2NO+{ O }_{ 2 }$$

C
$$2NO+{ O }_{ 2 }\longrightarrow 2N{ O }_{ 2 }$$

D
$$N{ H }_{ 4 }N{ O }_{ 2 }\longrightarrow { N }_{ 2 }+2{ H }_{ 2 }O$$
Solution
Correct Answer: Option D

Explanation:
- The rate law for the following reaction can numerically be stated as:
  $$aA + bB + cC \rightarrow dD + eE$$
  $$\dfrac{\mathrm{d} r}{\mathrm{d} t} \propto \left [ A \right ]^{\alpha}\left [ B \right ]^{\beta}\left [C \right ]^{\gamma}$$
  Or, $$ \dfrac{\mathrm{d} r}{\mathrm{d} t} = K \left [ A \right ]^{\alpha}\left [ B \right ]^{\beta}\left [C \right ]^{\gamma}$$ where,
  $$ \dfrac{\mathrm{d} r}{\mathrm{d} t} =$$ Rate of reaction
  $$K=$$ constant of proportionality or Rate constant
  $$ \left [ A \right ]^{\alpha},\left [ B \right ]^{\beta},\left [C \right ]^{\gamma}=$$ concentrations of reactants raised to power of order of reaction in respect to that particular reactant.
- For elementary or one-step reactions, the order of reaction with respect to each reactant is equal to their coefficients in a balanced equation.
- Since all given reactions are elementary, the order can be determined for each of these.
1. $$2HI \rightarrow H_{2} + I_{2}$$
$$\dfrac{\mathrm{d} r}{\mathrm{d} t} = K_{1}\left [ HI \right ]^{2}$$
Therefore, it’s a second order reaction.

2. $$2NO_{2} \rightarrow 2NO + O_{2}$$
$$\dfrac{\mathrm{d} r}{\mathrm{d} t} = K_{2}\left [ NO_{2} \right ]^{2}$$
Therefore, it’s a second order reaction.

3. $$2NO + O_{2} \rightarrow 2NO_{2}$$
$$\dfrac{\mathrm{d} r}{\mathrm{d} t} = K_{3}\left [ NO \right ]^{2} \left [ O_{2} \right ]^{1}$$
Therefore, it’s a third order reaction.

4. $$NH_{4}NO_{2} \rightarrow N_{2} + 2H_{2}O$$
$$\dfrac{\mathrm{d} r}{\mathrm{d} t} = K_{4} \left [ NH_{4}NO_{2} \right ]^{1}$$
Therefore, it’s a first order reaction.

Hence, $$NH_{4}NO_{2} \rightarrow N_{2} + 2H_{2}O$$ is a first order reaction.

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Re-Attempt Weekly Quiz Competition

Chemical Kinetics Test - 34

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Chemical Kinetics Test - 34

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SHARING IS CARING

Question 1

$$3$$ hrs

$$9$$ hrs

$$24$$ hrs

$$12$$ hrs

Solution

Question 2

less than $$\Delta H$$

zero

equal to $$\Delta H$$

more than $$\Delta H$$

Solution

Question 3

$$18.3$$%

$$31.8$$%

$$45.5$$%

$$68.2$$%

Solution

Question 4

zero order

first order

second order

third order

Solution

Question 5

$$C{ H }_{ 3 }COOC{ H }_{ 3 }+{ H }_{ 2 }O\xrightarrow [ ]{ \quad { H }^{ + }\quad } C{ H }_{ 3 }COOH+C{ H }_{ 3 }OH$$

$$C{ H }_{ 3 }COO{ C }_{ 2 }{ H }_{ 5 }+NaOH\longrightarrow C{ H }_{ 3 }COONa+{ C }_{ 2 }{ H }_{ 5 }OH$$

$$2{ H }_{ 2 }{ O }_{ 2 }\longrightarrow 2{ H }_{ 2 }O+{ O }_{ 2 }$$

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

Solution

Question 6

$${ t }_{ { 1 }/{ 2 } }\propto a$$

$${ t }_{ { 1 }/{ 2 } }\propto \dfrac { 1 }{ a } $$

$${ t }_{ { 1 }/{ 2 } }\propto { a }^{ 2 }$$

$${ t }_{ { 1 }/{ 2 } }\propto \dfrac { 1 }{ { a }^{ 2 } } $$

Solution

Question 7

concentration of the reactants

concentration of the products

time

temperature

Solution

Question 8

$$20kcal$$

$$-20kcal$$

$$18kcal$$

$$58kcal$$

Solution

Question 9

Zero

One

Two

Three

Solution

Question 10

$$2HI\longrightarrow { H }_{ 2 }+{ I }_{ 2 }$$

$$2N{ O }_{ 2 }\longrightarrow 2NO+{ O }_{ 2 }$$

$$2NO+{ O }_{ 2 }\longrightarrow 2N{ O }_{ 2 }$$

$$N{ H }_{ 4 }N{ O }_{ 2 }\longrightarrow { N }_{ 2 }+2{ H }_{ 2 }O$$

Solution

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