Chemical Kinetics Test

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Chemical Kinetics Test - 18

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Weekly Quiz Competition

Question 1
1 / -0

If a reaction obeys the following equation $$k=\frac{2.303}{t}log\frac{a}{a-x}$$ then the order is:

A
0

B
1

C
2

D
3

Solution

For the first order reaction, rate constant is given by following equation,
$$K=\frac{2.303}{t}log\frac{a}{a-x}$$
Hence, the reaction which obeys the given equation will be first order reaction.
Question 2
1 / -0

The rate of chemical reaction depends on the nature of reactants because:

A
the number of bonds broken in the reactant molecules and the number of bonds formed in product molecules changes

B
some of the reactants are solids at the room temperature

C
some of the reactants are coloured

D
some of rectants are liquid at room temperature

Solution

The rate of a chemical reaction depends on the nature of reactants because the number of bonds broken in the reactant molecules and the number of bonds formed in product molecules changes.

In other words, the ease with which bonds are formed and the ease with which bonds are broken determines the rate of the chemical reaction.

Option A is correct.
Question 3
1 / -0

The time for half change of a reactant in a zero order reaction is:

A
proportional to the initial concentration

B
proportional to the square root of the initial concentration

C
independent of initial concentration

D
inversely proportional to the initial concentration

Solution

For $$n^{th}$$ order, $$t_{\frac{1}{2}}=\dfrac{1}{a^{n-1}}$$.

For a zero order reaction, $$n=0$$,

$$t_{\frac{1}{2}} \propto \dfrac{1}{a^{-1}}$$

$$t_{\frac{1}{2}} \propto \ a$$
Question 4
1 / -0

If $$C_{o}=$$initial concentration of the reactant, $$C_{t}=$$ concentration of the reactant at time t and k$$=$$rate constant of the reaction, then the equation applicable for a first order reaction is:

A
$$C_{t}=C_{0}e^{-Kt}$$

B
$$C_{t}=C_{0}e^{Kt}$$

C
$$C_{0}=C_{t}e^{-Kt}$$

D
$$\displaystyle\frac{C_{0}}{C_{t}}=1$$

Solution

$$k=\frac{1}{t}ln \left [ \frac{C_0}{C_t} \right ]$$
$$kt=ln \frac{C_0}{C_t}$$
$$\frac{C_t}{C_0}=e^{-kt}$$
$$C_t=C_{0}e\ ^{-kt}$$
Thus, The integrated form for the first order reaction is $$C_{t}=C_{o}e^{-Kt}$$.
Here, $$C_{o}=$$ represents initial concentration of the reactant, $$C_{t}$$ represents the concentration of the reactant at time t and k is the rate constant of the reaction.
Other forms of the integrated rate law are $$ln \displaystyle \frac {C_{t}} {C_{o}}=-kt$$ and $$k=\displaystyle\frac {2.303} {t}log\displaystyle \frac {c} {c-x}$$.
Question 5
1 / -0

In a first order reaction, the concentration of product 'x' at time 't' is given by the expression:
(where,a $$=$$ initial concentration, k $$=$$ rate constant, n $$=$$ order)

A
$$x=a\left ( 1-e^{-kt} \right )$$

B
$$x=\dfrac{1}{\left ( a-x \right )}$$

C
$$x=\dfrac{1}{2^{n-1}}$$

D
$$x=\dfrac{a}{\left ( a-x \right )}$$

Solution

The integrated rate law for the first order reaction is $$[A]=[A]_0e^{-kt}$$.

Substituting the value of $$[A]=a-x$$ and $$[A]_0=a$$ in the above expression.

We get, $$a-x=ae^{-kt}$$.

Rearrange above expression, $$x=a\left ( 1-e^{-kt} \right )$$.
Question 6
1 / -0

If the activation energy of both the forward and the backward reactions are equal. Then $$\Delta $$H of the reaction is:

A
zero

B
+ve

C
-ve

D
cannot be predicted

Solution

$$A\rightleftharpoons B$$

If the activation energy of the forward and backward reactions are equal, then the energy of reactant and product become equal.

$$So,\ \Delta H =E_a-E_b= 0$$

Option A is correct.
Question 7
1 / -0

In a reaction between two gaseous reactants the number of binary collisions per second (Z) is given by:

A
$$Z=\pi \sigma ^{2}_{A-B}\sqrt{\frac{8KT}{\pi \mu }}n_{A}.n_{B}$$

B
$$Z=\pi \sigma _{A-B}\sqrt{\frac{8KT}{\pi \mu }}$$

C
$$Z= \sigma^{2} _{A-B}\sqrt{\frac{8KT}{\pi \mu }}n_{A}.n_{B}$$

D
$$Z= \sigma _{A-B}\sqrt{\frac{8KT}{\pi \mu }}n_{A}.n_{B}$$

Solution

$$Z=\pi \sigma ^{2}_{A-B}\sqrt{\frac{8KT}{\pi \mu }}n_{A}.n_{B}$$

Where,
$$Z=$$ number of binary collisions per second
$$\sigma ^{ }_{ A-B }=$$ Collision radius
$$n_{ A },n_{ B }=$$ Number of A and B molecules per unit volume.
$$K=$$ Boltzmann's constant
$$T=$$ Temperature
$$\mu =$$ Reduced mass
Question 8
1 / -0

The threshold energy of a chemical reaction depends upon:

A
nature of reacting species

B
temperature

C
concentration of species

D
number of collisions

Solution

The threshold energy of a chemical reaction doesn't depend on temperature, volume, concentration, and pressure but it depends upon the catalyst and nature of the reacting species that must have the correct spatial orientation.
Question 9
1 / -0

A zero-order reaction is one in which the rate of the reaction is independent of:

A
the temperature of the reaction

B
the concentration of the reactants

C
the concentration of the products

D
the volume of the vessel in which the reaction is carried out

Solution

The rate of a zero order reaction is independent of the concentration of the reactants. The reaction proceeds at a constant rate throughout.
Question 10
1 / -0

In a first order reaction, fraction of the total concentration of the reactant that varies with time 't' is equal to:

A
$$e^{+kt}$$

B
$$10^{+0.434kt}$$

C
$$\dfrac{1}{2^{-n}}$$

D
$$e^{-kt}$$

Solution

$$k = \dfrac { 2.303 }{ t } \times log \dfrac { { A }_{ 0 } }{ { A }_{ t } } $$
where,
$${ A }_{ 0 }=$$ is the initial concentration
$${ A }_{ t }=$$ is the concentration at time t

And k is the specific rate constant for the reaction,
$$\therefore kt= ln\dfrac { { A }_{ 0 } }{ { A }_{ t } }$$

Which can be written in the exponential form as,
$$\dfrac { { A }_{ 0 } }{ { A }_{ t } } = { e }^{ kt }$$

$$\dfrac { { A }_{ t } }{ { A }_{ 0 } } ={ e }^{ -kt }$$

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Chemical Kinetics Test - 18

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Chemical Kinetics Test - 18

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Question 1

0

1

2

3

Solution

Question 2

the number of bonds broken in the reactant molecules and the number of bonds formed in product molecules changes

some of the reactants are solids at the room temperature

some of the reactants are coloured

some of rectants are liquid at room temperature

Solution

Question 3

proportional to the initial concentration

proportional to the square root of the initial concentration

independent of initial concentration

inversely proportional to the initial concentration

Solution

Question 4

$$C_{t}=C_{0}e^{-Kt}$$

$$C_{t}=C_{0}e^{Kt}$$

$$C_{0}=C_{t}e^{-Kt}$$

$$\displaystyle\frac{C_{0}}{C_{t}}=1$$

Solution

Question 5

$$x=a\left ( 1-e^{-kt} \right )$$

$$x=\dfrac{1}{\left ( a-x \right )}$$

$$x=\dfrac{1}{2^{n-1}}$$

$$x=\dfrac{a}{\left ( a-x \right )}$$

Solution

Question 6

zero

+ve

-ve

cannot be predicted

Solution

Question 7

$$Z=\pi \sigma ^{2}_{A-B}\sqrt{\frac{8KT}{\pi \mu }}n_{A}.n_{B}$$

$$Z=\pi \sigma _{A-B}\sqrt{\frac{8KT}{\pi \mu }}$$

$$Z= \sigma^{2} _{A-B}\sqrt{\frac{8KT}{\pi \mu }}n_{A}.n_{B}$$

$$Z= \sigma _{A-B}\sqrt{\frac{8KT}{\pi \mu }}n_{A}.n_{B}$$

Solution

Question 8

nature of reacting species

temperature

concentration of species

number of collisions

Solution

Question 9

the temperature of the reaction

the concentration of the reactants

the concentration of the products

the volume of the vessel in which the reaction is carried out

Solution

Question 10

$$e^{+kt}$$

$$10^{+0.434kt}$$

$$\dfrac{1}{2^{-n}}$$

$$e^{-kt}$$

Solution

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