Chemical Kinetics Test

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Chemical Kinetics Test - 40

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Question 1
1 / -0

A first order reaction has a rate constant $$1.15\times 10^{-3}s^{-1}$$. How long will $$5\ g$$ of this reactant take to reduce to $$3\ g$$?

A
$$444\ s$$

B
$$400\ s$$

C
$$528\ s$$

D
$$669\ s$$

Solution

$$t = \dfrac {2.303}{k}\log \dfrac {[R_{0}]}{[R]} = \dfrac {2.303}{1.15\times 10^{-3}} \log\left( \dfrac {5}{3}\right)$$

$$= 2.00\times 10^{3} \log (1.667)$$

$$= 2\times 10^{3} \times 0.2219 = 444\ s$$.
Question 2
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The rate constant for a first order reaction is $$2\times 10^{-2} min^{-1}$$. The half-life period of reaction is ?

A
$$69.3\ min.$$

B
$$34.65\ min.$$

C
$$17.37\ min.$$

D
$$3.46\ min.$$

Solution

$$t_{1/2} = \dfrac {0.693}{k} = \dfrac {0.693}{2\times 10^{-3}} = 34.65\ min.$$
Question 3
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If hydrogen and oxygen are mixed and kept in the same vessel at room temperature, the reaction does not take place to form water because:

A
activation energy for the reaction is very high at room temperature

B
molecules have no proper orientation to react to form water

C
the frequency of collisions is not high enough for the reaction to take place

D
no catalyst is present in the reaction mixture

Solution

Activation energy = It is defined as the least possible amount of energy (minimum) which is required to start a reaction or the amount of energy available in a chemical system for a reaction to take place.

The activation energy for the water formation reaction is very high at room temperature due to this reaction does not take place, So the correct option is [A]
Question 4
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The potential energy diagram for a reaction $$X\rightarrow Y$$ is given. $$A$$ and $$C$$ in the graph corresponds to:

A
$$A \rightarrow$$ activation energy, $$C\rightarrow \triangle H^{\circ}$$

B
$$A\rightarrow$$ energy of reactants, $$C \rightarrow$$ energy of products

C
$$A\rightarrow \triangle H^{\circ}, C\rightarrow$$ activation energy

D
$$A\rightarrow$$ activation energy, $$C\rightarrow$$ threshold energy

Solution

Activation energy is defined as the least possible amount of energy (minimum) which is required to start a reaction or the amount of energy available in a chemical system for a reaction to take place $$\to A$$.

$$C \to \triangle { H }^{ o }$$ is the difference of the forward and backward activation energies.

Hence, the correct answer is option $$\text{A}$$.
Question 5
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Graph 9c) An endothermic reaction with high activation energy for the forward reaction can be shown by the figure:

A

B

C

D

Solution

In an endothermic reaction, the energy of the reactant is less than the energy of the product.
Here only in graph C, the energy of the product is greater than the energy of the reactant.
Thus, option C is correct.
Question 6
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What will be the half-life of the first order reaction for which the value of rate constant is $$200\ s^{-1}$$?

A
$$3.46\times 10^{-2} s$$

B
$$3.46\times 10^{-3} s$$

C
$$4.26\times 10^{-2} s$$

D
$$4.26\times 10^{-3} s$$

Solution

$$Given:k=200s^{-1}$$

$$t_{1/2} = \dfrac {0.693}{k} = \dfrac {0.693}{200} = 3.46\times 10^{-3} s$$.
Question 7
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The activation energy in a chemical reaction is defined as ?

A
The difference in energies of reactants and products

B
The sum of energies of reactants and products

C
The difference in energy of intermediate complex with the average energy of reactants and products

D
The difference in energy of intermediate complex and the average energy of reactants

Solution

Activation energy is the difference in energy of intermediate activated complex and the average energy of reactants.
Question 8
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In a first order reaction, the concentration of reactant is reduced to $$1/8$$ of the initial concentration in $$75$$ minutes at $$298\ K$$. What is the half-life period of the reaction in minutes?

A
$$50\ min$$

B
$$15\ min$$

C
$$30\ min$$

D
$$25\ min$$

Solution

Given:

$$a = 1, a - x = 1/8, t = 75\ min$$

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

$$= \dfrac {2.303\times 0.903}{75} \ min.^{-1}$$

For first order reaction,

$$t_{1/2} = \dfrac {0.693}{k} = \dfrac {0.693\times 75}{2.303\times 0.903} = 25\ min.$$

Hence, the correct answer is option $$\text{D}$$.
Question 9
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The decomposition of a hydrocarbon follows the equation $$k = (4.5\times 10^{11}s^{-1})e^{-28000/ K/T}$$. What will be the value of activation energy?

A
$$669\ kJ\ mol^{-1}$$

B
$$232.79\ kJ\ mol^{-1}$$

C
$$4.5\times 10^{11}\ kJ\ mol^{-1}$$

D
$$28000\ kJ\ mol^{-1}$$

Solution

Arrhenius equation, $$k = Ae^{-E_{a}/RT}$$

Given equation is

$$k = (4.5\times 10^{11}s^{-1})e^{-28000\ K/T}$$

Comparing both the equations, we get

$$-\dfrac {E_{a}}{RT} = -\dfrac {28000\ K}{T}$$

$$E_{a} = 28000\ K\times R = 28000\ K\times 8.314\ J\ K^{-1} mol^{-1}$$

$$= 232.79\ kJ\ mol^{-1}$$.
Question 10
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Activation energy of a chemical reaction can be determined by _________.

A
determining the rate constant at standard temperature

B
determining the rate constants at two temperatures

C
determining probability of collision

D
using catalyst

Solution

We know that,

$$ \log { \dfrac { { k }_{ 2 } }{ { k }_{ 1 } } } =\dfrac { { E }_{ a } }{ 2.303R } \left[ \dfrac { { T }_{ 2 }-{ T }_{ 1 } }{ { T }_{ 1 }{ T }_{ 2 } } \right]$$

From this equation, we observe that $${ E }_{ a } $$ at temperature$$ { T }_{ 1 }$$ and $${ T }_{ 2 }$$ can be determined by determining rate of constant$$ { K }_{ 1 }$$ and$$ { K }_{ 2 }$$

Option B is correct.