Chemical Kinetics Test

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Chemical Kinetics Test - 29

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Weekly Quiz Competition

Question 1
1 / -0

A first order reaction is given as $$A\rightarrow$$ products. Its integrated equation is:

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

B
$$k=\dfrac{1}{t}log\dfrac{a}{a-x}$$

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

D
$$-k=\dfrac{1}{t}log\dfrac{a-x}{a}$$

Solution

A first-order reaction is given as $$A \longrightarrow \text{products}$$.

$$\dfrac{d[A]}{dt} = -k[A]$$

On solving differential equation,

$$ln \dfrac{a-x}{a} = -kt$$

$$kt = ln \dfrac{a}{a-x}$$

$$\displaystyle k=\frac{1}{t}ln\frac{a} {a-x}$$

$$k$$ is the rate constant, $$t$$ is the time, $$a$$ is the initial concentration of A and $$(a-x)$$ is the concentration of A at time $$t$$. When natural logarithm is changed to logarithm at base 10, the equation becomes :

$$\displaystyle k=\frac{2.303}{t}log _{10}\frac{a} {a-x}$$

Hence, option $$C$$ is correct.
Question 2
1 / -0

In the first order reaction, 75% of the reactant gets disappeared in 1.386 h. The rate constant of the reaction is:

A
$$3.0 \times { 10 }^{ -3 }{ s }^{ -1 }$$

B
$$2.8 { \times 10 }^{ -4}{ s }^{ -1 }$$

C
$$17.2 { \times 10 }^{ -3 }{ s }^{ -1 }$$

D
$$1.8 { \times 10 }^{ -3 }{ s }^{ -1 }$$

Solution

For a first-order reaction, the rate constant will be

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

where,
K is rate constant.
t is time taken for x amount of reactant to convert to products.
a is initial concentration of reactant.

$$t=1.386\times60\times60sec$$
$$x=0.75a$$

$$k=\dfrac { 2.303 }{ 1.386\times60\times60 } \log { \dfrac { a }{ a-0.75a } }$$

$$\Longrightarrow k= \dfrac { 2.303 }{ 1.386\times3600 } \times\log { 4 } =2.8\times10^{ -4 }$$
Question 3
1 / -0

The first order integrated rate equation is:

A
$$k\, =\, \displaystyle \frac {x}{t}$$

B
$$k\, =\, -\displaystyle \frac {2.303}{t}\, log\, \displaystyle \frac {\alpha}{\alpha\, -\, x}$$

C
$$k\, =\, \displaystyle \frac {1}{t}\, ln\, \displaystyle \frac {\alpha}{\alpha\, -\, x}$$

D
$$k\, =\, \displaystyle \frac {1}{t}\, ln \displaystyle \frac {x}{\alpha(\alpha\, -\, x)}$$

Solution

For first order,
rate $$=\, \displaystyle \frac {d[R]}{dt}\, =\, k[R]$$
or $$=\, \displaystyle \frac {d[R]}{[R]}\, =\, kdt[R]$$ ....(i)
On integrating Eq. (i)
$$\displaystyle \int \displaystyle \frac {d[R]}{[R]}\, =\, k\,\int dt$$
In [R] = - kt + C ....(ii)
At t = 0, [R] = [R$$_0$$]
[where, R = final concentration, ie, $$\alpha\, -\, x$$ and $$R_0$$ is the initial concentration, ir, $$\alpha$$].
ln $$[R_0]$$ = C
On putting the value of C in Eq. (ii), we get
ln[R] = - kt + ln$$[R_0]$$
- kt = ln [R] - ln$$[R_0]$$
kt = ln$$[R_0]$$ - ln [R]
or $$k\, =\, \displaystyle \frac {1}{t}\, ln\, \displaystyle \frac {[R_0]}{[R]}$$
$$k\, =\, \displaystyle \frac {1}{t}\, ln\, \displaystyle \frac {\alpha}{\alpha\, -\, x}$$
Question 4
1 / -0

Found the enthalpy change of the forward reaction ?

A
B

B
C

C
D

D
E

Solution

The magnitude of the enthalpy change of the forward reaction is represented by "B".
The enthalpy change is the difference in the enthalpy of products and the enthalpy of reactants.
Question 5
1 / -0

Assertion: Reactions happen faster at higher temperatures.
Reason: As temperatures increase, there is also an increase in the number of collisions with the required activation energy for a reaction to occur.

A
Both Assertion and Reason are true and Reason is the correct explanation of Assertion

B
Both Assertion and Reason are true but Reason is not the correct explanation of Assertion

C
Assertion is true but Reason is false

D
Assertion is false but Reason is true

E
Both Assertion and Reason are false

Solution

When you raise the temperature of a system, the molecules bounce around a lot more. They have more energy. When they bounce around more, they are more likely to collide.
That fact means they are also more likely to combine. When you lower the temperature, the molecules are slower and collide less.

Reason is the correct explanation for assertion.
Question 6
1 / -0

The minimum energy required for molecules to react and form compounds can be defined as :

A
Reduction potential

B
Ionization energy

C
Electronegativity

D
Heat of formation

E
Activation energy

Solution

Activation energy may also be defined as the minimum energy required to start a chemical reaction. The activation energy of a reaction is usually denoted by E_a and given in units of kilojoules per mole $$(kJ/mol)$$ or kilocalories per mole $$(kcal/mol).$$

The standard enthalpy of formation is defined as the change in enthalpy when one mole of a substance in the standard state

The ionization energy (IE) is qualitatively defined as the amount of energy required to remove the most loosely bound electron, the valence electron, of an isolated gaseous atom to form a cation.
Question 7
1 / -0

The time required for $$100\%$$ completion of zero order reaction is:

A
$$\displaystyle\frac{a}{k}$$

B
$$\displaystyle\frac{a}{2k}$$

C
$$\displaystyle\frac{2a}{k}$$

D
$$\displaystyle\frac{k}{a}$$

Solution

$$Explanation: $$
For a zero-order reaction,
$$x=kt$$
Where, $$x =$$ Amount of substance
$$k =$$ Rate constant
$$t = $$ Time

For 100% completion of the reaction, x = a

Therefore, $$a = kt$$

or $$t=\cfrac ak$$

Hence, the correct option is $$\text{A}$$
Question 8
1 / -0

In a reaction, the potential energy of the reactants is 40 kJ/mol, the potential energy of the products is 10 kJ/mol and the potential energy of the activated complex is 55 kJ/mol. What is the activation energy for the reverse reaction?

A
45 kJ/mol

B
-30 kJ/mol

C
15 kJ/mol

D
35 kJ/mol

E
-55 kJ/mol

Solution

In a reaction the potential energy of the reactants is 40 kJ/mol, the potential energy of the products i 10 kJ/mol and the potential energy of the activated complex is 55 kJ/mol. What is the activation energy for the reverse reaction?
A
45 kJ/mol
The activation energy for the reverse reaction is the difference between the potential energy of the activated complex and the potential energy of the product of forward reaction. It is
$$\displaystyle 55 - 10 = 45 \: kJ/mol$$
Question 9
1 / -0

How an increase in concentration is related to number of collisions?

A
Directly

B
Inversely

C
Has no effect

D
None

Solution

The increase in concentration is directly related to number of collisions.
For higher concentrations, the average distance between molecules is small. Due to this, the number of collisions increases.
For lower concentrations, the average distance between molecules is large. Due to this, the number of collisions decreases.
Question 10
1 / -0

The addition of a catalyst to a reaction changes which of the following?

A
Activation energy

B
Entropy

C
Enthalpy

D
Nature of the products

Solution

Activation energy changes when there is addition of a catalyst to a reaction.
A catalyst provides an alternate path of lower activation energy.

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Chemical Kinetics Test - 29

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Chemical Kinetics Test - 29

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Question 1

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

$$k=\dfrac{1}{t}log\dfrac{a}{a-x}$$

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

$$-k=\dfrac{1}{t}log\dfrac{a-x}{a}$$

Solution

Question 2

$$3.0 \times { 10 }^{ -3 }{ s }^{ -1 }$$

$$2.8 { \times 10 }^{ -4}{ s }^{ -1 }$$

$$17.2 { \times 10 }^{ -3 }{ s }^{ -1 }$$

$$1.8 { \times 10 }^{ -3 }{ s }^{ -1 }$$

Solution

Question 3

$$k\, =\, \displaystyle \frac {x}{t}$$

$$k\, =\, -\displaystyle \frac {2.303}{t}\, log\, \displaystyle \frac {\alpha}{\alpha\, -\, x}$$

$$k\, =\, \displaystyle \frac {1}{t}\, ln\, \displaystyle \frac {\alpha}{\alpha\, -\, x}$$

$$k\, =\, \displaystyle \frac {1}{t}\, ln \displaystyle \frac {x}{\alpha(\alpha\, -\, x)}$$

Solution

Question 4

B

C

D

E

Solution

Question 5

Both Assertion and Reason are true and Reason is the correct explanation of Assertion

Both Assertion and Reason are true but Reason is not the correct explanation of Assertion

Assertion is true but Reason is false

Assertion is false but Reason is true

Both Assertion and Reason are false

Solution

Question 6

Reduction potential

Ionization energy

Electronegativity

Heat of formation

Activation energy

Solution

Question 7

$$\displaystyle\frac{a}{k}$$

$$\displaystyle\frac{a}{2k}$$

$$\displaystyle\frac{2a}{k}$$

$$\displaystyle\frac{k}{a}$$

Solution

Question 8

45 kJ/mol

-30 kJ/mol

15 kJ/mol

35 kJ/mol

-55 kJ/mol

Solution

Question 9

Directly

Inversely

Has no effect

None

Solution

Question 10

Activation energy

Entropy

Enthalpy

Nature of the products

Solution

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