Equilibrium Test

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Equilibrium Test - 14

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Question 1
1 / -0

In a reversible reaction $$H_{2} + I_{2} \leftrightharpoons 2HI$$, if the concentration of $$H_{2}$$ and $$I_{2}$$ are increased, the value of $$K_{c}$$:

A
decreases

B
increases

C
remains the same

D
changes exponentially

Solution

The magnitude of the equilibrium constant is not affected by the changes in concentrations of reactants and products, pressure and volume. Thus, when the concentrations of hydrogen and iodine are increased, the value of the equilibrium constant remains the same.
Question 2
1 / -0

Solubility of a substance which dissolves with a decrease in volume and absorption of heat will be favoured by:

A
high P and high T

B
low P and low T

C
high P and low T

D
low P and high T

Solution

During dissolution of a substance, the volume decreases. When pressure is increased, the reaction will proceed in the forward direction which will decrease the volume and nullify the effect of increased pressure.
During the dissolution of a substance, heat is absorbed. Hence, the forward reaction will be favored by high temperature as the effect of increased temperature will be nullified by the absorption of heat.
Hence, the solubility of a substance which dissolves with a decrease in volume and absorption of heat will be favored by high pressure and high temperature.
Question 3
1 / -0

For an exothermic reaction, the equilibrium constant :

A
increases with increase of temperature

B
decreases with increase of temperature

C
increases with increase of pressure

D
decreases with increase of pressure

Solution

When the temperature of an exothermic reaction is increased, the reaction will proceed in the backward direction, so as to absorb heat and nullify the effect of increasing temperature.

This will decrease the value of the equilibrium constant $$K_c$$. In other words, the equilibrium will shift to left.

Hence, $$B$$ is the correct option.
Question 4
1 / -0

At constant pressure, the presence of inert gases:

A
reduces the dissociation of $$PCl_{5}$$

B
increases the dissociation of $$PCl_{5}$$

C
does not affect the degree of dissociation of $$PCl_{5}$$

D
steps up the formation of $$PCl_{5}$$

Solution

For the reaction $$PCl_{5} \leftrightharpoons PCl_{3} + Cl_{2}$$, the forward reaction occurs with an increase in the number of moles from 1 to 2. Also, the reverse reaction occurs with a decrease in the number of moles from 2 to 1.

When inert gas is added at constant pressure, the equilibrium will shift in the direction in which there is an increase in the number of moles of gases.

Thus, the addition of inert gas at constant pressure will increase the dissociation of $$PCl_5$$.

Option B is correct.
Question 5
1 / -0

An increase in pressure would favor the following reaction:

A
$$N_{2(g)} + O_{2(g)}\Leftrightarrow 2NO_{(g)}$$

B
$$2SO_{2}(g) + O_{2} (g) \Leftrightarrow 2SO_{3}(g)$$

C
$$PCl_{5}(g) \Leftrightarrow PCl_{3} (g) + Cl_{2}(g) $$

D
$$H_{2} (g) + I_{2} (g) \Leftrightarrow 2HI$$

Solution

The equilibrum reaction is $$2SO_{2}(g) + O_{2} (g) \Leftrightarrow 2SO_{3}(g)$$. The number of moles decreases in the forward reaction from 3 to 2. Also the number of moles increases in the reverse reaction form 2 to 3. When the pressure is increased by decreasing the volume at constant temperature, the forward reaction will be favored as the forward reaction occurs with decrease in the number of moles.
Question 6
1 / -0

The degree of dissociation of an electrolyte depends on:

A
nature of the electrolyte

B
nature of solvent

C
temperature

D
all the above

Solution

Degree of dissociation depends on all the above factors
1. Nature of electrolyte:- Weak or Strong
(incomplete dissociation) or (Complete dissociation)
2. As the temperature increases, the degree of dissociation increases.
3. The nature of solvent polar or non-polar also affects degree dissociation.
Option D is correct.
Question 7
1 / -0

Assertion A: A catalyst has no effect on the state of equilibrium
Reason R: A catalyst influences the rates of both forward and backward reactions to the same extent.

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

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

C
Assertion is true Reason is false

D
Reason is true Assertion is false

Solution

When a catalyst is present, the equilibrium is achieved faster but its magnitude remains unaffected. Thus the value of the equilibrium constant in presence of a catalyst is same as in absence of catalyst.
This is because, the catalyst influence the rate of forward reaction and reverse reaction to same extent.
Question 8
1 / -0

Cottrell precipitator acts on which of the following principle?

A
Hardy-Schulze rule

B
Distribution law

C
Le Chatelier's principle

D
Neutralization of charge on the colloidal particles

Solution

it is related with neutralisation of charge on colloidal particlr
Question 9
1 / -0

In the equilibrium reaction, $$N_{2} + 3H_{2} \Leftrightarrow 2NH_{3}$$, the sign of $$\Delta H$$ accompanying the reaction is;

A
positive

B
negative

C
may be positive or negative

D
cannot be predicted

Solution

The reaction for the formation of ammonia is $$N_{2} + 3H_{2} \Leftrightarrow 2NH_{3}+92.2 kJ$$. Thus the reaction is exothermic and the sign of $$\Delta H$$ is negative, i.e $$\Delta H=-92.2kJ$$.
Question 10
1 / -0

Which of the following is not affected by change in pressure?

A
$$N_{2}(g) + O_{2}(g)\Leftrightarrow 2NO(g)$$

B
$$N_{2} (g) + 3H_{2}(g) \Leftrightarrow 2NH_{3}(g) $$

C
$$PCl_{5} (g) \Leftrightarrow PCl_{3} (g) + Cl_{2}$$

D
$$2SO_{2} (g) + O_{2}(g) \Leftrightarrow 2SO_{3}(g)$$

Solution

For the reaction, $$N_{2}(g) + O_{2}(g)\Leftrightarrow 2NO(g)$$, the value of $$\Delta n$$ is $$2-(1+1)=0$$. When there is no change in the number of moles of reactants and products, there is no effect on the equilibrium when pressure is changed.
In all other option, the number of mole of reactant and products are different,
Hence, if pressure is changed the equilibrium will be affected.

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Equilibrium Test - 14

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Equilibrium Test - 14

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Question 1

decreases

increases

remains the same

changes exponentially

Solution

Question 2

high P and high T

low P and low T

high P and low T

low P and high T

Solution

Question 3

increases with increase of temperature

decreases with increase of temperature

increases with increase of pressure

decreases with increase of pressure

Solution

Question 4

reduces the dissociation of $$PCl_{5}$$

increases the dissociation of $$PCl_{5}$$

does not affect the degree of dissociation of $$PCl_{5}$$

steps up the formation of $$PCl_{5}$$

Solution

Question 5

$$N_{2(g)} + O_{2(g)}\Leftrightarrow 2NO_{(g)}$$

$$2SO_{2}(g) + O_{2} (g) \Leftrightarrow 2SO_{3}(g)$$

$$PCl_{5}(g) \Leftrightarrow PCl_{3} (g) + Cl_{2}(g) $$

$$H_{2} (g) + I_{2} (g) \Leftrightarrow 2HI$$

Solution

Question 6

nature of the electrolyte

nature of solvent

temperature

all the above

Solution

Question 7

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

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

Assertion is true Reason is false

Reason is true Assertion is false

Solution

Question 8

Hardy-Schulze rule

Distribution law

Le Chatelier's principle

Neutralization of charge on the colloidal particles

Solution

Question 9

positive

negative

may be positive or negative

cannot be predicted

Solution

Question 10

$$N_{2}(g) + O_{2}(g)\Leftrightarrow 2NO(g)$$

$$N_{2} (g) + 3H_{2}(g) \Leftrightarrow 2NH_{3}(g) $$

$$PCl_{5} (g) \Leftrightarrow PCl_{3} (g) + Cl_{2}$$

$$2SO_{2} (g) + O_{2}(g) \Leftrightarrow 2SO_{3}(g)$$

Solution

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