Equilibrium Test

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Equilibrium Test - 16

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Weekly Quiz Competition

Question 1
1 / -0

Milk of magnesia used as a medicine for treating indigestion is a substance that -

A
Helps in disintegration of food products leading to their facile metabolism.

B
Combines with gastric hydrochloric acid thereby enhancing the latter's efficiency

C
Improves the enzymatic activities inside the stomach.

D
Neutralises excess acidity, providing a buffered medium inside the stomach.

Solution

Milk of magnesia used as a medicine for treating indigestion is a substance that neutralises excess acidity, providing a buffered medium inside the stomach.
Question 2
1 / -0

Which is not a buffer solution:

A
$$NH_{4}Cl+NH_{4}OH$$

B
$$CH_{3}COOH+CH_{3}COONa$$

C
$$CH_{3}COONH_{4}$$

D
$$NH_{4}NO_{3}$$

Solution

A buffer solution either is a mixture of a weak acid and its salt with strong base or a mixture of a weak base and its salt with strong acid. Hence, clearly $$CH_{3}COONH_{4}$$ is not a buffer solution.
Question 3
1 / -0

For a reversible reaction if the concentrations of the reactants are doubled then the equilibrium:

A
be halved

B
be doubled

C
remain the same

D
become $$1/4^{th}$$

Solution

For a given reaction , at equilibrium $$K_{c} = \dfrac{K_{f}}{K_{r}} = \dfrac{concentration \ of \ products}{concentration \ of \ reactants}$$

Therefore $$K_{c}$$ is independent of change in concentration.
Question 4
1 / -0

The unit of $$K_w$$ is:

A
$$mol^2/L^2$$

B
$$mol/L^2$$

C
$$mol^2/L$$

D
$$mol/L$$

Solution

$${ K }_{ w }=\left[ { H }^{ + } \right] \left[ { OH }^{ - } \right] =[mol/L]\left[ mol/L\right] $$

$${ K }_{ w }=\dfrac { { mol }^{ 2 }}{ {L }^{ 2 } } $$
Question 5
1 / -0

The degree of dissociation of weak electrolyte increases by :

A
increasing the concentration of weak electrolyte solution

B
adding a common ion

C
dissolution in solvent of low dielectric constant

D
adding more amount of solvent into the solution

Solution

When more amount of solvent is added to the solution, it helps in dissociating the molecules into ions of a weak electrolyte. Thus, degree of dissociation of weak electrolyte increases upon dilution.
Question 6
1 / -0

For the reversible reaction, if a concentration of reactants are doubled at a definite temperature, then the value of equilibrium constant will:

A
be reduced to half

B
be reduced to one fourth

C
also be doubled

D
remains constant

Solution

Doubling the initial concentrations of reactants in an equilibrium does not result in change in $$K_{c}$$ value. $$K_{c}$$ remains the same at a particular temperature.
Question 7
1 / -0

$$A(s) \rightleftharpoons 2B(g) + C(g)$$
The above equilibrium was established by initially taking $$A(s)$$ only. At equilibrium, $$B$$ is removed so that its partial pressure at new equilibrium becomes 1/3$$^{rd}$$ of original total pressure. The ratio of total pressure at a new equilibrium and at initial equilibrium will be:

A
2/3

B
14/13

C
5/3

D
17/19

Solution

$$A(s) \rightleftharpoons 2B(g) + C(g)$$
eq. 1 2p p
eq. 2 p p'
$$P_{T_1} = 3p$$ $$K_p = 4p^3$$, $$(p)^2 p'= 4p^3$$
$$p' = 4p,$$ $$P_{T_2 } = p + 4p = 5p$$
$$\displaystyle \frac{P_{T_2}}{P_{T_1}} = \frac{5p }{3p} = 5/3$$
Question 8
1 / -0

What will be the correct order of vapour pressure of water,acetone and ether at $$\displaystyle 30^{\circ}C$$. Given that among these compounds, water has maximum boiling point and ether has minimum boiling point?

A
Water < ether < acetone

B
Water < acetone < ether

C
Ether < acetone < water

D
Acetone < ether < water

Solution

The compound with higher boiling point will have lower vapour pressure and the compound with lower boiling point will have higher vapour pressure.
The correct order of boiling points is $$ Water > acetone > ether$$.
The correct order of vapour pressure is $$ Water < acetone < ether$$.
Question 9
1 / -0

The solubility of $$AgI$$ in $$Nal$$ solution is less than that in pure water because:

A
$$AgI$$ forms complex with $$NaI$$

B
le-chatelier's principle

C
solubility product of $$AgI$$ is less than that of $$Nal$$

D
the temperature of the solution decreases

Solution

The reaction for the dissociation of silver iodide is,
$$AgI(s) \rightleftharpoons Ag^+(aq)+I^-(aq)$$
The expression for the dissociation of silver iodide is, $$NaI \rightarrow Na^++I^-$$.
Thus iodide ion acts as common ion. Due to presence of the common ion, the equilibrium for the dissociation of AgI is shifted backwards.
This is because, the increase in $$[I^-]$$ brings in an increase in the solubility product for AgI which is $$[Ag^+][I^-]$$.
This is in accordance with Le-chatlier's principle.
Question 10
1 / -0

The yield of product in the reaction $$2A(g)+B(g)\rightleftharpoons 2C(g)+Q\ kJ$$ would be lower at

A
low temperature and low pressure

B
high temperature and high pressure

C
low temperature and to high pressure

D
high temperature and low pressure

Solution

The forward reaction is an exothermic reaction. Whereas the reverse reaction is an endothermic reaction. At high temperature, reverse reaction is favored so that more heat is absorbed which will nullify the effect of increased temperature.
Due to this the yield of $$C$$ will be lower. The number of moles for the reverse reaction increases from 2 to 3. At low pressure, reverse reaction is favored so that the number of moles increases which will nullify the effect of reduced pressure.
Due to this the yield of $$C$$ will be lower.

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Equilibrium Test - 16

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Equilibrium Test - 16

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Question 1

Helps in disintegration of food products leading to their facile metabolism.

Combines with gastric hydrochloric acid thereby enhancing the latter's efficiency

Improves the enzymatic activities inside the stomach.

Neutralises excess acidity, providing a buffered medium inside the stomach.

Solution

Question 2

$$NH_{4}Cl+NH_{4}OH$$

$$CH_{3}COOH+CH_{3}COONa$$

$$CH_{3}COONH_{4}$$

$$NH_{4}NO_{3}$$

Solution

Question 3

be halved

be doubled

remain the same

become $$1/4^{th}$$

Solution

Question 4

$$mol^2/L^2$$

$$mol/L^2$$

$$mol^2/L$$

$$mol/L$$

Solution

Question 5

increasing the concentration of weak electrolyte solution

adding a common ion

dissolution in solvent of low dielectric constant

adding more amount of solvent into the solution

Solution

Question 6

be reduced to half

be reduced to one fourth

also be doubled

remains constant

Solution

Question 7

2/3

14/13

5/3

17/19

Solution

Question 8

Water < ether < acetone

Water < acetone < ether

Ether < acetone < water

Acetone < ether < water

Solution

Question 9

$$AgI$$ forms complex with $$NaI$$

le-chatelier's principle

solubility product of $$AgI$$ is less than that of $$Nal$$

the temperature of the solution decreases

Solution

Question 10

low temperature and low pressure

high temperature and high pressure

low temperature and to high pressure

high temperature and low pressure

Solution

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