Equilibrium Test

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Equilibrium Test - 31

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Question 1
1 / -0

Two systems, $$P{ Cl }_{ 5 }(g)\rightleftharpoons P{ Cl }_{ 3 }(g)+{ Cl }_{ 2 }(g)$$ and $$CO{ Cl }_{ 2 }(g)\rightleftharpoons CO(g)+{ Cl }_{ 2 }(g)$$ are simultaneously in equilibrium in a vessel at constant volume. If some $$CO(g)$$ is introduced in the vessel at constant volume, then at new equilibrium, the concentration of:

A
$$P{Cl}_{5}$$ is greater

B
$$P{Cl}_{3}$$ remains unchanged

C
$$P{Cl}_{5}$$ is less

D
$${Cl}_{2}$$ greater

Solution

Explanation :
Given equation:
$$PCl_5(g)\rightleftharpoons PCl_3(g)+Cl_2(g)$$
$$COCl_2(g)\rightleftharpoons CO(g)+Cl_2(g)$$
It is give that both the above reactions run simultaneously in a vessel at constant volume.
If we add some amount of $$CO(g)$$ in the vessel at constant volume,
then at new equilibrium the concentration is $$=? $$

When $$CO$$ is added to the above system the second reaction will go backward by Le Chatelier's principle and Hence, the concentration of the common molecule $$Cl_2$$ will also reduce.
So, to counteract the decreased concentration of chlorine, first equilibrium will go forward and hence, the concentration of $$PCl_5$$ will decrease.
$$\therefore$$ when $$CO(g)$$ is introduced in the vessel at constant volume, then at a new equilibrium, the concentration of $$PCl_5$$ get lesser.
Hence the correct answer is option $$C$$.
Question 2
1 / -0

For the reaction, $${ H }_{ 2 }(g)+{ I }_{ 2 }(g)\rightleftharpoons 2HI(g)$$, the equilibrium constant $${K}_{p}$$ changes with:

A
total pressure

B
catalyst

C
the amount of $${H}_{2}$$ and $${I}_{2}$$

D
temperature

Solution

The given reaction is : $${ H }_{ 2\left( g \right) }+{ I }_{ 2\left( g \right) }\rightleftharpoons 2{ HI }_{ \left( g \right) }$$
The value of $${ K }_{ P }$$ depends only on temperature; so it will vary as the temperature varies.
Van't Hoff equation : $$dln{ K }_{ P }/dT=\Delta H/{ RT }^{ 2 }$$
Question 3
1 / -0

In which of the following reactions, the increase in volume of the container will favour the formation of products?

A
$$C(s)+{ H }_{ 2 }O(g)\rightleftharpoons CO(g)+{ H }_{ 2 }(g)$$

B
$${ H }_{ 2 }(g)+{ I }_{ 2 }(g)\rightleftharpoons 2HI(g)$$

C
$$4{ NH }_{ 3 }(g)+5{ O }_{ 2 }(g)\rightleftharpoons 4NO(g)+6{ H }_{ 2 }O(l)$$

D
$$3{ O }_{ 2 }(g)\rightleftharpoons 2{ O }_{ 3 }(g)$$

Solution

increasing the volume of the container will decrease the concentration of all the gaseous states in the reaction.
By le chatelier's principle, the reaction will move in the direction of increasing no of moles that is the option $$A$$ .
Question 4
1 / -0

Given the equilibrium system, $${ NH }_{ 4 }Cl(s)\rightleftharpoons { NH }_{ 4 }^{ + }(aq.)+{ Cl }^{ - }(aq.);\left( \Delta { H }^{ o }=+3.5kcal/mol \right) $$.
What change will shift the equilibrium to the right?

A
Decreasing the temperature

B
Increasing the temperature

C
Dissolving $$NaCl$$ crystals in equilibrium mixture

D
Dissolving $${ NH }_{ 4 }{NO}_{3}$$ crystals in the equilibrium mixture

Solution

The given reaction is :-
$${ NH }_{ 4 }Cl\left( s \right) \rightleftharpoons { NH }_{ 4 }^{ + }\left( aq \right) +{ Cl }^{ - }\left( aq \right) $$
$$\triangle H=+3.5Kcal/mol.$$
Since the given reaction is endothermic reaction so by le chatelier's principle with the increase of temperature the reaction proceeds in forward direction.
Question 5
1 / -0

Consider this equilibrium, for which $$\Delta H< 0$$.
$$HgO(s)+4{ I }^{ - }(aq.)+{ H }_{ 2 }O(l)\rightleftharpoons {Hg{ I }_{ 4 }}^{ 2- }+2{ OH }^{ - }$$
Which changes will increase the equilibrium concentration $${Hg{ I }_{ 4} }^{ 2- }$$?
$$I$$. Increasing the mass of $$HgO(s)$$ present
$$II$$. Increasing $$\left[ { I }^{ - } \right] $$
$$III$$. Adding $$1\ M$$ $$HCl$$

A
$$I$$ only

B
$$II$$ only

C
$$II$$ and $$III$$ only

D
$$I,II$$ and $$III$$

Solution

We have to increase the con. of $$HgI_4$$ then according to le chandelier principal
1- increase the con. of HgO is wrong because Hgo is solid then no. the effect in equ.
2 - increase the con. of iodine ion
3 - and adding 1 M of HCL because it's will decrease the con. of OH ion on neutralizing it.
Question 6
1 / -0

An increase in temperature increases which of the following?
1. The rate constant of a reaction
2. The ionic product of water
3. The equilibrium constant of an exothermic reaction
Select the correct answer using the code given below.

A
1 and 2 only

B
1 and 3 only

C
2 and 3 only

D
1,2 and 3 only

Solution

By Arrhenius equation,
$$K=A{ e }^{ -Ea/RT }\quad \rightarrow \left( I \right) $$
By Van't Hoff's equation :-
$$\dfrac { dln{ K }_{ eq } }{ dT } =\dfrac { \triangle H }{ { RT }^{ 2 } } \quad \rightarrow \left( II \right) $$
By (I) & (II), we can say that rate constant of reaction increases with increase of temperature and equilibrium constant of reaction that is exothermic decreases with increase of temperature.
Question 7
1 / -0

Water gas, an individual fuel, consisting $$CO$$ and $${H}_{2}$$ in equimolar amounts is obtained by passing steam over red-hot carbon in accordance with the reaction:
$$C(s)+{ H }_{ 2 }O(g)\rightleftharpoons CO(g)+{ H }_{ 2 }(g);\Delta { H }=+130.5kJ\quad $$
The yield of water gas can be increased by:
(1) Reducing the total pressure of the system
(2) Increasing pressure of steam
(3) Raising the temperature
(4) Introducing hot carbon

A
1, 2 and 3 are correct

B
1 and 2 are correct

C
2 and 4 are correct

D
1 and 3 are correct

Solution

The given reaction is :-
$$C\left( s \right) +{ H }_{ 2 }{ O }_{ \left( g \right) }\rightleftharpoons CO\left( g \right) +{ H }_{ 2\left( g \right) }\quad ;\quad \triangle H=+130.5KJ/mol$$
(I) Low pressure favours the reaction which are accompanied by increase of total no. of moles. So, reducing pressure of the system favours forward reaction.
(II) If the pressure of steam is increased then, by Le Chatelier's principle, the forward reaction is favoured.
(III) Since the endothermic in nature, the increase of temperature favours the forward reaction as predicted by Le Chatelier's principle.
Question 8
1 / -0

The reaction, $$2S{ O }_{ 2 }(g)+{ O }_{ 2 }(g)\rightleftharpoons 2{ SO }_{ 3 }(g)$$ (Exothermic), will shift in forward direction by:

A
adding $${SO}_{3}$$ at constant volume

B
increasing volume of container

C
adding $${SO}_{2}$$ at constant volume

D
adding inert gas at constant volume

Solution

The given reaction is : $$2{ SO }_{ 2\left( g \right) }+{ O }_{ 2\left( g \right) }\rightleftharpoons 2{ SO }_{ 3\left( g \right) }$$ (Exothermic) by Le Chatelier's principle, addition of $${ SO }_{ 2 }$$ at constant volume will favour the forward reaction.
Question 9
1 / -0

The addition of $$HCl$$ will not suppress the ionisation of:

A
Acetic acid

B
Sulphuric acid

C
$${H}_{2}S$$

D
Benzoic acid

Solution

The addition of $$HCl$$ will not suppress the ionisation of $$H_2SO_4$$ because $$H_2SO_4$$ is stronger acid than $$HCl$$.
Question 10
1 / -0

A monoprotic acid in $$1.00M$$ solution is $$0.001$$% ionised. The dissociation constant of acid is:

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

B
$$1\times { 10 }^{ -6 }$$

C
$$1\times { 10 }^{ -8 }$$

D
$$1\times { 10 }^{ -10 }$$

Solution

According to Ostwald's dilution law
$$K=\cfrac { { \alpha }^{ 2 }C }{ \left( 1-\alpha \right) } $$
As degree of ionisation is very less than 1 so above equation reduces to
$$K=C\alpha^2$$
$$K=1(10^{-5})^2$$
$$K=10^{-10}$$

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Equilibrium Test - 31

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Equilibrium Test - 31

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Question 1

$$P{Cl}_{5}$$ is greater

$$P{Cl}_{3}$$ remains unchanged

$$P{Cl}_{5}$$ is less

$${Cl}_{2}$$ greater

Solution

Question 2

total pressure

catalyst

the amount of $${H}_{2}$$ and $${I}_{2}$$

temperature

Solution

Question 3

$$C(s)+{ H }_{ 2 }O(g)\rightleftharpoons CO(g)+{ H }_{ 2 }(g)$$

$${ H }_{ 2 }(g)+{ I }_{ 2 }(g)\rightleftharpoons 2HI(g)$$

$$4{ NH }_{ 3 }(g)+5{ O }_{ 2 }(g)\rightleftharpoons 4NO(g)+6{ H }_{ 2 }O(l)$$

$$3{ O }_{ 2 }(g)\rightleftharpoons 2{ O }_{ 3 }(g)$$

Solution

Question 4

Decreasing the temperature

Increasing the temperature

Dissolving $$NaCl$$ crystals in equilibrium mixture

Dissolving $${ NH }_{ 4 }{NO}_{3}$$ crystals in the equilibrium mixture

Solution

Question 5

$$I$$ only

$$II$$ only

$$II$$ and $$III$$ only

$$I,II$$ and $$III$$

Solution

Question 6

1 and 2 only

1 and 3 only

2 and 3 only

1,2 and 3 only

Solution

Question 7

1, 2 and 3 are correct

1 and 2 are correct

2 and 4 are correct

1 and 3 are correct

Solution

Question 8

adding $${SO}_{3}$$ at constant volume

increasing volume of container

adding $${SO}_{2}$$ at constant volume

adding inert gas at constant volume

Solution

Question 9

Acetic acid

Sulphuric acid

$${H}_{2}S$$

Benzoic acid

Solution

Question 10

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

$$1\times { 10 }^{ -6 }$$

$$1\times { 10 }^{ -8 }$$

$$1\times { 10 }^{ -10 }$$

Solution

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