Equilibrium Test

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Equilibrium Test - 54

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Weekly Quiz Competition

Question 1
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For which of the following reactions at equilibrium at constant temperature doubling the volume will cause a shift to the right?

A
$${ N }_{ 2 }{ O }_{ 4 }(g)\rightleftharpoons 2{ NO }_{ 2 }(g)$$

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

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

D
$${ N }_{ 2 }(g)+ { O }_{ 2 }(g)\leftrightharpoons 2NO(g)$$

Solution

For the reaction $${ N }_{ 2 }{ O }_{ 4 }(g)\rightleftharpoons 2{ NO }_{ 2 }(g)$$ at equilibrium at constant temperature doubling the volume will cause a shift to the right
The number of moles of gaseous products is more than the number of moles of gaseous reactants. Doubling the volume will reduce the pressure to one half. The equilibrium will shift to right (to more number of moles of gaseous species) to increase the pressure. This will nullify the effect of decrease in the pressure.
Question 2
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On heating a mixture of $${ SO }_{ 2 }{ Cl }_{ 2 }$$ and $$CO$$, two equilibria are simultaneously established:
$${ SO }_{ 2 }{ Cl }_{ 2 }\rightleftharpoons { SO }_{ 2 }+{ Cl }_{ 2 }$$
$$CO(g)+{ Cl }_{ 2 }(g)\rightleftharpoons CO{ Cl }_{ 2 }(g)$$
On adding more $${SO}_{2}$$ at equilibrium what will happen?

A
Amount of $$CO$$ will decrease

B
Amount of $${ SO }_{ 2 }{ Cl }_{ 2 }$$ and $$CO{ Cl }_{ 2 }$$ will increase

C
Amount of $$CO$$ will remain unaffected

D
Amount of $${ SO }_{ 2 }{ Cl }_{ 2 }$$ and $$CO$$ will increase

Solution

The given equilibrium are :-
$${ SO }_{ 2 }{ Cl }_{ 2 }\rightleftharpoons { SO }_{ 2 }+{ Cl }_{ 2 }\quad \rightarrow \left( I \right) $$
$$CO\left( g \right) +{ Cl }_{ 2 }\left( g \right) \rightleftharpoons CO{ Cl }_{ 2 }\left( g \right) $$
If more $${ SO }_{ 2 }$$ is added at equilibrium then by Le Chatelier's principle reaction $$(I)$$ will proceed backwards and $${ SO }_{ 2 }{ Cl }_{ 2 }$$ formation will be favoured and $${ Cl }_{ 2 }$$ formation is less. Now, Since $${ Cl }_{ 2 }$$ formation is reduced so in equation $$(II)$$ by Le Chatelier's principle, the reaction proceeds backwards side and enhances the formation of $$CO$$.
Question 3
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In which of the following equilibrium, change in the volume of the system does not alter the number of moles?

A
$${ N }_{ 2 }(g)+{ O }_{ 2 }(g)\leftrightharpoons 2NO(g)$$

B
$$P{ Cl }_{ 5 }(g)\rightleftharpoons P{ Cl }_{ 3 }(g)+{ Cl }_{ 2 }(g)$$

C
$${ N }_{ 2 }(g)+3{ H }_{ 2 }(g)\rightleftharpoons 2{ NH }_{ 3 }(g)\quad $$

D
$${ SO }_{ 2 }{ Cl }_{ 2 }(g)\rightleftharpoons { SO }_{ 2 }(g)+{ Cl }_{ 2 }(g)$$

Solution

Hint: According to Le Chatlier's principle if some changes are made in a system then the system will move in that direction where it can get released from any stress.

Explanation:

If in a reaction number of moles remains the same in reactants and products means the change in the number of moles is zero then the change in volume will not affect the system.

$$N_2(g)+O_2(g)\rightarrow 2NO(g)$$

In this reaction, $$\Delta n=2-2=0$$.

So, here change in volume will not affect the system.

$$PCl_5(g)\rightarrow PCl_3(g)+Cl_2(g)$$

In this reaction, $$\Delta n=2-1=1$$.
So, here change in volume will affect the system.

$$N_2(g)+3H_2(g)\rightarrow 2NH_3(g)$$

In this reaction, $$\Delta n=2-4=-2$$.

So, here change in volume will affect the system.

$$SO_2Cl_2(g) \rightarrow SO_2(g)+Cl_2(g)$$

In this reaction, $$\Delta n=2-1=1$$.

So, here change in volume will affect the system.

Final Answer: Correct option $$(A).$$
Question 4
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The % yield of ammonia as a function of time in the reaction, $${ N }_{ 2 }(g)+3{ H }_{ 2 }(g)\rightleftharpoons 2{ NH }_{ 3 }(g)$$ [$$\Delta H< 0$$] at $$\left( P,{ T }_{ 1 } \right) $$ is given above.
If the reaction is conducted at $$\left( P,{ T }_{ 2 } \right) $$ with $${ T }_{ 2 }>{ T }_{ 1 }$$, the % yield of ammonia as a function of time is represented by:

A

B

C

D

Solution

The given reaction is exothermic hence the yield will be low, at higher temperature. At initial stage rate of forward reaction would be higher at higher temperature, the % yield of $${NH}_{3}$$ should be more at $${T}_{2}$$ than at $${T}_{1}$$.
Question 5
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The equilibrium constants for the reaction, $${Br}_{2}\rightleftharpoons 2Br$$ at $$500K$$ and $$700K$$ are $$1\times { 10 }^{ -10 }$$ and $$1\times { 10 }^{ -5 }$$ respectively. The reaction is:

A
Endothermic

B
Exothermic

C
Fast

D
Slow

Solution

The equilibrium constants for the reaction, $${Br}_{2}\rightleftharpoons 2Br$$ at $$500K$$ and $$700K$$ are $$1\times { 10 }^{ -10 }$$ and $$1\times { 10 }^{ -5 }$$ respectively. The reaction is endothermic.
An endothermic reaction is characterized by absorption of heat. When heat is added, the position of the equilibrium shifts to forward direction. With increase in temperature, the value of the equilibrium constant increases for an endothermic reaction and decreases for exothermic reaction (Refer image).
Question 6
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Le Chatelier's principle is not applicable to:

A
$$Fe(s)+S(s)\rightleftharpoons FeS(s)$$

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

C
$${ N }_{ 2 }(g)+3{ H }_{ 2 }(g)\rightleftharpoons 2{ NH }_{ 3 }(g)$$

D
$${ N }_{ 2 }(g)+{ O }_{ 2 }(g)\rightleftharpoons 2NO(g)$$

Solution

Hint: Le Chatelier’s principle is used to determine the changes in a system.

Explanation:

Le Chatelier’s principle states that when some variables like temperature, pressure, concentration, etc. are changed in a system at equilibrium then the system will go in that direction where it can get relief from those changes.

So, Le Chatelier’s principle is applicable to all gaseous systems chemical or physical at the equilibrium.

It is not applicable for solids.

In the reaction $$Fe(s)+S(s) \rightleftharpoons FeS(s)$$ all the reactants and products are solids. So, Le Chatlier’s principle is not applicable to this reaction.

Final Answer: Hence the correct answer is option $$(A).$$
Question 7
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For the reaction,
$${ N }_{ 2 }(g)+3{ H }_{ 2 }(g)\rightleftharpoons 2{ NH }_{ 3 }(g)+Heat$$
When temperature increases, then:

A
the formation of $${NH}_{3}$$ increases

B
the formation of $${NH}_{3}$$ decreases

C
the concentration of $${N}_{2}$$ decreases

D
the concentration of $${H}_{2}$$ decreases

Solution

For the reaction,
$${ N }_{ 2 }(g)+3{ H }_{ 2 }(g)\rightleftharpoons 2{ NH }_{ 3 }(g)+Heat$$
When temperature increases formation of $${NH}_{3}$$ decreases.

The forward reaction is an exothermic reaction. With the increase in temperature, heat is added to the reaction. The equilibrium will shift in the reverse direction so that added heat is consumed during the reverse reaction. This will nullify the effect of added heat.
Question 8
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Select the correct statements about the following reaction:
$$N_2(g)+3H_2(g)\rightleftharpoons 2NH_3(g); \Delta H=-22.4 \, kcal \, mol^{-1}$$

A
Increase in pressure will favour forward reaction

B
Addition of inert gas forms more $$NH_3$$

C
At low temperature, there is forward shill

D
Catalyst will increase the amount of $$NH_3$$

Solution

$$N_2(g)+3H_2(g)\rightleftharpoons 2NH_3(g); \Delta H=-22.4 \, kcal \, mol^{-1}$$
According to lechatliers principle the reaction is carried in the direction where the volume is decreased when the pressure increased.
So in the above reaction 4 volumes of reactants formed 2 volumes of product in the forward direction.So forward reaction is favored.
Hence option A is correct.
Question 9
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(A) pH of $$10^{-7}$$ M NaOH solution exists between 7 to 7.3 at $$25^o C$$.
(R) Due to common ion effect ionization of water is suppressed.

A
Both (R) and (A) are true and reason is the correct explanation of assertion

B
Both (R) and (A) are true but reason is not correct explanation of assertion

C
Assertion (A) is true but reason (R) is false

D
Assertion (A) and reason (R) both are false

E
Assertion (A) is false but reason (R) is true

Solution

A) $$pH$$ of $${ 10 }^{ -7 }$$ $$M$$ $$NaOH$$ will be slightly greater than $$7$$. It is between $$7$$ to $$7.3$$ at $${ 25 }^{ 0 }C$$ due to presence of very small quantity of excess $${ OH }^{ - }$$.
R) Due to common ion effect ionisation of water is suppressed.
$${ H }_{ 2 }O\rightleftharpoons { H }^{ + }+{ OH }^{ - }$$
So in presence of common ion $${ H }^{ + }$$ or $${ OH }^{ - }$$ the ionisation of $${ H }_{ 2 }O$$ will be decreased.
So both $$(R)$$ and $$(A)$$ are true and reason is the correct explanation of assertion.
Question 10
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Equilibrium constants are given for the following two equilibria.
$$(i) A_2(g)+ B_2 (g)\rightleftharpoons 2AB (g);$$ $$K=2 \times 10^{-4}$$
$$(ii) 2AB(g)+ C_2(g) \rightleftharpoons 2ABC (g);$$ $$K= 2\times 10^{-2}\, L \, mol^{-1}$$

Calculate the equilibrium constant for the following equilibrium.
$$ABC (g)\rightleftharpoons \dfrac {1}{2}A_2 (g)+\dfrac {1}{2}B_2(g)+\dfrac {1}{2}C_2 (g)$$

A
$$500\ M^{1/2}$$

B
$$4 \times 10^{-8} \ M^{1/2}$$

C
$$500\, M^{-1/2}$$

D
$$200\, M^{-1/2}$$

E
$$500\, M^{1/2}$$

Solution

$$A_2(g)+B_2(g)\rightleftharpoons 2AB(g)$$

$$K_1 = 2\times 10^{-4}$$

$$K_1=\dfrac {[AB]^2}{[A_2][B_2]}$$ .... (i)

$$2AB (g)+ C_2(g)\rightleftharpoons 2ABC(g)$$

$$K_2 = 2 \times 10^{-2}L/mol$$

So, $$K_2 =\dfrac {[ABC]^2}{[AB]^2[C_2]}$$ .... (ii)

$$ABC \rightleftharpoons \dfrac {1}{2}A_2(g)+\dfrac {1}{2}B_2 (g)+\dfrac {1}{2}C_2(g)$$

So, $$K_3 =\dfrac {[A_2]^{1/2}[B]^{1/2}[C_2]^{1/2}}{[ABC]}$$

Eq. (i) and Eq. (ii) are added, reversed and divided by 2 then we get Eq. (iii).

Therefore, $$K_3 =\dfrac {1}{\sqrt{K_1 \times K_2}}$$

$$=\dfrac {1}{\sqrt{2\times 10^{-4}\times 2 \times 10^{-2}}}$$

$$=\dfrac {1000}{2} = 500 \, M^{1/2}$$