Equilibrium Test

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Equilibrium Test - 17

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Weekly Quiz Competition

Question 1
1 / -0

$$pK_a$$ for the acid $$HA$$ is $$6$$. The value of $$K$$ for the reaction $$A^-+H_3O^+\rightleftharpoons HA+H_2O$$ is:

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

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

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

D
$$1\times 10^6$$

Solution

The acid dissociation equilibrium is as shown below
$$HA+H_2O\rightleftharpoons H_3O^++A^-$$
The acid dissociation constant is
$$K_a=10^{-6}$$
The reaction $$HA+H_2O\rightleftharpoons H_3O^++A^-$$ is the reverse of the reaction for the acid dissociation equilibrium.
Hence, $$K= \dfrac {1} {K_a}\dfrac {1}{10^{-6}}=10^6$$
Question 2
1 / -0

In which of the following reactions, increase in the pressure at constant temperature does not affect the moles at equilibrium?

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

B
$$C(g) +\frac{1}{2}O_{2}(g)\rightleftharpoons CO(g)$$

C
$$H_{2}(g)+\frac{1}{2}O_{2}(g)\rightleftharpoons H_{2}O(g)$$

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

Solution

In the reaction of option D, the total number of moles of reactants is equal to the number of moles of product. For such reactions, when pressure is changed, there is no change in equilibrium.
Question 3
1 / -0

The equilibrium $$SO_{2}Cl_{2}(g)\rightleftharpoons SO_{2}(g)+Cl_{2}(g)$$ is attained at $$25^{\circ}$$C in a closed rigid container and an inert gas, helium is introduced. Which of the following statement(s) is/are correct?

A
Concentrations of $$SO_{2},Cl_{2}$$ and $$SO_{2}Cl_{2}$$ do not change.

B
More chlorine is formed.

C
Concentration of $$SO_{2}$$ is reduced.

D
More $$SO_{2}Cl_{2}$$ is formed.

Solution

The volume of closed rigid container remains constant. The addition of the inert gas at constant volume, has no effect on the equilibrium of the reaction. This is because, the number of moles of reactants and products is same. Thus when helium is introduced, the concentrations of the reactants and products do not change.
Question 4
1 / -0

The conditions favourable for the reaction: $$\displaystyle 2SO_{2}(g)+O_{2}(g)\rightleftharpoons 2SO_{3}(g);\triangle H^{\circ}=-198 KJ$$ are

A
low temperature,high pressure

B
any value of T and P

C
low temperature and low pressure

D
high temperature and high pressure

Solution

The negative value of enthalpy change indicates that the forward reaction is exothermic.
When the temperature is lowered, the reaction will proceed in the forward direction so that heat is liberated which will nullify the effect of lowered temperature.
The forward reaction proceeds with decrease in the number of moles, the increased pressure will shift the equilibrium towards right which will decrease the number of moles and nullify the effect of increased pressure. This will increase the yield of the forward reaction.
Thus the formation of $$SO_{3}(g)$$ will be favored at low temperature and high pressure.
Question 5
1 / -0

Change in volume of the system does not alter the number of moles in which of the following equilibrium?

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

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

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

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:
The change in volume will not affect the system if the change in the number of moles is zero.

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

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

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

$$PCl_5\rightarrow PCl_3+Cl_2$$

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

So, here change in volume will affect the system.

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

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 is (A).
Question 6
1 / -0

Solubility of $$AgCN$$ is maximum in :

A
acidic buffer solution

B
basic buffer solution

C
in pure water

D
equal in all solution

Solution

Reaction:

$$AgCN+HA\rightarrow HCN$$ due to this reaction, reaction shifts forward and solubility of AgCN increases.

So AgCN has maximum solubility in buffer solution.
Question 7
1 / -0

Directions For Questions

Equilibrium constants are given ( in atm ) for the following reactions at $$\displaystyle 0^{\circ}C$$ :

$$\displaystyle SrCl_{2} \: . \: 6H_{2}O(s)\rightleftharpoons SrCl_{2} \: . \: 2H_{2}O(s)+4H_{2}O(g)$$ $$\displaystyle K_{p}=5\times 10^{-12}$$

$$\displaystyle Na_{2}HPO_{4}\: .\: 12H_{2}O(s)\rightleftharpoons Na_{2}HPO_{4} \: .\: 7H_{2}O(s)+5H_{2}O(g)$$ $$\displaystyle K_{p}=2.43\times 10^{-13}$$

$$\displaystyle Na_{2}SO_{4}\: .\: 10H_{2}O(s)\rightleftharpoons Na_{2}SO_{4} \: (s)+10 H_{2}O(g)$$ $$\displaystyle K_{p}=1.024\times 10^{-27}$$

The vapor pressure of water at $$\displaystyle 0^{\circ}C\ is \ 4.56\ torr $$ ?

...view full instructions

Whaich is the most effective drying agent at $$0^{\circ}C$$?

A
$$SrCl_{2}\cdot 2H_{2}O$$

B
$$Na_{2}HPO_{4}\cdot 7 H_{2}O$$

C
$$Na_{2}SO_{4}$$

D
$$All\ equally$$
Question 8
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Statement -1: Total number of moles in a closed system at new equilibrium is less than the old equilibrium if some amount of a substance is removed from a system (consider a reaction $$A(g) \rightleftharpoons B(g))$$ at equillibrium.
Statement -2: Equilibrium constant of the reaction is changed by removing some amount of a substance at equilibrium.

A
Statement-1 is true, Statement -2 is true and Statement-2 is correct explanation for Statement-1

B
Statement-1 is true, Statement-2 is true and Statement -2 is NOT the correct explanation for Statement-1

C
Statement-1 is false, Statement-2 is true

D
Statement-1 is true, Statement-2 is false

Solution

When some amount of the substance is removed, the total number of moles decreases as the mass decreases. But the relative concentrations of products and reactants will remain same so that the value of the equilibrium constant is not affected.
Question 9
1 / -0

A $$0.10\space M$$ solution of $$HF$$ is $$8.0\mbox{%}$$ ionized. What is the $$K_a$$?

A
$$6.4\times10^{-4}$$

B
$$8.8\times10^{-4}$$

C
$$6.95\times10^{-4}$$

D
$$7.6\times10^{-4}$$

Solution

As we know, $$C\alpha^2 = K_a = 0.1\times(0.08)^2 = 6.4\times10^{-4}$$.
Question 10
1 / -0

At $${90}^{o}C$$ pure water has $$[{H}_{3}{O}^{+}]={10}^{-6}$$ $$mol/litre$$. The value of $${K}_{w}$$ at $${90}^{o}C$$ is

A
$${10}^{-6}$$

B
$${10}^{-12}$$

C
$${10}^{-8}$$

D
$${10}^{-14}$$

Solution

For pure water, $$[{H}^{+}]=[{OH}^{-}]$$
and
$${K}_{w}=[{H}^{+}][{OH}^{-}]={ \left[ { H }^{ + } \right] }^{ 2 }={ \left[ { 10 }^{ -6 } \right] }^{ 2 }={10}^{-12}$$

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Equilibrium Test - 17

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Equilibrium Test - 17

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Question 1

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

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

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

$$1\times 10^6$$

Solution

Question 2

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

$$C(g) +\frac{1}{2}O_{2}(g)\rightleftharpoons CO(g)$$

$$H_{2}(g)+\frac{1}{2}O_{2}(g)\rightleftharpoons H_{2}O(g)$$

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

Solution

Question 3

Concentrations of $$SO_{2},Cl_{2}$$ and $$SO_{2}Cl_{2}$$ do not change.

More chlorine is formed.

Concentration of $$SO_{2}$$ is reduced.

More $$SO_{2}Cl_{2}$$ is formed.

Solution

Question 4

low temperature,high pressure

any value of T and P

low temperature and low pressure

high temperature and high pressure

Solution

Question 5

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

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

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

$$SO_{2}Cl_{2}(g)\rightleftharpoons SO_{2}(g)+Cl_{2}(g)$$

Solution

Question 6

acidic buffer solution

basic buffer solution

in pure water

equal in all solution

Solution

Question 7

$$SrCl_{2}\cdot 2H_{2}O$$

$$Na_{2}HPO_{4}\cdot 7 H_{2}O$$

$$Na_{2}SO_{4}$$

$$All\ equally$$

Question 8

Statement-1 is true, Statement -2 is true and Statement-2 is correct explanation for Statement-1

Statement-1 is true, Statement-2 is true and Statement -2 is NOT the correct explanation for Statement-1

Statement-1 is false, Statement-2 is true

Statement-1 is true, Statement-2 is false

Solution

Question 9

$$6.4\times10^{-4}$$

$$8.8\times10^{-4}$$

$$6.95\times10^{-4}$$

$$7.6\times10^{-4}$$

Solution

Question 10

$${10}^{-6}$$

$${10}^{-12}$$

$${10}^{-8}$$

$${10}^{-14}$$

Solution

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