Equilibrium Test

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Equilibrium Test - 47

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Weekly Quiz Competition

Question 1
1 / -0

For the reaction $$PCl_{5}(g) \rightleftharpoons PCl_{3}(g)+Cl_{2}(g)$$, the forward reaction at constant temperature is favoured by:

A
introducing an inert gas at constant volume

B
introducing chlorine gas at constant volume

C
introducing an inert gas at constant pressure

D
increasing the volume of the container

Solution

When an inert gas is introduced at constant pressure, the equilibrium will shift in a direction in which the number of moles increases. This is true for the forward reaction in case of dissociation equilibrium of $$PCl_5$$.

Hence, option C is the right answer.
Question 2
1 / -0

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

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

B
Of common ion effect

C
Solubility product of $$AgI$$ is less than that of $$NaI$$

D
The temperature of the solution decreases

Solution

The common ion presence in the solution decrease the solubility of a given sparingly soluble compound. So, solubility of AgI in NaI solution is less.
Question 3
1 / -0

The exothermic formation of $$ ClF_{3}$$ is represented by the equation: $$\displaystyle Cl_{2}(g)+3F_{2}(g)\rightleftharpoons 2ClF_{3}(g), \Delta H=-329$$ kJ
Which of the following will increase the quantity of $$ClF_{3}$$ in an equilibrium mixture of $$Cl_{2},F_{2}$$ and $$ClF_{3}$$?

A
Increasing the temperature

B
Removing $$Cl_{2}$$

C
Increasing the volume of container

D
Adding $$F_{2}$$

Solution

The equilibrium reaction is $$\displaystyle Cl_{2}(g)+3F_{2}(g)\rightleftharpoons 2ClF_{3}(g) \Delta H=-329$$
When fluorine is added, more and more fluorine will react with chlorine to form more $$ClF_{3}$$ and the equilibrium will shift towards right. This will increase the quantity of $$ClF_{3}$$ in the equilibrium mixture.
Question 4
1 / -0

What is the percent ionization $$(\alpha)$$ of a $$0.01M$$ $$HA$$ solution?
(Given that $$K_a = 10^{-4}$$)

A
$$9.5\mbox{%}$$

B
$$1\mbox{%}$$

C
$$10.5\mbox{%}$$

D
$$10\mbox{%}$$

Solution

As we know, $$\dfrac{c\alpha^2}{1-\alpha} = K_a$$, where $$\alpha$$ is the degree of dissociation and $$K_a$$ is the dissociation constant.

Given,
$$c = 0.01M$$ and $$K_a = 10^{-4}$$.

Therefore, $$k_a=\dfrac{0.01\times \alpha^2}{1-\alpha} = 10^{-4}$$,

$$0.01\alpha^2+10^{-4}\alpha-10^{-4}=0$$

or, $$\alpha = 0.095$$.

$$\mbox{%}$$ dissocation is $$0.095\times100 = 9.5\mbox{%}$$.
Question 5
1 / -0

What is the effect of the reduction of the volume of the system for the following equilibrium?
$$2C(s)+O_{2}(g)\rightleftharpoons 2CO(g)$$

A
The equilibrium will be shifted to the left by the increased pressure caused by the reduction in volume

B
The equilibrium will be shifted to the right by the decreased pressure caused by the reduction in volume

C
The equilibrium will be shifted to the left by the increased pressure caused by the increase in volume

D
The equilibrium will be shifted to the right by the increased pressure caused by the reduction in volume
Question 6
1 / -0

The value of the ion product constant for water, $$(K_w)$$ at $$60^{\small\circ}C$$ is $$9.6\times10^{-14}\, M^2$$. What is the $$[H_3O^+]$$ of a neutral aqueous solution at $$60^{\small\circ}C$$ and the nature of an aqueous solution with a $$pH = 7.0$$ at $$60^{\small\circ}C$$ are respectively?

A
$$3.1\times10^{-8}, \space acidic$$

B
$$3.1\times10^{-7}, \space neutral$$

C
$$3.1\times10^{-8}, \space basic$$

D
$$3.1\times10^{-7}, \space basic$$

Solution

As we know,
$$K_w = [H^+][OH^-]; [H_3O^+] = \sqrt {K_w} = 3.1\times10^{-7}$$.

$$pH= -log(3.1\times10^{-7})=6.5$$

Neutral solution have pH < 7

So, the solution with pH= 7 will be basic in nature.
Question 7
1 / -0

Following two equilibrium is simultaneously established in a container: $$\displaystyle PCl_{5}(g)\rightleftharpoons PCl_{3}(g)+ Cl_{2}(g)$$ and $$CO(g)+ Cl_{2}(g) \rightleftharpoons COCl_{2}(g)$$
If some Ni(s) is introduced in the container forming $$Ni(CO)_{4}$$(g), then at new equilibrium:

A
$$PCl_{2}$$ concentration will increase

B
$$PCl_{3}$$ concentration will decrease

C
$$Cl_{2}$$ concentration will remain same

D
$$CO$$ concentration will remain same

Solution

Ni reacts with CO. Hence, the equilibrium for the reaction $$CO(g)+ Cl_{2}(g) \rightleftharpoons COCl_{2}(g)$$ will shift to left so that more and more CO is formed. Due to this, the concentration of chlorine increases. Hence, the equilibrium for the reaction $$\displaystyle PCl_{5}(g)\rightleftharpoons PCl_{3}(g)+ Cl_{2}(g)$$ will shift to left so that more and more chlorine is consumed. Hence, the concentration of $$PCl_3$$ will decrease.
Question 8
1 / -0

If first dissociation of $$X(OH)_3$$ is $$100\mbox{%}$$ whereas second dissociation is $$50\mbox{%}$$ and third dissociation is negligible then the $$pH$$ of $$4\times10^{-3}\space M\space X(OH)_3$$ is :

A
$$11.78$$

B
$$10.78$$

C
$$2.5$$

D
$$2.22$$

Solution

For polyprotic base:

$$[OH^-] = C\alpha_1+C\alpha_2 = C(\alpha_1+\alpha_2)=4\times10^{-3} \times 1.5 = 6\times 10^{-3}$$

So, $$[H^+] = \dfrac{10^{-14}}{6\times 10^{-3}}={1.67\times 10^{-12}}$$

$$ pH = -log({1.67\times 10^{-12}})=11.78$$
Question 9
1 / -0

Which of the following expressions for $$\mbox{%}$$ ionization of a monoacidic base $$(BOH)$$ in aqueous solution is not correct at appreciable concentration?

A
$$100\times\sqrt{\displaystyle\frac{K_b}{c}}$$

B
$$\displaystyle\frac{1}{1+10(pK_b - pOH)}$$

C
$$\displaystyle\frac{K_w[H^+]}{K_b+K_w}$$

D
$$\displaystyle\frac{K_b}{K_b+[OH^-]}$$

Solution

The expression for the dissociation of the base is $$BOH(aq) \rightleftharpoons B^+(aq) +OH^-(aq)$$.
Let $$h$$ be the degree of dissociation.
At equilibrium, the concentrations of $$BOH, B^+ and OH^-$$ are $$c(1-h), ch$$ and $$ch $$ respectively.
$$K_b=\dfrac {[B^+][OH^-]}{[BOH]}=\dfrac {ch^2}{1-h}$$
Since $$h$$ is small, $$(1-h) \approx 1$$.
Hence, $$h= \sqrt{{\dfrac{k_b}{C}}}$$.
or, $$h= 100 \times \sqrt{{\dfrac{k_b}{C}}}\:\:\%$$.
Since total concentration of base is the sum of the concentration of $$BOH$$ and $$B^+$$.
Hence, $$h=\dfrac {[B^+]}{[B^+]+[BOH]}=\dfrac {1}{1+\dfrac {[BOH]}{[B^+]}}=\dfrac {1}{1+\dfrac {[OH^-]}{K_b}}.$$
or, $$\dfrac {K_b}{K_b+[OH^-]}=\dfrac {K_b[H^+]}{K_b[H^+]+K_w}$$

Also, $$pOH=-log[OH^-]$$.
$$[OH^-]=10^{-pOH}$$
$$K_b=10^{-pK_b}$$
$$h= \dfrac {1}{1+\dfrac {10^{-pOH}}{10^{-pK_b}}}=\dfrac {1}{1+10^{pK_b-pOH}}$$
Question 10
1 / -0

For the equilibrium,
$$LiCl\cdot { 3NH }_{ 3 }(s)\rightleftharpoons LiCl\cdot { NH }_{ 3 }(s)+2{ NH }_{ 3 }(g)\quad $$ $${ K }_{ p }=9{ atm }^{ 2 }$$ at $${ 37 }^{ o }C$$.
A $$5$$ litre vessel contains $$0.1$$ mole of $$LiCl\cdot { 3NH }_{ 3 }$$. How many moles of $${ NH }_{ 3 }$$ should be added to the flask at this temperature to derive the backward reaction for completion?
$$R=0.082 \ atm.L/molK$$

A
$$0.2$$

B
$$0.59$$

C
$$0.69$$

D
$$0.79$$

Solution

Ammonia is the only gaseous species in the reaction. Hence, its pressure will appear in the expression for the equilibrium constant.
$$k_p=P_{NH_3}^2=9 atm^2$$
Hence, $$P_{NH_3}=3 atm$$
Using ideal gas equation, calculate the number of moles corresponding to 3 atm pressure.
$$n=\dfrac {3 atm \times 5 L} {0.082 L atm /mol K \times 310 K}=0.59 mole.$$
The initial number of moles of $$LiCl\cdot { 3NH }_{ 3 }(s), LiCl\cdot { NH }_{ 3 }(s) \ and \ { NH }_{ 3 }$$ are 0.1, 0 and 0 respectively.
At equilibrium, the number of moles of $$LiCl\cdot { 3NH }_{ 3 }(s), LiCl\cdot { NH }_{ 3 }(s) \ and \ { NH }_{ 3 }$$ are $$0.1-x,\ x$$ and $$2x$$ respectively.
But $$2x=0.59$$
Since 0.1 mole of $$NH_3$$ of $$Licl_3.NH_3$$ added .
$$\therefore$$ Twice $$NH_3$$ requires for backward reaction.
$$\therefore$$ $$0.1\times 2 = 0.2$$
total $$NH_3 = 0.59 + 0.2 = 0.79 \ mole$$.

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Equilibrium Test - 47

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Equilibrium Test - 47

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Question 1

introducing an inert gas at constant volume

introducing chlorine gas at constant volume

introducing an inert gas at constant pressure

increasing the volume of the container

Solution

Question 2

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

Of common ion effect

Solubility product of $$AgI$$ is less than that of $$NaI$$

The temperature of the solution decreases

Solution

Question 3

Increasing the temperature

Removing $$Cl_{2}$$

Increasing the volume of container

Adding $$F_{2}$$

Solution

Question 4

$$9.5\mbox{%}$$

$$1\mbox{%}$$

$$10.5\mbox{%}$$

$$10\mbox{%}$$

Solution

Question 5

The equilibrium will be shifted to the left by the increased pressure caused by the reduction in volume

The equilibrium will be shifted to the right by the decreased pressure caused by the reduction in volume

The equilibrium will be shifted to the left by the increased pressure caused by the increase in volume

The equilibrium will be shifted to the right by the increased pressure caused by the reduction in volume

Question 6

$$3.1\times10^{-8}, \space acidic$$

$$3.1\times10^{-7}, \space neutral$$

$$3.1\times10^{-8}, \space basic$$

$$3.1\times10^{-7}, \space basic$$

Solution

Question 7

$$PCl_{2}$$ concentration will increase

$$PCl_{3}$$ concentration will decrease

$$Cl_{2}$$ concentration will remain same

$$CO$$ concentration will remain same

Solution

Question 8

$$11.78$$

$$10.78$$

$$2.5$$

$$2.22$$

Solution

Question 9

$$100\times\sqrt{\displaystyle\frac{K_b}{c}}$$

$$\displaystyle\frac{1}{1+10(pK_b - pOH)}$$

$$\displaystyle\frac{K_w[H^+]}{K_b+K_w}$$

$$\displaystyle\frac{K_b}{K_b+[OH^-]}$$

Solution

Question 10

$$0.2$$

$$0.59$$

$$0.69$$

$$0.79$$

Solution

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