Equilibrium Test

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Equilibrium Test - 57

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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 $Cl_2$ at constant volume

C
introducing $PCl_5$ at constant volume

D
reducing the volume of the container.

Solution

$PCl_5(g)\rightleftharpoons PCl_3(g) + Cl_2(g)$
In order for a forward reaction to take place, increase in the concentration of reactants is necessary. Here more concentration of $PCl_5$ leads to faster dissociation of it.
Question 2
1 / -0

When $0.1\ mol$ arsenic acid, $H_{2}AsO_{4}$ is dissolved in $1L$ buffer solution of $pH = 4$ , which of the following hold good? $K_{1} = 2.5\times 10^{-4}, K_{2} = 5\times 10^{4}, K_{3} = 2\times 10^{-23}$ for arsenic acid $['< <'$ sign denotes that the higher concentration is at least $100$ times more than the lower one].

A
$[H_{2}AsO_{4}] < < [H_{2}AsO_{4}^{-}]$

B
$[H_{2}AsO_{4}] < < [HAsO_{4}^{2-}]$

C
$[HAsO_{4}^{2-}] < < [HAsO_{4}^{-}]$

D
$[AsO_{4}^{2-}] < < [HAsO_{4}^{2-}]$

Solution

$\begin{array}{l} p H=4 \\ -\log \left[H^{+}\right]=4 \\ \text { } \quad\left[H^+\right]=10^{-4} \end{array}$
$\frac{\left[\mathrm{H}_{3} \mathrm{AsO}_{4}\right]}{\left[\mathrm{H}_{2} \mathrm{AsO}_{4}^{-}\right]}=\frac{\left[\mathrm{H}^{+}\right]}{\mathrm{K}_{1}}=\frac{10^{-4}}{2.5 \times 10^{-4}}$
$\therefore \frac{1}{2.5}$
$\Rightarrow \quad\left[H_{3} As O_{4}\right]<\left[H_{2} As O_{4}^{-}\right]$
$\begin{aligned} \frac{\left[H_{2} A_{S} O_{4}^{-}\right]}{\left[H As O_{4}^{2-}\right]}=\frac{\left[H^{+}\right]}{K_{2}} &=\frac{10^{-4}}{5 \times 10^{4}} \\ &=\frac{10^{-4-4}}{5} \\ &=\frac{1}{5 \times 10^{8}} \end{aligned}$
$\Rightarrow\left[\mathrm{H}_{2} \mathrm{~A}{\mathrm{s}} \mathrm{O}_{4}^{-}\right]<<\left[\mathrm{H} \mathrm{A}{\mathrm{s}} \mathrm{O}_{4}^{2-}\right]$
$\begin{array}{c} {\left[\mathrm{HAsO}_{4}{ }^{2-}\right]<<\left[\mathrm{H}_{2} \mathrm{~A}{\mathrm{s}} \mathrm{O}_{4}^{-}\right]} \\ \text {not correct proof above. } \end{array}$
$\frac{\left[As O_{4}^{2-}\right]}{\left[H As O_{4}^{2-}\right]}=\frac{K_{3}}{\left[H^{+}\right]}=\frac{2 \times 10^{-23}}{10^{-4}}$
$=\frac{2}{10 \times 10^{18}}$
$=\frac{1}{5 \times 10^{18}}$
$\begin{array}{l} \Rightarrow\left[\mathrm{A}s \mathrm{O}_{4}{ }^{2-}\right]<<\left[\mathrm{H} \mathrm{A}_{\mathrm{S}} \mathrm{O}_{4}{ }^{2-}\right] \\ \end{array}$
Question 3
1 / -0

The volume of the reaction vessel containing an equilibrium mixture is increased in the following reation.
$SO_2Cl_{2(g)} \, \rightleftharpoons \, SO_{2(g)} \, + \, Cl_{2(g)}$ .
When equilibrium is re-established :

A
the amount of $Cl_{2(g)}$ remains unchanged

B
the amount $Cl_{2(g)}$ increases

C
the amount of $SO_2Cl_{2(g)}$ increases

D
the amount of $SO_{2(g)}$ drcreases

Solution

$S{ O }_{ 2 }{ Cl }_{ 2\left( g \right) }\rightleftharpoons S{ O }_{ 2\left( g \right) }+{ Cl }_{ 2\left( g \right) }$
When the volume of the vessel increases pressure decreases. Hence according to Le-Chatlier principle the equilibrium shift in a direction to increase the presure i.e., towards right as more gas molecules are produced.
So, the amount of ${ Cl }_{ 2\left( g \right) }$ increases.
Question 4
1 / -0

For the given endothermic reaction, $A(g)\rightleftharpoons 2B(g)$ . The variation in concentration due to different changes is plotted.
Neglect the slope of change in concentration.

Given the correct order of initials $T(true)$ or $F(false)$ for following statements:
P) Effect $-I$ is decreasing in temperature at constant volume
Q) Effect $-II$ is a decrease in total equilibrium pressure by changing the volume
R) Effect $- III$ is the addition of $B$ only at constant volume

A
$FTT$

B
$TFT$

C
$TFF$

D
$TTT$

Solution
Question 5
1 / -0

For a chemical reaction $3 \left( g \right) +Y(g)\leftrightharpoons { X }_{ 3 }Y_{ (g) }$ , the amount of ${ X }_{ 3 }Y$ , at equilibrium is affected by:

A
temperature and pressure

B
temperature only

C
pressure only

D
temperature, pressure and catalyst

Solution

Given reaction-

$3\times (g)+Y(g)\rightleftharpoons { X }_{ 3 }Y(g)$ is a synthesis reaction.

no. moles reactant $=(3+1)=4$

no. moles product $=1$

According to Lechatelier's Principle-

$Pressure$ : Increasing pressure shifts equilibrium to the side of reaction with fever moles of gaseous molecules. In this case, increasing pressure will shift the equilibrium to right.

$Temperature$ : The given reaction is a synthesis reaction. Synthesis reactions release energy, so they are exothermic.
Increasing the temperature will cause the equilibrium to shift towards the product side.

$Catalyst$ : catalysts speed up the reactions, but have no effect on the equilibrium.

Hence, the correct option is A
Question 6
1 / -0

The osmotic pressure of equimolar solutions of a) $Al_2(SO_4)_3$ , b) $KCl$ and c) $sugar$ will be:

A
$b>a>c$

B
$a>b>c$

C
$c>b>a$

D
$b>c>a$

Solution

Solution:-
As we now that osmotic pressure is given as-
$\pi = iCRT$
As $'i'$ increases, $′\pi′$ also increases.
$i = 5$ for ${Al}_{2} {\left( S{O}_{4} \right)}_{3}$
$i = 2$ for $KCl$
$i = 1$ for glucose
Therefore, osmotic pressure of given solutions will be in order-
$\left( a \right) > \left( b \right) > \left( c \right)$
Question 7
1 / -0

Which of the following is a true statement:

A
The ionisation constant and ionic product of water are same.

B
Water is a strong electrolyte.

C
The value of ionic product of water is less than that of its ionisation constant

D
At $298\ K$ , the number of ${ H }^{ + }$ ions in a litre of water is $6.023\times { 10 }^{ 16 }$ .

Solution

Let's check the options one by one:
(1) False. They are not the same.
Ionisation constant of water is
${ K }_{ a\left( { H }_{ 2 }O \right) }=\dfrac { \left[ { H }^{ + } \right] \left[ { OH }^{ - } \right] }{ \left[ { H }_{ 2 }O \right] }$
Ionic product of water is
${ K }_{ w }={ K }_{ a\left( { H }_{ 2 }O \right) }\times \left[ { H }_{ 2 }O \right] \\ { K }_{ w }=\left[ { H }^{ + } \right] \left[ { OH }^{ - } \right]$

(2) False. Water is a weak electrolyte. In fact, pure water is a non-electrolyte.

(3) False. ${ K }_{ w }={ K }_{ a\left( { H }_{ 2 }O \right) }\times \left[ { H }_{ 2 }O \right]$
Clearly, ${ K }_{ w }>{ K }_{ a\left( { H }_{ 2 }O \right) }$
value of ionic product is more than value of ionisation constant.

(4) At 298K, pH of water is 7
$\therefore conc\quad of\quad \left[ { H }^{ + } \right] ={ 10 }^{ -7 }$ moles per litre
1 mole contains $6.023\times { 10 }^{ 23 }$ ions
$\therefore conc\left[ { H }^{ + } \right] ={ 10 }^{ -7 }\times 6.023\times { 10 }^{ 23 }=6.023\times { 10 }^{ 16 }$
Question 8
1 / -0

One litre of water contains $1.0\times { 10 }^{ -7 }$ moles of ${ H }^{ + }$ ions. The degree of ionization of water is:

A
$1.8\times { 10 }^{ -9}%$

B
$0.8\times { 10 }^{ -9 }%$

C
$3.6\times { 10 }^{ -9 }%$

D
$3.6\times { 10 }^{ -7 }%$

Solution

Volume= $1L$
Density of water= $1g/ml$
Mass of water= $1000g$
Moles of water = $\dfrac{1000}{18}$
$H_2O\rightleftharpoons H^{+}+OH^-$
$t=O$ $C$ $O$ $O$
$t.teq$ $c(1-\alpha)$ $e\alpha$ $e\alpha$
$c\alpha=10^{-7}$
$\Rightarrow \dfrac{1000}{8}\times \alpha=10^{-2}$
$\Rightarrow \alpha=1.8\times 10^{-9}$
Question 9
1 / -0

The ionization constant of benzoic acid is $6.46 \times 10^{-5}$ and $K_{sp}$ for silver benzoate is $2.5\times 10^{-13}$ . How many times is silver benzoate more soluble in a buffer of $pH = 3.19$ compared to its solubility in pure water?

A
$4$

B
$3.32$

C
$3.01$

D
$2.5$

Solution

Since $pH=3.19$
$[H_3O^+]=6.46\times10^{-4}\ M$
$C_6H_5COOH+H_2O\leftrightharpoons C_6H_5COO^-+H_3O$
$K_a=\dfrac{[C_6H_5COO^-][H_3O^+]}{[C_6H_5COOH]}$
$\dfrac{[C_6H_5COOH]}{[C_6H_5COO^-]}=\dfrac{[H_3O^+]}{K_a}=\dfrac{6.46\times10^{-4}}{6.46\times10^{-5}}=10$
Let the solubility of $C_6H_5COOAg$ be $x\ mol/L$
Then,
$[Ag^+]=x$
$[C_6H_5COOH]+[C_6H_5COO^-]=x$
$10[C_6H_5COO^-]+[C_6H_5COO^-]=x$
$[C_6H_5COO^-]=\dfrac{x}{11}$
$K_{sp}[AG^+][C_6H_5COO^-]$
$2.5\times10^{-13}=x\left(\dfrac{x}{11}\right)$
$x=1.66\times10^{-6}\ mol/L$
Thus, the solubility of silver benzoate in a pH 3.19 solution is $1.66\times10^{-6}\ mol/L$
Now, let the solubility of $C_6H_5COOAg$ be $x'\ M$
Then, $[Ag^+]=x' \ M$ and $[C_6H_5COO^-]=x'\ M$
$K_{sp}=[Ag^+][C_6H_5COO^-]$
$K_{sp}=(x')^2$
$x'=\sqrt{K_{sp}}=\sqrt{2.5\times10^{-13}}=5\times10^{-7}\ mol/L$
$\therefore \dfrac{x}{x'}=\dfrac{1.66\times10^{-6}}{5\times10^{-7}}=3.32$
Hence, $C_6H_5COOAg$ is approximately $3.32$ times more soluble in low pH solution.
Question 10
1 / -0

$H_2(g)+I_2(g)\leftrightharpoons 2HI(g)$
The correct statement(s) regarding the above reaction is/are:

A
With increase of pressure amount of $HI$ increases

B
With increase of the volume of container amount of $HI$ product will decrease

C
Addition of $Ne$ gas at constant pressure will have no effect on equilibrium of formation

D
All of these

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Equilibrium Test - 57

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Equilibrium Test - 57

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Question 1

introducing an inert gas at constant volume

introducing Cl2Cl_2Cl2​ at constant volume

introducing PCl5PCl_5PCl5​ at constant volume

reducing the volume of the container.

Solution

Question 2

[H2AsO4]<<[H2AsO4−][H_{2}AsO_{4}] < < [H_{2}AsO_{4}^{-}][H2​AsO4​]<<[H2​AsO4−​]

[H2AsO4]<<[HAsO42−][H_{2}AsO_{4}] < < [HAsO_{4}^{2-}][H2​AsO4​]<<[HAsO42−​]

[HAsO42−]<<[HAsO4−][HAsO_{4}^{2-}] < < [HAsO_{4}^{-}][HAsO42−​]<<[HAsO4−​]

[AsO42−]<<[HAsO42−][AsO_{4}^{2-}] < < [HAsO_{4}^{2-}][AsO42−​]<<[HAsO42−​]

Solution

Question 3

the amount of Cl2(g)Cl_{2(g)}Cl2(g)​ remains unchanged

the amount Cl2(g)Cl_{2(g)}Cl2(g)​ increases

the amount of SO2Cl2(g)SO_2Cl_{2(g)}SO2​Cl2(g)​ increases

the amount of SO2(g)SO_{2(g)}SO2(g)​ drcreases

Solution

Question 4

FTTFTTFTT

TFTTFTTFT

TFFTFFTFF

TTTTTTTTT

Solution

Question 5

temperature and pressure

temperature only

pressure only

temperature, pressure and catalyst

Solution

Question 6

b>a>cb>a>cb>a>c

a>b>ca>b>ca>b>c

c>b>ac>b>ac>b>a

b>c>ab>c>ab>c>a

Solution

Question 7

The ionisation constant and ionic product of water are same.

Water is a strong electrolyte.

The value of ionic product of water is less than that of its ionisation constant

At 298 K298\ K298 K, the number of H+{ H }^{ + }H+ ions in a litre of water is 6.023×10166.023\times { 10 }^{ 16 }6.023×1016.

Solution

Question 8

1.8×10−91.8\times { 10 }^{ -9}%1.8×10−9

0.8×10−90.8\times { 10 }^{ -9 }%0.8×10−9

3.6×10−93.6\times { 10 }^{ -9 }%3.6×10−9

3.6×10−73.6\times { 10 }^{ -7 }%3.6×10−7

Solution

Question 9

444

3.323.323.32

3.013.013.01

2.52.52.5

Solution

Question 10

With increase of pressure amount of HIHIHI increases

With increase of the volume of container amount of HIHIHI product will decrease

Addition of NeNeNe gas at constant pressure will have no effect on equilibrium of formation

All of these

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introducing $Cl_2$ at constant volume

introducing $PCl_5$ at constant volume

$[H_{2}AsO_{4}] < < [H_{2}AsO_{4}^{-}]$

$[H_{2}AsO_{4}] < < [HAsO_{4}^{2-}]$

$[HAsO_{4}^{2-}] < < [HAsO_{4}^{-}]$

$[AsO_{4}^{2-}] < < [HAsO_{4}^{2-}]$

the amount of $Cl_{2(g)}$ remains unchanged

the amount $Cl_{2(g)}$ increases

the amount of $SO_2Cl_{2(g)}$ increases

the amount of $SO_{2(g)}$ drcreases

$FTT$

$TFT$

$TFF$

$TTT$

$b>a>c$

$a>b>c$

$c>b>a$

$b>c>a$

At $298\ K$ , the number of ${ H }^{ + }$ ions in a litre of water is $6.023\times { 10 }^{ 16 }$ .

$1.8\times { 10 }^{ -9}%$

$0.8\times { 10 }^{ -9 }%$

$3.6\times { 10 }^{ -9 }%$

$3.6\times { 10 }^{ -7 }%$

$4$

$3.32$

$3.01$

$2.5$

With increase of pressure amount of $HI$ increases

With increase of the volume of container amount of $HI$ product will decrease

Addition of $Ne$ gas at constant pressure will have no effect on equilibrium of formation