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 $$Cl_2$$ at constant volume

introducing $$PCl_5$$ at constant volume

reducing the volume of the container.

Solution

Question 2

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

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

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

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

Solution

Question 3

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

Solution

Question 4

$$FTT$$

$$TFT$$

$$TFF$$

$$TTT$$

Solution

Question 5

temperature and pressure

temperature only

pressure only

temperature, pressure and catalyst

Solution

Question 6

$$b>a>c$$

$$a>b>c$$

$$c>b>a$$

$$b>c>a$$

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

Solution

Question 8

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

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

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

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

Solution

Question 9

$$4$$

$$3.32$$

$$3.01$$

$$2.5$$

Solution

Question 10

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

All of these

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