Equilibrium Test

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Equilibrium Test - 50

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Question 1
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Directions For Questions

Physical and chemical equilibrium can respond to a change in their pressure, temperature, and concentration of reactants and products. To describe the change in the equilibrium we have a principle named Le Chatelier's principle. According to this principle, even if we make some changes in equilibrium, then also the system even re-establishes the equilibrium by undoing the effect.

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Consider the following equilibrium: $$2NO_2\rightleftharpoons 2NO_3; \Delta H=-ve$$ If $$O_2$$ is added and volume of there action vessel is reduced, the equilibrium

A
shifts in the product side

B
shifts in the reactant side

C
cannot be predicted

D
remains unchanged

Solution

Consider the following equilibrium: $$\displaystyle 2NO_2 + O_2 \rightleftharpoons 2NO_3$$; $$\displaystyle \Delta H =−ve$$ If $$O_2$$ is added and volume of there action vessel is reduced, the equilibrium shifts in the product side.

Since the enthalpy change $$\displaystyle \Delta H$$ is negative, the reaction is exothermic.

Also the reaction proceeds with decrease in the number of moles of gaseous species. Volume is reduced, the pressure increases and the reaction will shift in a direction in which there is decrease in number of moles.

Also when one of the reactants (oxygen) is added, the reaction will shift to product side.
Question 2
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If the equilibrium constant of $$BOH \rightleftharpoons B^{\oplus}+\overset{\circleddash}{O}H$$ at $$25^oC$$ is $$2.5\times 10^{-6}$$, then equilibrium constant for $$BOH + H^{\oplus} \rightleftharpoons B^{\oplus}+H_2O$$ at the same temperature is

A
$$4.0\times 10^{-9}$$

B
$$4.0\times 10^{5}$$

C
$$2.5\times 10^{8}$$

D
$$2.5\times 10^{-6}$$

Solution

If the equilibrium constant of $$BOH \rightleftharpoons B^{\oplus}+\overset{\circleddash}{O}H$$ at $$25^oC$$ is $$2.5\times 10^{-6}$$, then equilibrium constant for $$BOH + H^{\oplus} \rightleftharpoons B^{\oplus}+H_2O$$ at the same temperature is $$2.5\times 10^{8}$$
$$BOH \rightleftharpoons B^{\oplus}+\overset{\circleddash}{O}H$$. $$K = 2.5\times 10^{-6}$$.....(1)
$$\displaystyle H^+ + OH^- \rightleftharpoons H_2O$$ $$\dfrac {1} {K_w} = 10^{14}$$ ......(2)
---------------------------------------------------------------------------------------------------
$$BOH + H^{\oplus} \rightleftharpoons B^{\oplus}+H_2O$$ $$K' = ?$$......(3)
The reaction (3) is obtained by adding reactions (1) and (2),
$$\displaystyle K' = K \times \frac {1}{K_w} = 2.5 \times 10^{-6} \times 10^{14} = 2.5 \times 10^8 $$
Question 3
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$$16.6 mg$$ of solid silver chromate (molar mass = $$332$$) when put into $$500 mL$$ water, silver ion and chromate ion are obtained. On adding more solid, the concentration of ions does not increase. If $$100 mg$$ of solid is put into 2 litres of water, then the amount of solid remained undissociated would be :

A
$$33.6 mg$$

B
$$3.36 mg$$

C
$$66.4 mg$$

D
$$6.64 mg$$

Solution

16.6 mg of solid silver chromate (molar mass = 332) when put into 500 mL water, silver ion and chromate ion are obtained.
On adding more solid, the concentration of ions does not increase. If 100 mg of solid is put into 2 litre of water, then the amount of solid remained undissociated would be $$\displaystyle 100 \: \text {mg} - \frac {2000 \: \text {mL}}{500 \: \text {mL}} \times 16.6 \: \text {mg} = 33.6 \: \text {mg}$$
Note: The solubility of silver chromate is 16.6 mg in 500 mL water.
In 2 L or 2000 mL water, the solubility will be $$\displaystyle \frac {2000 \:\text {mL}}{ 500 \: \text {mL}} \times 16.6 \: \text {mg}$$
Question 4
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Which statement/relationship is correct?

A
Upon hydrolysis of salt of a strong base and weak acid gives a solution with $$pH < 7$$

B
$$pH = -\log { \displaystyle\frac { 1 }{ \left[ { H }^{ + } \right] } } $$

C
Only at $$25^{0}C$$ the $$pH$$ of water is 7

D
The value of $$p{ K }_{ w }$$ at $$25^{0}C$$ is 7

Solution

As we know, $$K_w = [H^+][OH^-] = 10^{-14}$$

So value of $$pK_w$$ at $$25^{0}C$$ is $$14$$.

At neutral point, $$[H^+] =\sqrt {K_w} = 10^{-7} M$$

$$pH = -log[H^+] = 7$$

Also, upon hydrolysis of salt of a strong base and weak acid gives a solution with $$pH > 7$$
Question 5
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Which of the following factor shifted the reaction, $$PCl_3 + Cl_2\rightleftharpoons PCl_5$$ at left side?

A
Addition of $$PCl_5$$

B
Increase pressure

C
Constant temperature

D
Catalyst

Solution

If a chemical reaction is at equilibrium and experiences a change in pressure, temperature, or concentration of products or reactants, the equilibrium shifts in the opposite direction to offset the change.

By using the above principle, Le chatelier's principle, it is clear that adding $$PCl_5$$ will shift the reaction to the left side.

Using catalyst doesn't affect the state of equilibrium. Also by keeping the temperature constant, the equilibrium state is not affected.

If you increase the pressure, the position of equilibrium will move in such a way as to decrease the pressure again - if that is possible. It can do this by favouring the reaction which produces fewer molecules. If there are the same number of molecules on each side of the equation, then a change of pressure makes no difference to the position of equilibrium. So, the equilibrium shifts to the right by increasing pressure.

Hence, option A is correct.
Question 6
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If $$0.00050$$ mol $$NaH{CO}_{3}$$ is added to $$1$$ litre of a buffered solution of $$pH\ 8$$, then how much material will exist in each of the three forms $${H}_{2}{CO}_{3},H{CO}_{3}^{-}$$ and $${CO}_{3}^{2-}$$?
For $${H}_{2}{CO}_{3}$$, $${K}_{1}=5\times {10}^{-7}$$; $${K}_{2}=5\times {10}^{-13}$$

A
$$[{H}_{2}{CO}_{3}]=9.85\times {10}^{-6}M$$, $$[H{CO}_{3}^{-}]=4.9\times {10}^{-4}M$$, $$[{CO}_{3}^{-2}]=2.45\times {10}^{-8}M$$

B
$$[{H}_{2}{CO}_{3}]=4.9\times {10}^{-6}M$$, $$[H{CO}_{3}^{-}]=9.85\times {10}^{-4}M$$, $$[{CO}_{3}^{-2}]=2.45\times {10}^{-8}M$$

C
$$[{H}_{2}{CO}_{3}]=9.85\times {10}^{-6}M$$, $$[H{CO}_{3}^{-}]=2.45\times {10}^{-4}M$$, $$[{CO}_{3}^{-2}]=4.9\times {10}^{-8}M$$

D
$$[{H}_{2}{CO}_{3}]=92.45\times {10}^{-6}M$$, $$[H{CO}_{3}^{-}]=4.9\times {10}^{-4}M$$, $$[{CO}_{3}^{-2}]=9.85\times {10}^{-8}M$$

Solution

Given, $$pH=8, [{H}^{+}]={10}^{-8}, [{OH}^{-}={10}^{-6}$$

$$H{CO}_{3}^{-}\rightleftharpoons {H}^{+}+{CO}_{3}^{-2}$$ $$K=5\times {10}^{-13}$$
$$0.0005$$
$$0.0005-y-z$$ $${10}^{-8}$$ $$y$$
$$H{CO}_{3}^{-}+{H}_{2}O\rightleftharpoons {H}_{2}{CO}_{3}+{OH}^{-}$$ $$K=\cfrac{{K}_{w}}{{K}_{1}}=2\times {10}^{-8}$$
$$0.0005$$
$$0.0005-y-z$$ $$z$$ $${10}^{-6}$$
Since equilibrium constant for first reaction is very less, $$y<<z$$
$$2\times {10}^{-8}=\cfrac{z({10}^{-6})}{0.0005-z}$$
$$51z=0.0005,\Rightarrow z=9.8\times {10}^{-6}$$
$$[{H}_{2}{CO}_{3}]=9.8\times {10}^{-6}M$$
$$[H{CO}_{3}^{-}]=0.0005-9.8\times {10}^{-6}=4.9\times {10}^{-4}M$$
$$5\times {10}^{-3}=\cfrac{{10}^{-8}\times y}{4.9\times {10}^{-4}}$$
$$[{CO}_{3}^{2-}]=y=2.45\times {10}^{-8}M$$
Question 7
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The equilibrium, $$SO_2Cl_2 \:(g)\rightleftharpoons SO_2 \:(g) + Cl_2 \:(g)$$ is attained at $$25 ^{\circ}\! C$$ in a closed container and an inert gas, helium, is introduced. Which of the following statements is correct?

A
Concentrations of $$SO_2, \: Cl_2 $$ and $$SO_2Cl_2$$ are changed

B
No effect on equilibrium

C
Concentration of $$SO_2$$ is reduced

D
$$K_p$$ of reaction is increasing

Solution

$$SO_2Cl_2 \:(g)\rightleftharpoons SO_2 \:(g) + Cl_2 \:(g)$$

The reaction will go to forward direction by adding reactants and go to backward by adding products.

Also, no effect on equilibrium concentration by adding inert gas helium at constant pressure.
Question 8
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Calculate the $${OH}^{-}$$ concentration and the $${H}_{3}{PO}_{4}$$ concentration of a solution prepared by dissolving $$0.1$$ mol of $${Na}_{3}{PO}_{4}$$ in sufficient water to make $$1L$$ of solution.
$${K}_{1}=7.1\times {10}^{-3}$$, $${K}_{2}=6.3\times {10}^{-8}$$; $${K}_{3}=4.5\times {10}^{-13}$$

A
$$[{OH}^{-}]=1.36\times {10}^{-2}M$$, $$[{H}_{3}{PO}_{4}]=2.96\times {10}^{-18}M$$

B
$$[{OH}^{-}]=3.73\times {10}^{-2}M$$, $$[{H}_{3}{PO}_{4}]=5.93\times {10}^{-18}M$$

C
$$[{OH}^{-}]=5.9\times {10}^{-2}M$$, $$[{H}_{3}{PO}_{4}]=6.45\times {10}^{-18}M$$

D
$$[{OH}^{-}]=6.45\times {10}^{-2}M$$, $$[{H}_{3}{PO}_{4}]=5.9\times {10}^{-18}M$$

Solution

$${Na}_{3}{PO}_{4}+{H}_{2}O\longrightarrow {Na}_{2}H{PO}_{4}+NaOH$$
$$K=\cfrac{{K}_{w}}{{K}_{3}}=0.0222$$
$${Na}_{2}H{PO}_{4}+{H}_{2}O\longrightarrow Na{H}_{2}{PO}_{4}+NaOH$$
$$K=\cfrac{{K}_{w}}{{K}_{2}}=1.58\times {10}^{-7}$$
$$Na{H}_{2}{PO}_{4}+{H}_{2}O\longrightarrow {H}_{3}{PO}_{4}+NaOH$$
$$K=\cfrac{{K}_{w}}{{K}_{3}}=1.4\times {10}^{-12}$$
Since equilibrium constant of 2nd and 3rd reaction is very less, $$[{OH}^{-}]$$ will mainly come from 1st reaction.
$${Na}_{3}{PO}_{4}+{H}_{2}O\longrightarrow {Na}_{2}H{PO}_{4}+NaOH$$
$$0.1$$
$$0.1-x$$ $$x$$ $$x$$
$$\cfrac{{x}^{2}}{0.1-x}=0.0222$$ $$\Rightarrow$$ $$45{x}^{2}+x-0.1=0$$
$$x=3.73\times {10}^{-2}$$
$$[{OH}^{-}]=x=3.73\times {10}^{-2}M$$
$${Na}_{2}H{PO}_{4}+NaOH\longrightarrow Na{H}_{2}{PO}_{4}+NaOH$$
$$x$$
$$x-y$$ $$y$$ $$y+x$$
$$x-y \simeq x$$
$$y+x\simeq x$$
$$1.58\times {10}^{-7}=\cfrac{(x+y)}{(x-y)}y=y$$
$$Na{H}_{2}{PO}_{4}+{H}_{2}O\longrightarrow {H}_{3}{PO}_{4}+NaOH$$
$$y$$
$$y-z$$ $$z$$ $$z+x$$
$$y-z\simeq y$$, $$z+x\simeq x$$
$$1.4\times {10}^{-12}=\cfrac{z(x+z)}{(y-z)}=\cfrac{z\times x}{y}=\cfrac{z\times 3.73\times {10}^{2}}{1.58\times {10}^{-7}}$$
$$z=5..93\times {10}^{-18}$$
$$[{H}_{3}{PO}_{4}]=5.93\times {10}^{-18}M$$
Question 9
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Ammonia is the weak base that reacts with water according to the equation:

$$NH_3(aq)+H_2O(l)⇌NH_4^+(aq)+OH^-(aq) $$

Will any of the following increase the per cent of ammonia that is converted to the ammonium ion in water?

A
Addition of $$NaOH$$

B
Addition of $$HCl$$

C
Addition of $$NH_4Cl$$

D
None of these

Solution

$$NH_3 (aq) + H_2O (l) \rightleftharpoons NH_4^+ (aq) + OH^ (aq)$$

When HCl is added, the hydrogen ions will neutralize hydroxide ions to form water. The equilibrium will shift in the forward direction so that more hydroxide ions are obtained. This will increase the percentage of ammonia that is converted to ammonium ions in water.

Hence, option B is correct.
Question 10
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$$EDTA,$$ often abbreviated as $${H}_{4}Y$$, forms very stable complexes with almost all metal ions. Calculate the fraction of EDTA in the fully protonated form, $${H}_{4}Y$$ in a solution obtained by dissolving $$0.1$$ mol $${Na}_{4}Y$$ in $$1$$ litre.
Given, the acid dissociation constants of $${H}_{4}Y$$ are as follows:
$${k}_{1}=1.02\times {10}^{-2}$$, $${k}_{2}=2.13\times {10}^{-3}$$, $${k}_{3}=6.92\times {10}^{-7}$$; $${k}_{4}=5.50\times {10}^{-11}$$

A
$$3.82\times {10}^{-26}$$

B
$$3.82\times {10}^{-6}$$

C
$$3.82\times {10}^{-20}$$

D
$$3.82\times {10}^{-30}$$

Solution

$${Na}_{4}Y+{H}_{2}O\rightleftharpoons {Na}_{3}HY+NaOH\quad \cfrac{{k}_{w}}{{k}_{4}}=1.818\times {10}^{-4}$$
$$0.1$$ Initial
$$0.1-x$$ $$x$$ $$x$$ Final
$$\cfrac{{x}^{2}}{0.1-x}=1.818\times {10}^{-4}$$ $$\Rightarrow$$ $$5500.55{x}^{2}+x-0.1=0$$
$$x=4.17\times {10}^{-3}$$
$${Na}_{3}HY+{H}_{2}O\rightleftharpoons {Na}_{2}{H}_{2}Y+NaOH\cfrac{{k}_{w}}{{k}_{3}}=1.445\times {10}^{-8}$$
$$x$$
$$x-y$$ $$y$$ $$y+x$$
Since $$y<<x$$
$$x-y\simeq x$$
$$x+y\simeq x$$
$$1.445\times {10}^{-8}=\cfrac{y\times x}{x}=y$$
$${Na}_{2}{H}_{2}Y+{H}_{2}O\rightleftharpoons Na{H}_{3}Y+NaOH\quad \cfrac {{k}_{w}}{{k}_{2}}=4.7\times {10}^{-12}$$
$$y$$
$$y-z$$ $$z$$ $$z+x$$
$$y-z\simeq y$$, $$z+x\simeq x$$
$$4.7\times {10}^{-12}=\cfrac{z\times x}{y}$$
$$z=1.628\times {10}^{-17}$$
$$Na{H}_{3}Y+{H}_{2}O\rightleftharpoons {H}_{4}Y+NaOH$$ $$\cfrac {{k}_{w}}{{k}_{1}}=9.8\times {10}^{-13}$$
$$z$$
$$z-t$$ $$t$$ $$t+x$$
$$z-t\simeq z$$
$$t+x\simeq x$$
$$9.8\times {10}^{-13}=\cfrac{t.x}{z}$$
$$t=3.82\times {10}^{-26}$$

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Equilibrium Test - 50

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Equilibrium Test - 50

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SHARING IS CARING

Question 1

shifts in the product side

shifts in the reactant side

cannot be predicted

remains unchanged

Solution

Question 2

$$4.0\times 10^{-9}$$

$$4.0\times 10^{5}$$

$$2.5\times 10^{8}$$

$$2.5\times 10^{-6}$$

Solution

Question 3

$$33.6 mg$$

$$3.36 mg$$

$$66.4 mg$$

$$6.64 mg$$

Solution

Question 4

Upon hydrolysis of salt of a strong base and weak acid gives a solution with $$pH < 7$$

$$pH = -\log { \displaystyle\frac { 1 }{ \left[ { H }^{ + } \right] } } $$

Only at $$25^{0}C$$ the $$pH$$ of water is 7

The value of $$p{ K }_{ w }$$ at $$25^{0}C$$ is 7

Solution

Question 5

Addition of $$PCl_5$$

Increase pressure

Constant temperature

Catalyst

Solution

Question 6

$$[{H}_{2}{CO}_{3}]=9.85\times {10}^{-6}M$$, $$[H{CO}_{3}^{-}]=4.9\times {10}^{-4}M$$, $$[{CO}_{3}^{-2}]=2.45\times {10}^{-8}M$$

$$[{H}_{2}{CO}_{3}]=4.9\times {10}^{-6}M$$, $$[H{CO}_{3}^{-}]=9.85\times {10}^{-4}M$$, $$[{CO}_{3}^{-2}]=2.45\times {10}^{-8}M$$

$$[{H}_{2}{CO}_{3}]=9.85\times {10}^{-6}M$$, $$[H{CO}_{3}^{-}]=2.45\times {10}^{-4}M$$, $$[{CO}_{3}^{-2}]=4.9\times {10}^{-8}M$$

$$[{H}_{2}{CO}_{3}]=92.45\times {10}^{-6}M$$, $$[H{CO}_{3}^{-}]=4.9\times {10}^{-4}M$$, $$[{CO}_{3}^{-2}]=9.85\times {10}^{-8}M$$

Solution

Question 7

Concentrations of $$SO_2, \: Cl_2 $$ and $$SO_2Cl_2$$ are changed

No effect on equilibrium

Concentration of $$SO_2$$ is reduced

$$K_p$$ of reaction is increasing

Solution

Question 8

$$[{OH}^{-}]=1.36\times {10}^{-2}M$$, $$[{H}_{3}{PO}_{4}]=2.96\times {10}^{-18}M$$

$$[{OH}^{-}]=3.73\times {10}^{-2}M$$, $$[{H}_{3}{PO}_{4}]=5.93\times {10}^{-18}M$$

$$[{OH}^{-}]=5.9\times {10}^{-2}M$$, $$[{H}_{3}{PO}_{4}]=6.45\times {10}^{-18}M$$

$$[{OH}^{-}]=6.45\times {10}^{-2}M$$, $$[{H}_{3}{PO}_{4}]=5.9\times {10}^{-18}M$$

Solution

Question 9

Addition of $$NaOH$$

Addition of $$HCl$$

Addition of $$NH_4Cl$$

None of these

Solution

Question 10

$$3.82\times {10}^{-26}$$

$$3.82\times {10}^{-6}$$

$$3.82\times {10}^{-20}$$

$$3.82\times {10}^{-30}$$

Solution

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