Equilibrium Test

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Equilibrium Test - 48

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Weekly Quiz Competition

Question 1
1 / -0

What concentration of $$F-CH_2COOH, (K_a = 2.6\times10^{-3})$$ is needed so that $$[H^+] = 2\times10^{-3}$$?

A
$$1.53 \times10^{-3}\space M$$

B
$$2.6\times10^{-3}\space M$$

C
$$5.2\times10^{-3}\space M$$

D
$$3.53\times10^{-3}\space M$$

Solution

As we know,
$$\alpha = \sqrt{\dfrac{K_a}{C}}$$
$$[H^+] = C\alpha = \sqrt{C.K_a} = 2\times10^{-3}$$
Putting the value of C $$=2\times10^{-3}\space M$$ and $$K_a = 2.6 \times 10^{-3}$$,
C = $$\dfrac {(2 \times 10^{-3})^2}{2.6 \times 10^{-3}} = 1.53 \times 10^{-3} M.$$
Question 2
1 / -0

Directions For Questions

For a reversible reaction at a certain temperature when it is at equilibrium or equilibrium has been attained whether physical or chemical, a change in certain variables might change the state of equilibrium. These variables include pressure, volume, concentration and temperature. Due to these changes, a system under equilibrium changes its state in such a manner, i.e., the equilibrium moves in forward direction or backward direction, so that the effect of change is annuled. For a gaseous phase endothermic decomposition of phosphorus pentachloride, can be made spontaneously by increasing concentration of $$PC{l}_{5}$$, lowering the pressure and increasing temperature of the system.

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The equilibrium, solid $$\rightleftharpoons$$ liquid $$\rightleftharpoons$$ gas, will shift in forward direction when :

A
temperature is raised

B
temperature is constant

C
temperature is lowered

D
pressure is increased

Solution

According to Le Chatelier's principle adding heat to a solid and liquid in equilibrium with endothermic nature will cause the mass of solid to decrease.
Increase in temperature favours forward reaction. In endothermic reactions, heat is absorbed with the reactants.
Question 3
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Which can act as buffer?

A
$${NH}_{4}Cl+{NH}_{4}OH$$

B
$${CH}_{3}COOH+{CH}_{3}COONa$$

C
$$40$$ mL of $$0.1M$$ $$NaCN+10$$ mL of $$0.1M$$ $$HCl$$

D
All of the above

Solution

Hint: Recollect the preparation of buffers.
Correct Answer: Option (D).

Explanation:
Buffer is a solution that resists the change in $$pH$$ when an acid or a base is added to it.
Types of buffers:
$$(i)$$ Basic buffer is a mixture of a weak base and its salt with a strong acid.
$$(ii)$$ Acidic buffer is a mixture of a weak acid and its salt with a strong base.

Option (A): $$NH_4OH$$ is a weak base and $$NH_4Cl$$ is its salt with strong acid $$HCl$$.

Option (B): $$CH_3COOH$$ is a weak acid and $$CH_3COONa$$ is its salt with strong base $$NaOH$$.

Option (C): Let us first look at the reaction that takes place:
$$NaCN+HCl\rightarrow NaCl+HCN$$
Now, initial moles of $$HCl$$ taken $$= 10 \times 0.1=1\ millimol$$
Whereas, initial moles of $$NaCN$$ taken $$=40 \times 0.1 = 4\ millimol$$

Thus, $$NaCN$$ is taken in far excess compared to $$HCl$$. So, after the reaction is over, much of $$NaCN$$ remains unreacted. So, the resulting solution contains $$HCN$$ which is a weak acid, and $$NaCN$$ which is its salt with a strong base $$NaOH$$. So, it is also a buffer.
Question 4
1 / -0

Which of the following reactions proceed in forward direction with an increase in temperature?

A
$${H}_{2}\left(g\right) + {I}_{2}\left(g\right) \rightleftharpoons 2HI\left(g\right) + 3000 cal$$

B
$${N}_{2}\left(g\right) + {O}_{2}\left(g\right) \rightleftharpoons 2NO\left(g\right) - 43200 cal$$

C
$${N}_{2}\left(g\right) + 3{H}_{2}\left(g\right) \rightleftharpoons 2N{H}_{3}\left(g\right) + 22400 cal$$

D
$$C\left(g\right) + {O}_{2}\left(g\right) \rightleftharpoons C{O}_{2}\left(g\right) + 94300 cal$$

Solution

Option (B) is correct.
Endothermic reactions are favoured with the increase in temperature. In an endothermic reaction, an increase in temperature favours the reaction to occur in the forward direction. It means heat is absorbed by the reactants. When the temperature of an equilibrium mixture is increased, the rate of both reactions increases but the rate of the endothermic reaction (the reaction that absorbs the added energy) is increased more.
Question 5
1 / -0

Simultaneous solubility of $$AgCNS\ (a)$$ and $$AgBr\ (b)$$ in a solution of water will be
$${K}_{{sp}_{(AgBr)}}=5\times {10}^{-13}$$ and $${K}_{{sp}_{(AgCNS)}}={10}^{-12}$$

A
$$a=4.08\times {10}^{-7}mol$$ $${litre}^{-1}$$; $$b=8.16\times {10}^{-7}$$ $$mol$$ $${litre}^{-1}$$

B
$$a=4.08\times {10}^{-7}mol$$ $${litre}^{-1}$$; $$b=4.08\times {10}^{-7}$$ $$mol$$ $${litre}^{-1}$$

C
$$a=8.16\times {10}^{-7}mol$$ $${litre}^{-1}$$; $$b=4.08\times {10}^{-7}$$ $$mol$$ $${litre}^{-1}$$

D
None of these

Solution

Suppose solubility of $$AgCNS$$ and $$AgBr$$ in a solution are $$a$$ and $$b$$ respectively.

$$AgCNS(s)\rightleftharpoons \underset { a+b }{ { Ag }^{ + }(aq) } +\underset { a }{ { CNS }^{ - }(aq) } $$

$$AgBr(s)\rightleftharpoons \underset { a+b }{ { Ag }^{ + } } +\underset { b }{ { Br }^{ - } } $$

$$\therefore$$ $$[{Ag}^{2+}]=(a+b); [CN{S}^{-}]=a$$ and $$[{Br}^{-}]=b$$

For $$AgCNS:{ K }_{ { sp }_{ AgCNS }\quad }=\left[ { Ag }^{ + } \right] \left[ { CNS }^{ - } \right] $$

or $$1\times {10}^{-12}=(a+b)(a)......(i)$$

For $$AgBr: { K }_{ { sp }_{ AgBr }\quad }=\left[ { Ag }^{ + } \right] \left[ { Br }^{ - } \right] $$

or $$5\times {10}^{-13}=(a+b)(b)........(ii)$$

By Eqs $$(i)$$ and $$(ii)$$, we get

$$\cfrac{a}{b}=\cfrac{{10}^{-12}}{5\times {10}^{-13}}=2$$ ($$a=2b$$)

$$\therefore $$ By Eq $$(i)$$

$$(2b+b)(2b)=1\times {10}^{-12}$$

$$\therefore$$ $$6{b}^{2}=1\times {10}^{-12}$$

or $$b=4.08\times {10}^{-7}$$ $$mol$$ $${litre}^{-1}$$

Similarly by Eq $$(ii)$$

$$[a+(a/2)]a=1\times {10}^{-12}$$

$$a=8.16\times {10}^{-7}$$ $$mol$$ $${litre}^{-1}$$
Question 6
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For the chemical reaction, $$3X\left(g\right) + Y\left(g\right) \longrightarrow {X}_{3}Y\left(g\right)$$, 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

$$3X\left(g\right) + Y\left(g\right) \longrightarrow {X}_{3}Y\left(g\right)$$
3 1 1
$$\Delta n = 1-4 = -3$$

The reaction shows a change in mole during the course of the reaction and thus, an increase in pressure will favour forward reaction.

Also, $${K}_{p}$$ changes with temperature. Catalyst has no effect on $${K}_{p}$$. Thus, $$P$$ and $$T$$ influence the equilibrium concentrations.
Question 7
1 / -0

Consider the reversible reaction,

$$HCN\left(aq\right) \rightleftharpoons {H}^{+}\left(aq\right) + C{N}^{-}\left(aq\right)$$

At equilibrium, the addition of $$C{N}^{-}\left(aq\right)$$ would :

A
reduce $$HCN\left(aq\right)$$ concentration

B
decrease the $${H}^{+}\left(aq\right)$$ ion concentration

C
increase the equilibrium constant

D
decrease the equilibrium constant

Solution

At equilibrium, the addition of $$C{N}^{-}\left(aq\right)$$ would decrease the $${H}^{+}\left(aq\right)$$ ion concentration.
$${K}_{a} =
\displaystyle\frac{\left[{H}^{+}\right]\left[C{N}^{-}\right]}{\left[HCN\right]}$$

An increase in $$\left[C{N}^{-}\right]$$ will decrease $$\left[{H}^{+}\right]$$ to maintain $${K}_{a}$$ constant.
In a general way: Le Chatelier's principle states that if the system is changed in a way that increases the concentration of one of the reacting species, it must favour the reaction in which that species is consumed. In other words, if there is an increase in products, the reaction quotient is increased, making it greater than the equilibrium constant.
Question 8
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An aqueous solution of hydrogen sulphide shows the equilibrium,
$${H}_{2}S \rightleftharpoons {H}^{+} + H{S}^{-}$$
If dilute hydrochloric acid is added to an aqueous solution of hydrogen sulphide without any change in temperature, then :

A
the equilibrium constant will change

B
the concentration of $$H{S}^{-}$$ will increase

C
the concentration of undissociated hydrogen sulphide will decrease

D
the concentration of $$H{S}^{-}$$ will decrease

Solution

Reaction:
$${H}_{2}S \rightleftharpoons {H}^{+} + H{S}^{-}$$

As we know, for this reaction
$${K}_{a}$$ for $${H}_{2}S = \displaystyle\frac{\left[{H}^{+}\right]\left[H{S}^{-}\right]}{\left[{H}_{2}S\right]}$$

An increase in $$\left[{H}^{+}\right]$$ will show a decrease in $$\left[H{S}^{-}\right]$$ to maintain constant $${K}_{a}$$ value.
Question 9
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In lime kiln, the reversible reaction,

$$CaC{O}_{3}\left(s\right) \rightleftharpoons CaO\left(s\right) + C{O}_{2}\left(g\right)$$

It proceeds to completion because :

A
of high temperature

B
$$C{O}_{2}$$ escapes out

C
$$CaO$$ is removed

D
of low temperature

Solution

Removal of the gaseous product always favours forward reaction. The equilibria that involve gases - that is, those equilibria that have different numbers of gaseous molecules on the left and right sides of the equilibrium equation.

Increased pressure favours the reaction that decreases the number of gaseous molecules. Increased pressure decreases the volume available to this gaseous equilibrium and favours the forward reaction.

Hence, option B is correct.
Question 10
1 / -0

For the reaction,
$$PC{l}_{5}\left(g\right) \rightleftharpoons PC{l}_{3}\left(g\right) + C{l}_{2}\left(g\right)$$

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
None of the above

Solution

According to Le Chatelier's principle, if a system in equilibrium is disturbed by changes in determining factors, such as temperature, pressure, and concentration of components then the system will tend to shift its equilibrium position in such a way so as to counteract the effect of the disturbance.
At constant pressure, there will be increase in total volume. Therefore, there will be decrease in number of moles per unit volume of each reactant and product. Hence, the equilibrium will shift towards the side where number of moles are increased.
Thus introducing constant pressure will favour the forward reaction at constant pressure.

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Equilibrium Test - 48

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Equilibrium Test - 48

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SHARING IS CARING

Question 1

$$1.53 \times10^{-3}\space M$$

$$2.6\times10^{-3}\space M$$

$$5.2\times10^{-3}\space M$$

$$3.53\times10^{-3}\space M$$

Solution

Question 2

temperature is raised

temperature is constant

temperature is lowered

pressure is increased

Solution

Question 3

$${NH}_{4}Cl+{NH}_{4}OH$$

$${CH}_{3}COOH+{CH}_{3}COONa$$

$$40$$ mL of $$0.1M$$ $$NaCN+10$$ mL of $$0.1M$$ $$HCl$$

All of the above

Solution

Question 4

$${H}_{2}\left(g\right) + {I}_{2}\left(g\right) \rightleftharpoons 2HI\left(g\right) + 3000 cal$$

$${N}_{2}\left(g\right) + {O}_{2}\left(g\right) \rightleftharpoons 2NO\left(g\right) - 43200 cal$$

$${N}_{2}\left(g\right) + 3{H}_{2}\left(g\right) \rightleftharpoons 2N{H}_{3}\left(g\right) + 22400 cal$$

$$C\left(g\right) + {O}_{2}\left(g\right) \rightleftharpoons C{O}_{2}\left(g\right) + 94300 cal$$

Solution

Question 5

$$a=4.08\times {10}^{-7}mol$$ $${litre}^{-1}$$; $$b=8.16\times {10}^{-7}$$ $$mol$$ $${litre}^{-1}$$

$$a=4.08\times {10}^{-7}mol$$ $${litre}^{-1}$$; $$b=4.08\times {10}^{-7}$$ $$mol$$ $${litre}^{-1}$$

$$a=8.16\times {10}^{-7}mol$$ $${litre}^{-1}$$; $$b=4.08\times {10}^{-7}$$ $$mol$$ $${litre}^{-1}$$

None of these

Solution

Question 6

temperature and pressure

temperature only

pressure only

temperature, pressure and catalyst

Solution

Question 7

reduce $$HCN\left(aq\right)$$ concentration

decrease the $${H}^{+}\left(aq\right)$$ ion concentration

increase the equilibrium constant

decrease the equilibrium constant

Solution

Question 8

the equilibrium constant will change

the concentration of $$H{S}^{-}$$ will increase

the concentration of undissociated hydrogen sulphide will decrease

the concentration of $$H{S}^{-}$$ will decrease

Solution

Question 9

of high temperature

$$C{O}_{2}$$ escapes out

$$CaO$$ is removed

of low temperature

Solution

Question 10

introducing an inert gas at constant volume

introducing chlorine gas at constant volume

introducing an inert gas at constant pressure

None of the above

Solution

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