Equilibrium Test

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Equilibrium Test - 49

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Question 1
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The thermal dissociation equilibrium of $$CaC{O}_{3}\left(s\right)$$ is studied under different conditions.
$$CaC{O}_{3}\left(s\right) \rightleftharpoons CaO\left(s\right) + C{O}_{2}\left(g\right)$$
For this equilibrium, the correct statement(s) is (are) :

A
$$\Delta H$$ is dependent on $$T$$

B
$$K$$ is independent of the initial amount of $$CaC{O}_{3}$$

C
$$K$$ is dependent on the pressure of $$C{O}_{2}$$ at a given $$T$$

D
$$\Delta H$$ is independent of the catalyst, if any

Solution

Option $$(C)$$ is correct.
$$K$$ is dependent on the pressure of $$C{O}_{2}$$ and $$CaCO_3$$ at a given $$T$$.
When a system at equilibrium undergoes a change in pressure, the equilibrium of the system will shift to offset the change and establish a new equilibrium. The system can shift in one of two ways:

Toward the reactants (i.e. in favour of the reverse reaction)
Toward the products (i.e. in favour of the forward reaction)
The effects of changes in pressure can be described as follows (this only applies to reactions involving gases):
When there is an increase in pressure, the equilibrium will shift towards the side of the reaction with fewer moles of gas.
When there is a decrease in pressure, the equilibrium will shift towards the side of the reaction with more moles of gas.
Question 2
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Assertion: On mixing equal volumes of 1 M $$HCl$$ and of 2 M $$CH_3COONa$$, an acidic buffer solution is formed.
Reason: The resultant mixture contains $$CH_3COOH$$ and $$CH_3COONa$$ which are parts of acidic buffer.

A
Both Assertion and Reason are correct and Reason is the correct explanation for Assertion

B
Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion

C
Assertion is correct but Reason is incorrect

D
Assertion is incorrect but Reason is correct

E
Both Assertion and Reason are incorrect

Solution

A solution that resists change in pH value upon addition of small amount of strong acid or base (less than 1 %) or when solution is diluted is called buffer solution.
Acidic buffer solution : An acidic buffer solution consists of solution of a weak acid and its salt with strong base. The best-known example is a mixture of solution of acetic acid and its salt with strong base $$(CH_3COONa)$$.
Reaction:
$$CH_3COONa + HCl \rightarrow CH_3COONa + CH_3COOH$$
2 1 - -
1 - 1 1
which are parts of acidic buffer
Question 3
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The reaction which proceeds in the forward direction is:

A
$$F{e}_{2}{O}_{3} + 6HCl \rightarrow 2FeC{l}_{3} + 3{H}_{2}O$$

B
$$SnC{l}_{4} + H{g}_{2}C{l}_{2} \rightarrow SnC{l}_{2} + 2HgC{l}_{2}$$

C
$$N{H}_{3} + {H}_{2}O + NaCl \rightarrow N{H}_{4}Cl + NaOH$$

D
$$2CuI + {I}_{2} + 4{K}^{+} \rightarrow 2{Cu}^{2+} + 3KI$$

Solution
Question 4
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Directions For Questions

The dissolution of ammonia gas in water does not obey Henry's law. On dissolving, a major portion of ammonia molecules reacts with $${H}_{2}O$$ to form $${NH}_{4}OH$$ molecules.

$${NH}_{4}OH$$ again dissociates into $${NH}_{4}^{+}$$ and $${OH}^{-}$$ ions. In solution therefore, we have $${NH}_{3}$$ molecules, $${NH}_{4}OH$$ molecules and $${NH}_{4}^{+}$$ ions and the following equilibrium exist:

$${NH}_{3}(g)$$ $$ \rightleftharpoons{NH}_{3}(l)+{H}_{2}O{(l)}\rightleftharpoons {NH}_{4}OH{(aq)}\rightleftharpoons {NH}_{4}^{+}{(aq)}+{OH}^{-}{(aq)}$$

Let, $${c}_{1}$$ $${(mol/L)}$$ of $${NH}_{3}$$ pass in liquid state which on dissolution in water forms $${c}_{2}$$ $${(mol/L)}$$ of $${NH}_{4}OH$$. The solution contains $${c}_{3}$$ $${(mol/L)}$$ of $${NH}_{4}^{+}$$ ions

...view full instructions

The dissociation constant of $${NH}_{4}OH$$ can be given as

A
$${ K }_{ b }=\cfrac { { \left( { c }_{ 3 } \right) }^{ 2 } }{ \left( { c }_{ 2 }-{ c }_{ 3 } \right) } $$

B
$${ K }_{ b }=\cfrac { { \left( { c }_{ 3 } \right) }^{ 2 } }{ { c }_{ 2 } } $$

C
$${ K }_{ b }=\cfrac { { c }_{ 3 } }{ \left( { c }_{ 2 }-{ c }_{ 3 } \right) } $$

D
$${ K }_{ b }=\cfrac { { c }_{ 3 } }{ \left( { c }_{ 1 }-{ c }_{ 2 } \right) } $$

Solution

As given,

$$\underset { { c }_{ 2 }\\ ({ c }_{ 2 }-{ c }_{ 3 }) }{ { NH }_{ 4 }OH } \rightleftharpoons \underset { 0\\ { c }_{ 3 } }{ { NH }_{ 4 }^{ + } } +\underset { 0\\ { c }_{ 3 } }{ { OH }^{ - } } $$

$$K_b = [NH_4^+][OH^-]/[NH_4OH]$$

$$ \quad { K }_{ b }=\cfrac { { { c }_{ 3 } }^{ 2 } }{ \left( { c }_{ 2 }-{ c }_{ 3 } \right) } $$
Question 5
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The equilibrium constant for the reaction, $${N}_{2}\left(g\right) + {O}_{2}\left(g\right) \rightleftharpoons 2NO\left(g\right)$$ is $$4\times {10}^{-4}$$ at $$2000 K$$. In presence of a catalyst, the equilibrium is attained $$10$$ times faster. Therefore, the equilibrium constant in the presence of the catalyst at $$2000 \:K$$ is:

A
$$4\times{10}^{-5}$$

B
$$4\times{10}^{-4}$$

C
$$4\times{10}^{-3}$$

D
$$2 \times 10^{-4}$$

Solution

The equilibrium constant in the presence of the catalyst at $$2000\:K$$ is $$4\times{10}^{-4}$$.

$${K}_{p}$$ and $${K}_{c}$$ values do not change with catalyst. They only change with temperature.
With catalyst equilibrium is attained faster.

Hence, the value of equilibrium constant remains the same.
Question 6
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Given a reaction:
$$C{l}_{2}\left(g\right) + 3{F}_{2}\left(g\right) \rightleftharpoons 2Cl{F}_{3}\left(g\right); \Delta H = -329 kJ$$
Which of the following will increase the quantity of $$Cl{F}_{3}$$ in an equilibrium mixture of $$C{l}_{2}, {F}_{2}$$ and $$Cl{F}_{3}$$?

A
Increase in temperature

B
Removal of $$C{l}_{2}$$

C
Increase in volume of container

D
Addition of $${F}_{2}$$

Solution

Favourable conditions for forward reaction according to Le Chatelier's principle are the decrease in temperature, increase in concentration of reactants and increase in pressure.
Le Chatelier's principle states that if a dynamic equilibrium is disturbed by changing the conditions, the position of equilibrium shifts to counteract the change to re-establish an equilibrium.
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.
Le Chatelier's principle implies that the addition of heat to a reaction will favor the endothermic direction of a reaction as this reduces the amount of heat produced in the system.
Increasing the concentration of reactants will drive the reaction to the right, while increasing the concentration of products will drive the reaction to the left.
Hence, addition of $$F_2$$ will increase the quantity of $$Cl{F}_{3}$$ in an equilibrium mixture of $$C{l}_{2}, {F}_{2}$$ and $$Cl{F}_{3}$$.
Question 7
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The volume of the reaction vessel containing an equilibrium mixture in the reaction,

$$S{O}_{2}C{l}_{2}\left(g\right) \rightleftharpoons S{O}_{2}\left(g\right) + C{l}_{2}\left(g\right)$$

is increased. When equilibrium is re-established :

A
the mass of $$S{O}_{2}\left(g\right)$$ will decrease

B
the mass of $$S{O}_{2}C{l}_{2}\left(g\right)$$ will increase

C
the mass of $$C{l}_{2}\left(g\right)$$ will increase

D
the mass of $$C{l}_{2}\left(g\right)$$ will remain unchanged

Solution

An increase in volume for the equilibrium having,

$${K}_{c} = \displaystyle\frac{\left[S{O}_{2}\right]\left[C{l}_{2}\right]}{\left[S{O}_{2}C{l}_{2}\right]} = \displaystyle\frac{\left[mole\ of\ S{O}_{2}\right]\left[mole\ of\ C{l}_{2}\right]}{\left[mole\ of\ S{O}_{2}C{l}_{2}\right]\times Volume}$$

will increase the mole of $$C{l}_{2}$$ or $$S{O}_{2}$$ to maintain $${K}_{c}$$ values constant.
Question 8
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Densities of diamond and graphite are $$3.5$$ and $$2.3 {g}/{mL}$$ respectively. Increase of pressure on the equilibrium:

$${C}_{\left(diamond\right)} \rightleftharpoons {C}_{\left(graphite\right)}$$

A
favours backward reaction

B
favours forward reaction

C
have no effect

D
increases the reaction rate

Solution

According to Le Chatlier's principle, if the pressure increases at equilibrium the reaction occurs in the direction where the volume decreases which is nothing but the density increased side.

As the density of the diamond is more than the graphite, the backward direction is favourable.

Hence option A is correct.
Question 9
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Acetyl salicylic acid (aspirin) ionises in water as: $$HC_9H_7O_4+H_2O\rightarrow H_3O^+ + C_9H_7O_4^-;$$ $$(K_a = 2.75\times 10^{-9})$$. If two tablets of aspirin each of 0.32 g is dissolved in water to produce 250 mL solution, calculate $$[\overset{\circleddash}{O}H].$$

A
$$1.61\times 10^{-9}M$$

B
$$1.61\times 10^{-7}M$$

C
$$1.61\times 10^{-3}M$$

D
$$1.61\times 10^{-5}M$$

Solution

$$[HC_9H_7O_4] = \displaystyle \frac{0.32\times 2\times 1000}{180\times 250} = 0.014\, M$$

$$\because \alpha = \displaystyle \sqrt{\frac{K_a}{C}} = \sqrt{\frac{2.75\times 10^{-9}}{0.014}} = 4.43\times 10^{-4}$$

$$\therefore [H^{\oplus}] = C.\alpha = 0.014 \times 4.43\times 10^{-4}=6.21\times 10^{-6}\, M$$

$$[C_9H_7O_4^{\ominus}]=C.\alpha = 6.21\times 10^{-6}\, M$$

$$[OH^{\circleddash}]=\displaystyle \frac{10^{-14}}{[H^{\oplus}]}=\frac{10^{-14}}{6.21\times 10^{-6}}= 1.61\times 10^{-9}M$$
Question 10
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Calculate the equilibrium constants for the reactions with water of $$H_2PO_4^{\circleddash}, HPO_4^{2-}$$ and $$PO_4^{3-}$$ as base. Comparing the relative values of two equilibrium constants of $$H_2PO_4^{\circleddash}$$ with water, deduce whether solutions of this ion in water are acidic or bases. Deduce whether solutions of $$HPO_4^{2-}$$ are acidic or bases. Given $$K_1, K_2$$ and $$K_3$$ for $$H_3PO_4$$ are $$7.1\times 10^{-3}, 6.3\times 10^{-8}$$ and $$4.5\times 10^{-13}$$ respectively .

A
$$1.4\times 10^{-12}, 1.6\times 10^{-7}, 2.2\times 10^{-2}, H_2PO_4^{\circleddash}$$ is acidic and $$HPO_4^{2-}$$ is basic

B
$$2.2\times 10^{-12}, 1.6\times 10^{-7}, 1.4\times 10^{-2}, H_2PO_4^{\circleddash}$$ is basic and $$HPO_4^{2-}$$ is basic

C
$$2.2\times 10^{-12}, 1.6\times 10^{-7}, 1.4\times 10^{-2}, H_2PO_4^{\circleddash}$$ is acidic and $$HPO_4^{2-}$$ is basic

D
$$1.4\times 10^{-12}, 1.6\times 10^{-7}, 2.2\times 10^{-2}, H_2PO_4^{\circleddash}$$ is basic and $$HPO_4^{2-}$$ is acidic

Solution

i. $$H_2PO_4^{\circleddash} + H_2O \rightleftharpoons H_3PO_4+\overset{\circleddash}{O}H$$
$$K_h = \displaystyle \frac{K_w}{K_a}=\frac{10^{-14}}{7.1\times 10^{-3}}=1.4\times 10^{-12}$$
ii. $$HPO_4^{2-} + H_2O \rightleftharpoons H_2PO_4^{\circleddash}+\overset{\circleddash}{O}H$$
$$K_h = \displaystyle \frac{K_w}{K_2} = \frac{10^{-14}}{6.3\times 10^{-8}}=1.6\times 10^{-7}$$
iii. $$PO_4^{3-}+H_2O \rightleftharpoons HPO_4^{2-}+\overset{\circleddash}{O}H$$
$$K_h = \displaystyle \frac{K_w}{K_3} = \frac{10^{-14}}{4.5\times 10^{-13}}=2.2\times 10^{-2}$$
$$H_2PO_4^{\circleddash}$$ can react with water to yield $$HPO_4^{2-}$$ and $$H_3O^{\oplus}$$ with a constant $$6.3\times 10^{-8}$$
Its reaction with water to yield $$H_3PO_4$$, $$\overset{\circleddash}{O}H$$ has a much smaller equilibrium constant, $$1.4 \times 10^{-12}$$, and so this ion in water is acidic comparision of the constants for $$HPO_4^{2-}(4.5\times 10^{-3}$$ as an acid, $$1.6\times 10^{-7}$$ as base $$)$$ leads to the conclusion that this ion in solution is basic.

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Equilibrium Test - 49

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Equilibrium Test - 49

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Question 1

$$\Delta H$$ is dependent on $$T$$

$$K$$ is independent of the initial amount of $$CaC{O}_{3}$$

$$K$$ is dependent on the pressure of $$C{O}_{2}$$ at a given $$T$$

$$\Delta H$$ is independent of the catalyst, if any

Solution

Question 2

Both Assertion and Reason are correct and Reason is the correct explanation for Assertion

Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion

Assertion is correct but Reason is incorrect

Assertion is incorrect but Reason is correct

Both Assertion and Reason are incorrect

Solution

Question 3

$$F{e}_{2}{O}_{3} + 6HCl \rightarrow 2FeC{l}_{3} + 3{H}_{2}O$$

$$SnC{l}_{4} + H{g}_{2}C{l}_{2} \rightarrow SnC{l}_{2} + 2HgC{l}_{2}$$

$$N{H}_{3} + {H}_{2}O + NaCl \rightarrow N{H}_{4}Cl + NaOH$$

$$2CuI + {I}_{2} + 4{K}^{+} \rightarrow 2{Cu}^{2+} + 3KI$$

Solution

Question 4

$${ K }_{ b }=\cfrac { { \left( { c }_{ 3 } \right) }^{ 2 } }{ \left( { c }_{ 2 }-{ c }_{ 3 } \right) } $$

$${ K }_{ b }=\cfrac { { \left( { c }_{ 3 } \right) }^{ 2 } }{ { c }_{ 2 } } $$

$${ K }_{ b }=\cfrac { { c }_{ 3 } }{ \left( { c }_{ 2 }-{ c }_{ 3 } \right) } $$

$${ K }_{ b }=\cfrac { { c }_{ 3 } }{ \left( { c }_{ 1 }-{ c }_{ 2 } \right) } $$

Solution

Question 5

$$4\times{10}^{-5}$$

$$4\times{10}^{-4}$$

$$4\times{10}^{-3}$$

$$2 \times 10^{-4}$$

Solution

Question 6

Increase in temperature

Removal of $$C{l}_{2}$$

Increase in volume of container

Addition of $${F}_{2}$$

Solution

Question 7

the mass of $$S{O}_{2}\left(g\right)$$ will decrease

the mass of $$S{O}_{2}C{l}_{2}\left(g\right)$$ will increase

the mass of $$C{l}_{2}\left(g\right)$$ will increase

the mass of $$C{l}_{2}\left(g\right)$$ will remain unchanged

Solution

Question 8

favours backward reaction

favours forward reaction

have no effect

increases the reaction rate

Solution

Question 9

$$1.61\times 10^{-9}M$$

$$1.61\times 10^{-7}M$$

$$1.61\times 10^{-3}M$$

$$1.61\times 10^{-5}M$$

Solution

Question 10

$$1.4\times 10^{-12}, 1.6\times 10^{-7}, 2.2\times 10^{-2}, H_2PO_4^{\circleddash}$$ is acidic and $$HPO_4^{2-}$$ is basic

$$2.2\times 10^{-12}, 1.6\times 10^{-7}, 1.4\times 10^{-2}, H_2PO_4^{\circleddash}$$ is basic and $$HPO_4^{2-}$$ is basic

$$2.2\times 10^{-12}, 1.6\times 10^{-7}, 1.4\times 10^{-2}, H_2PO_4^{\circleddash}$$ is acidic and $$HPO_4^{2-}$$ is basic

$$1.4\times 10^{-12}, 1.6\times 10^{-7}, 2.2\times 10^{-2}, H_2PO_4^{\circleddash}$$ is basic and $$HPO_4^{2-}$$ is acidic

Solution

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