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Which of the following relation holds true for an equilibrium reaction, of its reaction quotient $(Q) = 1$ ?

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Important Points to Remember in Chapter -1 - Equilibrium from Embibe Experts Gamma Question Bank for Engineering Chemistry Solutions

1. Chemical Equilibrium:

(i) Equilibrium is a dynamic process – equilibrium is established in a system when reactants combine to form products at the same rate in which products combine to form reactants.

${(\frac{d x}{d t})}_{f} = {(\frac{d x}{d t})}_{b} o r R_{f} = R_{b}$

(ii) Chemical equilibrium can be approached from either side. A catalyst can fasten the approach of equilibrium but does not alter the state of equilibrium.

(iii) System can be homogeneous or heterogeneous.

(iv) For a reaction in equilibrium

$a A + b B ⇌ c C + d D$

$K_{c} = \frac{{[C]}^{c} {[D]}^{d}}{{[A]}^{a} {[B]}^{b}}$ in terms of active mass

$K_{p} = \frac{[p_{C}]^{c} {[p_{D}]}^{d}}{{[p_{A}]}^{a} {[p_{B}]}^{b}}$ in terms of partial pressure

$K_{x} = \frac{{[X_{C}]}^{c} {[X_{D}]}^{d}}{{[X_{A}]}^{a} {[X_{B}]}^{b}}$ in terms of mole fraction

(v) Partial pressure of solid is taken as unity and in calculation of partial pressure of solids their number of moles are not considered.

(vi) $K_{p} = K_{c} {(R T)}^{{∆ n}_{g}}$

(a) When $Δ n_{g} = 0,$ then $K_{p} = K_{c}$

(b) When $Δ n_{g} > 0,$ then $K_{p} > K_{c}$

$Δ n_{g} = (c + d) - (a + b)$

Note: While determining $Δ n_{g}$ , take only gaseous species.

2. Properties of Equilibrium Constant:

For $A + B ⇌ C + D, K_{c} = K$

(i) $C + D ⇌ A + B, K_{c}^{'} = \frac{1}{K}$

(ii) $2 A + 2 B ⇌ 2 C + 2 D, K_{c}^{''} = K^{2}$

(iii) $\frac{A}{2} + \frac{B}{2} ⇌ \frac{C}{2} + \frac{D}{2}, K_{c}^{'''} = \sqrt{K}$

(iv) $A ⇌ B K_{c} = K_{1}$

$B ⇌ C K_{c} = K_{2}$

$C ⇌ D K_{c} = K_{3}$

Then $A ⇌ D, K_{c} = K_{1} K_{2} K_{3}$

(v) $A ⇌ B K_{c} = K_{1}$ $C ⇌ B K_{c} = K_{2}$

Then $A ⇌ C K_{c} = \frac{K_{1}}{K_{2}}$

3. Reaction Quotient for the Reversible Reaction(Q):

$A + B ⇌ C + D$

$Q = \frac{[C] [D]}{[A] [B]}$

$Q$ is taken at any condition of system.

$\Rightarrow$ If $Q = K_{e q},$ then system is in equilibrium.

$\Rightarrow$ If $Q > K_{e q,}$ system proceeds in backward side to attain equilibrium.

$\Rightarrow$ If $Q < K_{e q},$ system proceeds in forward side to attain equilibrium.

4. Degree of Dissociation:

Degree of dissociation $(α) = \frac{N u m b e r o f d i s s o c i a t e d m o l e s (x)}{I n i t i a l n u m b e r o f m o l e s (a)}$

Equilibrium constant $(K)$ depends upon the temperature and the way of writing the reaction.

5. Degree of Dissociation in terms of Vapor Density:

For the equilibrium: $A ⇌ n B$

$\propto = \frac{D - d}{d (n - 1)}$

Where $n$ is the number of moles of products from one mole of reactant, $D$ is the theoretical vapor density (if no dissociation takes place) and d is the observed-vapor density (due to dissociation or association)

Vapor density $\times 2 =$ Molecular weight.

6. Le-Chatelier's Principle:

If the system at equilibrium is subjected to change of any one of the factors such as concentration, temperature or pressure, the system adjusts itself in such a way as to nullify the effect of that change.

The following conclusions have been derived from this principle:

(i) Increases in concentration of any substance favors the direction of reaction in which it is used.

(ii) High pressure is favorable for the reaction in which there is a decrease in volume or the number of moles.

(iii) A rise in temperature favors the endothermic reaction.

Effect of temperature on equilibrium constant:

$Δ G^{0} = Δ H^{0} - T . Δ S^{0}$ and $Δ G^{0} = - R T l n (K)$

Assuming $∆ H^{0}$ and $Δ S^{0}$ are almost constant over a wide range of temperature.

$2.303 l o g K = - \frac{∆ H^{0}}{R T} + \frac{∆ S^{0}}{R}$

for two temperatures $T_{1}$ and $T_{2}$

$l o g \frac{K_{2}}{K_{1}} = \frac{∆ H^{0}}{2.303 R} [\frac{T_{2} - T_{1}}{T_{1} T_{2}}]$

This is Van't Hoff’s equation.

7. Application of Le-Chatelier's Principle:

(i) Ice water system (melting of ice):

lce $+$ Heat $⇌$ Water

It is an endothermic process and there is a decrease in volume. Thus, the favorable conditions for melting of ice are – $(a)$ High temperature and $(b)$ High pressure.

(ii) Solubility of gases in liquids:

When a gas dissolves in a liquid, there is a decrease in volume. Thus, increase in pressure will favor the dissolution of a gas in liquid.

Ionic Equilibrium:

8. Acid and Base:

(i) Arrhenius Concept: Acid ionizes in water to give $H_{3} O^{+}$ ion while base ionizes to give ${O H}^{-}$ ion.

(ii) Bronsted-Lowry's Protonic Concept: Acid is $H^{+}$ ion donor, and base is $H^{+}$ ion acceptor.

(iii) $H C l$ and chloride ion is a conjugate acid-base pair. If acid is weak, its conjugate base is strong and vice-versa.

(iv) A substance that can accept $H^{+}$ ion as well as can donate $H^{+}$ ion is called amphiprotic or amphoteric.

$H_{3} O^{+} ⇌ H_{2} O ⇌ H^{+} + {O H}^{-}$

(v) A Bronsted-Lowry acid-based reaction is always favored in the direction from the stronger to the weaker acid/base combinations.

9. Lewis Concept:

(i) Lewis acid is an electron-pair acceptor.

(ii) Lewis base is an electron-pair donor.

(iii) All Lewis bases are Bronsted-Lowry bases, but all the Lewis acid are not Bronsted acids.

10. Ionic Product of Water:

$K_{w} = [H_{3} O^{+}] [OH] = 1 \times 10^{- 14} a t 298 K$

$K_{w}$ is called an ionic product of water or autoionization or autoprotolysis constant.

11. Ionization of Acids and Bases:

(i) In a mixture of strong acids/bases:

$[H^{+}] = [H_{3} O^{+}] = \frac{Σ N V}{Σ V}, [{O H}^{-}] = \frac{Σ N V}{Σ V}$

(ii) In a mixture of acid and base, resultant is

(a) acidic mixture if $N_{1} V_{1}$ (acid) $> N_{2} V_{2}$ (base)

$[H_{3}^{+} O] = \frac{N_{1} V_{1} - N_{2} V_{2}}{V_{1} + V_{2}}$

(b) basic mixture if $N_{2} V_{2}$ (base) $> N_{1} V_{1}$ (acid)

$[{O H}^{-}] = \frac{N_{2} V_{2} - N_{1} V_{1}}{V_{1} + V_{2}}$

(iii) For conjugate acid-base pairs,

$K_{a} K_{b} = K_{w} = 1 \times 10^{14} a t 25^{°} C$

$K_{a} K_{b} = K_{w} = 10^{- 12} a t 90^{°} C$

Where $K_{a}$ is the ionization constant of acid and $K_{b}$ is the ionization constant of its conjugate base,

$H A + H_{2} O ⇌ H_{3} O^{+} + A^{-}, K_{a}$

$A^{-} + H_{2} O ⇌ H A + {O H}^{-}, K_{b}$

${p K}_{a} + {p K}_{b} = 14 = {p K}_{w}$

12. pH Scale:

$p H = - l o g [H_{3} O^{+}]$

$p O H = - l o g [{O H}^{-}]$

$p X = - l o g X$

$p H + p O H = {p K}_{w} = 14$

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13. Ostwald Dilution Law:

According to Ostwald dilution law $α \propto \sqrt{d i l u t i o n}$

At infinite dilution, $\propto = 100 %$

For a weak acid by Ostwald dilution law

$K_{a} = \frac{C α^{2}}{(1 - α)} \approx C α^{2}$

$[H^{+}] = \sqrt{K_{a} C}, p H = \frac{1}{2} [{p K}_{a} - l o g C]$

$[H^{+}] = C α$

and for a weak base:

$K_{b} = \frac{C α^{2}}{(1 - α)} \approx C α^{2} [{O H}^{-}] = C α$

$[{O H}^{-}] = \sqrt{K_{b} C}, p O H = \frac{1}{2} [{p K}_{b} - l o g C]$

Note: Ostwald dilution law is only applicable for weak electrolytes

14. Salt and Salt Hydrolysis:

$S a l t + W a t e r ⇌ A c i d + B a s e$

In the above step, hydrolysis is an endothermic step whereas neutralization is an exothermic step.

Types of Salt:

(i) General

(ii) Acidic

(iii) Basic

(iv) Mixed

(v) Double

(vi) Complex

Types of General Salts:

(i) $S A & S B$

(ii) $S A & W B$

(iii) $W A & S B$

(iv) $W A & W B$

Types of salt

Name of hydrolysis

Nature of aqueous solution and pH

$S A & S B$

$S A & W B$

$W A & S B$

Cationic

Anionic

Neutral $, p H = 7$

Acidic $, p H < 7$

Basic $, p H > 7$

	$K_{a} > K_{b}$	$K_{b} > K_{a}$	$K_{a} = K_{b}$
Hydrolysis	Cationic-anionic	Anionic-cationic	Neutral hydrolysis
Nature	Acidic	Basic	Neutral
$p H$	$p H < 7$	$p H > 7$	$p H = 7$

For WA - WB types of salt:

$K_{h} = \frac{K w}{l o n i z a t i o n c o n s t a n t o f w e a k a c i d o r b a s e}$

Summary:

Type of salts	$K_{h} = \frac{K w}{l o n i s a t i o n c o n s t a n t o f w e a k}$	$h = \sqrt{\frac{K_{h}}{C}}$	$[H^{+}]$	$p H$
SA & SB	-	-	-	$7$
WA & SB	$K_{h} = \frac{K_{W}}{K_{a}}$	$h = \sqrt{\frac{K_{w}}{K_{a} \times C}}$	$\sqrt{\frac{K_{w} \times K_{a}}{C}}$	$7 + \frac{1}{2} p K_{a} + \frac{1}{2} l o g C$
SA & WB	$K_{h} = \frac{K_{W}}{K_{b}}$	$h = \sqrt{\frac{K_{w}}{K_{b} \times C}}$	$\sqrt{\frac{K_{w} \times C}{K_{b}}}$	$7 - \frac{1}{2} p K_{b} - \frac{1}{2} l o g C$
WA & WB	$K_{h} = \frac{K_{w}}{K_{a} \times K_{b}}$	$h = \sqrt{\frac{K_{w}}{K_{a} \times K_{b}}}$	$\sqrt{\frac{K_{w} \times K_{a}}{K_{b}}}$	$7 + \frac{1}{2} p K_{a} - \frac{1}{2} p K_{b}$

Note: For amphiprotic anion (like ${H C O}_{3}^{-})$ :

$p H = \frac{{p K}_{1} + {p K}_{2}}{2}$

15. Buffer Solution:

(i) Buffer solutions are resistive in nature for $p H$ change.

(ii) On dilution, $p H$ of the buffer solution remains unchanged.

(iii) When small amount of $s t r o n g a c i d$ or $s t r o n g b a s e$ is mixed in buffer solution, pH of buffer solution remains almost unchanged.

16. Types of Buffer Solution:

(i) Simper buffer solution (aqueous solution of $w e a k a c i d + w e a k b a s e$ salts).

(ii) Mixed buffer solution:

(a) Acidic buffer solution $(w e a k a c i d + w e a k a c i d & s t r o n g b a s e s a l t s) .$

(b) Basic buffer solution $(w e a k b a s e + w e a k b a s e & s t r o n g a c i d s a l t s) .$

Henderson-Hasselbalch equation for buffer:

Acidic: $p H = {p K}_{a} + l o g \frac{[conjugate base]}{[weak acid]}$

Basic: $p O H = {p K}_{b} + l o g \frac{[conjugate acid]}{[weak base]}$

17. Solubility Equilibria of Sparingly Soluble Salts:

For a salt, $A_{x} B_{y} ⇌ x A^{y +} + y B^{x -}$

(i) $K_{s p} >$ Instantaneous ionic product $\Rightarrow$ unsaturated. (More salt can be dissolved)

(ii) $K_{s p} =$ Instantaneous ionic product $\Rightarrow$ saturated. (No more salt can be dissolved)

(iii) $K_{s p} <$ Instantaneous ionic product $\Rightarrow s u p e r$ saturated $\Rightarrow$ precipitation occurs.

18. Group Precipitation:

Group	Radicals	Condition for precipitation (Group reagent)	Forms of precipitation
Zero	${N a}^{+}, K^{+}, {N H}_{4}^{*}$	$1 - 2$ drops of ${C H}_{3} C O O H$	-
First	${P b}^{+ 2}, {H g}^{+ 1}$ $({H g}_{2}^{+ 2}), {A g}^{+}$	By mixing dilute HCl	Chloride $A g C l, {H g}_{2} {C l}_{2}, {P b C l}_{2}$
Second	${P b}^{+ 2}, {C u}^{+ 2}, {H g}^{+ 2}, {C d}^{+ 2},$ ${A s}^{+ 3},$ ${S b}^{+ 3}, {S n}^{+ 2}, {S n}^{+ 4}$ .	$H_{2} S$ gas passed in the presence of acidic medium	Sulfide $P b S, H g S, C u S, C d S,$ $S n S, {S n S}_{2}, {A s}_{2} S_{3},$ ${S b}_{2} S_{3}, {B i}_{2} S_{3}$ .
Third	${A l}^{+ 3}, {C r}^{+ 3}, {F e}^{+ 3}$	${N H}_{4} O H$ mixed in the presence of ${N H}_{4} C l$	Hydroxide $A l {(O H)}_{3}, F e {(O H)}_{3},$ $C r {(O H)}_{3}$
Fourth	${Z n}^{+ 2}, {N i}^{+ 2}, {M n}^{+ 2}, {C o}^{+ 2}$	$H_{2} S$ gas passed in the presence of basic medium	Sulfide $M n S, C o S,$ NiS, $Z n S$
Fifth	${B a}^{+ 2}, {S r}^{+ 2}, {C a}^{+ 2}$	${({N H}_{4})}_{2} {C O}_{3}$ mixed in the presence of ${N H}_{4} C l$	Carbonate ${B a C O}_{3}, {S r C O}_{3}, {C a C O}_{3}$
Sixth	${M g}^{+ 2}$	By mixing ${N a}_{2} {H P O}_{4}$	Hydrogen phosphate $({M g H P O}_{4})$

(i) For precipitation of common salt $(NaCl)$ , $H C l$ gas is passed $&$ for soap $(C_{17} H_{35} C O O N a), N a C l$ is mixed.

(ii) Ionization of weak electrolyte is decreased in the presence of common ion is called common-ion effect.

(iii) Solubility product of the sparingly soluble salt $A_{\times} B_{y}$ with solubility $(s)$ mole/liter in saturated solution

$A_{x} B_{y} ⇌ x A^{y +} + y B^{x -} K_{s p} = x^{x} {y^{y}}^{} {(s)}^{x + y}$

(iv) Salt analysis of inorganic mixture depends on the common-ion effect and values of solubility products.

(v) In the presence of common ion, solubility of weak electrolyte always decreases. Solute $A^{+} B^{-}$ is precipitated if $[A^{+}] [B^{-}] > K_{s p}$ .

19. Indicator:

(i) For acidic indicator $[HIn]$ , For basic indicator $(InOH)$ :

$H l n ⇌ H^{+} + {I n}^{-} I n O H ⇌ {I n}^{+} + {O H}^{-}$

$p H = {p K}_{I n} + l o g \frac{[{I n}^{-}]}{[HIn]} p O H = {p K}_{I n} + l o g \frac{[{I n}^{+}]}{[lnOH]}$

(ii) Color change of the indicator is explained by:

(a) Ostwald's theory

(b) Quinonoid theory

Name of indicator	Color in acidic medium	Color in basic medium	Working pH range of indicators
Methyl orange $(MeOH)$	Orange red	Yellow	$3.1$ to $4.5$
Methyl red	Red	Yellow	$4.2$ to $6.2$
Phenol red	Yellow	Red	$6.2$ to $8.2$
Phenolphthalein $(HPh)$	Colorless	Pink	$8.2$ to $10.2$

20. Acid-base Titration:

Type of titration	pH range of titration	Suitable indicators
$S t r o n g a c i d / s t r o n g b a s e$	$3 - 11$	All indicators ( $M e O H, H P h$ etc.)
$S t r o n g a c i d / w e a k b a s e$	$3 - 7$	Methyl orange $(M e O H)$ and methyl red
$W e a k a c i d / s t r o n g b a s e$	$7 - 11$	Phenolphthalein $(HPh)$
$W e a k a c i d / w e a k b a s e$	$6.5 - 7.5$	Phenol red

Which of the following relation holds true for an equilibrium reaction, of its reaction quotient Q=1?

Important Points to Remember in Chapter -1 - Equilibrium from Embibe Experts Gamma Question Bank for Engineering Chemistry Solutions

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Which of the following relation holds true for an equilibrium reaction, of its reaction quotient $(Q) = 1$ ?