NCERT Solutions for Class 11 Science Chemistry Chapter 7 - Equilibrium

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Answer:

We know that:

$∆{\mathrm{n}}_{\mathrm{g}}=[\mathrm{No}. \mathrm{of} \mathrm{gaseous} \mathrm{moles} \mathrm{of} \mathrm{product}-\mathrm{No}. \mathrm{of} \mathrm{gaseous} \mathrm{moles} \mathrm{of} \mathrm{reactants}]$
$$\begin{gathered} ∆{\mathrm{n}}_{\mathrm{g}}=2-0=2 \\ \mathrm{Therefore}, {\mathrm{K}}_{\mathrm{p}}={\mathrm{K}}_{\mathrm{c}}{\left(\mathrm{RT}\right)}^2 \end{gathered}$$

Hence, the correct answer is option (iv).

Answer:

We know that;

$$\begin{gathered} ∆\mathrm{G}°=-\mathrm{RTlnK} \\ \mathrm{Since}, ∆\mathrm{G}° \mathrm{is} \mathrm{positive}, \mathrm{this} \mathrm{is} \mathrm{possible} \mathrm{only} \mathrm{when} \mathrm{lnK} \mathrm{is} \mathrm{negative} \mathrm{or} \mathrm{K} \mathrm{is} \mathrm{less} \mathrm{that} 1. \end{gathered}$$

Hence, the correct answer is option (iv).

Answer:

All the physical processes don't stop at equilibrium. 

Hence, the correct answer is option (iii).

Answer:

The equilibrium constant may be calculated as follows:

 PCl₅ (g) ⇌ PCl₃ (g) + Cl₂ (g)

$$\begin{gathered} {\mathrm{K}}_{\mathrm{c}}=\frac{\left[{\mathrm{PCl}}_3\right]\left[{\mathrm{Cl}}_2\right]}{\left[{\mathrm{PCl}}_5\right]} \\ {\mathrm{K}}_{\mathrm{c}}=\frac{1.2\times {10}^{-3}\times 1.2\times {10}^{-3}}{0.8\times {10}^{-3}}=1.8\times {10}^{-3} \mathrm{mol} {\mathrm{L}}^{-1} \end{gathered}$$



Hence, the correct answer is option (ii).

Answer:

Fe(III) nitrate and SCN^- react with others as follows:

${\mathrm{Fe}}^{3+}+{\mathrm{SCN}}^{-}\to \underset{\mathrm{Red}}{{\left[\mathrm{Fe}\right(\mathrm{SCN}\left)\right]}^{2+}}$

When oxalic acid is added then it forms a stable complex with iron i.e. [Fe(C₂O₄)₃]³⁺ and therefore, decreases the concentration of free Fe³⁺ ions in the solution and result in the decrease of intensity of red colour.

Hence, the correct answer is option (ii).


 

Answer:

The reaction is endothermic in nature because on decreasing the temperature of the mixture, it proceeds in the backward direction according to Le Chatelier's principle. So, the enthalpy of the reaction is positive.

Hence, the correct answer is option (i).

Answer:

As the temperature increases, ionisation of water increases and pH of pure water becomes slightly lower than 7, even though it is still neutral.

Hence, the correct answer is option (iii).

Answer:

pH ∝ Concentration of H⁺ ions ∝ Ionisation of acid ∝ K_a
Since, the value of ionisation constants of acids follows the order: 1.8 × 10^–4 > 1.74 × 10^–5 > 3.0 × 10^–8 

Hence, the correct answer is option (iv).

Answer:

H₂S ⇌ H⁺ + HS^–              ......(1)
$K_{a_1}=\frac{\left[{\mathrm{H}}^{+}\right]\left[{\mathrm{HS}}^{-}\right]}{\left[{\mathrm{H}}_2\mathrm{S}\right]}$
HS^– â‡Œ H⁺ + S^2–               .......(2)
$K_{a_2}=\frac{\left[{\mathrm{H}}^{+}\right]\left[{\mathrm{S}}^{2-}\right]}{\left[{\mathrm{HS}}^{-}\right]}$
H₂S ⇌ 2H⁺ + S^2–             ......(3)
$K_{a_3}=\frac{{\left[{\mathrm{H}}^{+}\right]}^2\left[{\mathrm{S}}^{2-}\right]}{\left[{\mathrm{H}}_2\mathrm{S}\right]}$
Adding equations 1 and 2 to get equation 3, we get
$K_{a_3}=K_{a_1}\times K_{a_2}$

Hence, the correct answer is option (i).
 

Answer:

According to Lewis's concept, electron-deficient species are considered as acids.

Hence, the correct answer is option (iii).

Answer:

A basic buffer is obtained by mixing equal volumes of 0.1 mol dm^–3 NH₄OH and 0.05 mol dm^–3 HCl as 0.05 mol dm^–3 NH₄Cl (salt of a weak base) is formed and 0.05 mol dm^–3 NH₄OH (weak base) is left in the mixture.

Hence, the correct answer is option (iii).

Answer:

Silver chloride will be most soluble in aqueous ammonia as a highly soluble complex of [Ag(NH₃)₂]Cl is formed.

Hence, the correct answer is option (iv).

Answer:

The following proton transfer reactions are possible.

(i) CH₃COOH + H₂O  CH₃COO⁻ + H₃O⁺ ; K_a = 1.74 × 10⁻⁵

(ii) H₂O + H₂O  H₃O⁺ + OH⁻ ; K_a = 1.0 × 10⁻¹⁴

As K_a >> K_w, reaction (i) is the principle reaction.

The concentration of various species in reaction (i) can be shown as:

                             CH₃COOH + H₂O  CH₃COO⁻ + H₃O⁺

Initial Conc. (M)         0.01                                   0             0

At eq.                 (0.01 − x)                                     x              x

On substituting the equilibrium concentrations in the equilibrium reaction, we obtain

$K_a=\frac{x^2}{\left(0.01-x\right)}=1.74\times {10}^{-5}$

Since x << 0.01, 0.01 − x ≈ 0.01 
$$\begin{gathered} \frac{x^2}{0.01}=1.74\times {10}^{-5} \\ {x}^2=1.74\times {10}^{-7} \\ x=4.17\times {10}^{-4} \end{gathered}$$

Now, pH = −log [H₃O⁺]

pH = −log (4.17 × 10⁻⁴)

pH = −(−3.38) = 3.38 ≈ 3.4

Hence, the correct answer is option (i).

Answer:

$$\begin{gathered} \mathrm{pH}=7+\frac{1}{2}\left[-\log K_a-\left(-\log K_b\right)\right] \\ \mathrm{pH}=7+\frac{1}{2}\left[-{\mathrm{logK}}_{\mathrm{a}}+{\mathrm{logK}}_{\mathrm{b}}\right] \\ \mathrm{pH}=7+\frac{1}{2}\left[-\log 1.8\times {10}^{-5}+\log 1.8\times {10}^{-5}\right] \\ \mathrm{pH}=7+\frac{1}{2}\left[0\right]=7 \end{gathered}$$

Hence, the correct answer is option (iii).

Answer:

∆G^⊖ = – RT lnK
At the stage of half completion of reaction [A] = [B], therefore, K = 1.
Thus, ∆G^⊖ = 0.
Hence, the correct answer is option (i).

Answer:

According to Le-Chatelier’s principle, at constant temperature, the equilibrium composition will change but K remains the same because it depends only on the temperature.

Hence, the correct answer is option (i).

Answer:

Higher is the boiling point, lesser is the vapour pressure. Therefore, compound with higher boiling point has lowest vapour pressure and the compound with lower boiling point has the highest vapour pressure.
The correct order of vapour pressure is as follows:
Water < acetone < ether

Hence, the correct answer is option (ii).

Answer:

$\frac{1}{2}{\mathrm{H}}_2\left(g\right)+\frac{1}{2}{\mathrm{I}}_2\left(g\right) \rightleftharpoons \mathrm{HI}\left(g\right)$, K_c = 5
On reversing the above reaction, we get
$\mathrm{HI}\left(g\right) \rightleftharpoons \frac{1}{2}{\mathrm{H}}_2\left(g\right)+\frac{1}{2}{\mathrm{I}}_2\left(g\right), K_c=\frac{1}{5}=0.2$
On multiplying the above reaction by 2 times, we get
2HI(g) ⇌ H₂(g) + I₂(g), K_c = (0.2)² = 0.04

Hence, the correct answer is option (i).

Answer:

The addition of an inert gas such as argon at constant volume does not change the partial pressures or the molar concentrations of the substance involved in any reaction because it does not take part in the reaction and the equilibrium remains undisturbed.

Hence, the correct answer is option (iv).

Answer:

(i) K increases with increase in temperature which shows that reaction is endothermic in nature.
(iii) Q > K, Therefore, reaction proceeds in the backward direction.
(iv) ∆ n > 0. Therefore, ∆ S > 0.

Hence ,the correct answers are options (i), (iii) & (iv).

Answer:

Melting point and freezing point are the temperatures at which the solid and liquid phases of a pure substance exist in equilibrium.

Hence , the correct answers are options (i) & (iv).

Answer:

HCl          Cl^-
Acid         Conjugate base

H₂O         H₃O⁺
Base         Conjugate acid

Answer:

Sugar does not ionise in water but NaCl ionises completely in water and produces Na⁺ and Cl^- ions.
NaCl $\underset{}{\to }$ Na⁺ + Cl^-
Conductance increases with an increase in the concentration of salt as the number of ions increases. 
 

Answer:

BF₃ acts as a Lewis acid as it is electron deficient compound and coordinate bond is formed as given below :
      H₃N : → BF₃ 

Answer:

The extent of ionisation of a weak base is directly proportional to the value of ionisation constant (K_b). Therefore, the decreasing order of the extent of ionisation of bases at equilibrium is: Dimethylamine > Ammonia > Pyridine > Urea
Since dimethylamine will ionise to the maximum extent, it is the strongest base out of the four given bases. 


 

Answer:

OH^–,RO^– ,CH₃COO^– and Cl^– are conjugate bases of H₂O, ROH, CH₃COOH and HCl respectively.
The acidic strength of these acids follows the order: HCl > CH₃COOH > H₂O > ROH
Since, conjugate base of a weak acid is always stronger, therefore, the decreasing order of basic strength of the following conjugate bases follows the order: RO^– > OH^– > CH₃COO^– > Cl^–

Answer:

KNO₃(aq) is a neutral salt, CH₃COONa(aq) is a basic salt, NH₄Cl(aq) is an acidic salt and C₆H₅COONH₄(aq) is a neutral salt. But C₆H₅COONH₄(aq) has slightly higher pH than 7 because it is made from weak acid (C₆H₅COOH) and weak base (NH₄OH). Therefore, the pH of the following salt solutions follows the order: NH₄Cl(aq) < KNO₃(aq) < C₆H₅COONH₄(aq) < CH₃COONa(aq)

Answer:

For the reaction the reaction quotient Q_c is given by

$$\begin{gathered} Q_c=\frac{\left[{\mathrm{H}}_2\right]\left[{\mathrm{I}}_2\right]}{{\left[\mathrm{HI}\right]}^2} \\ \mathrm{Putting} \mathrm{the} \mathrm{given} \mathrm{values}, \mathrm{we} \mathrm{get} \\ {Q}_c=\frac{1\times {10}^{-5}\times 1\times {10}^{-5}}{{\left(2\times {10}^{-5}\right)}^2} \\ {Q}_c=0.25 \end{gathered}$$
Since  Q_c (0.25) > K_c (1 × 10^–4), therefore, the the reaction will proceed in backward direction.

Answer:

Concentration of 10^–8 mol dm^–3 indicates that the solution of HCl is very dilute. Hence, the contribution of H₃O⁺ concentration (10^–7) from water is significant and should also be included for the calculation of pH.
Therefore, pH = – log [10^–8 + 10^–7] 
pH = – log 10^–7[0.1 + 1] 
pH = – log 1.1$\times$10^–7
pH = 6.96

Answer:

pH = – log [H⁺]
Hence, the concentration of acid is 10^-5 M and when it is diluted 100 times, the concentration becomes 10^-7 M. This would give the pH of the solution to be 7 but it is not possible for an acidic solution to have pH equal to 7. So, the contribution of H₃O⁺ concentration (10^–7) from water is significant and should also be included for the calculation of pH.
Therefore, pH = – log [10^–7 + 10^–7] 
pH = – log 2$\times$10^–7
pH = 6.6990

Answer:

                                                                                  BaSO₄(s) â‡Œ Ba²⁺(aq) + SO₄^2–(aq)
At t = 0                                                                                        1              0                 0
At equilibrium in water                                                                  1−S             S                S
At equilibrium in the presence of sulphuric acid                             1−S             S                S + 0.01
K_sp for BaSO₄ in water = [Ba²⁺] [SO₄^2–] = (S)(S) = S²  
K_sp = (8 × 10^–4)² = 64 × 10^–8                 ... (1)
The expression for K_sp in the presence of sulphuric acid will be as follows :
K_sp = (S)(S + 0.01)                                 ... (2)
Since value of K_sp will not change in the presence of sulphuric acid, therefore from (1) and (2)
(S)(S + 0.01) = 64 × 10^–8
S² + 0.01S = 64 × 10^–8
S² + 0.01S – 64 × 10^–8 = 0
$$\begin{gathered} \mathrm{S}=\frac{-0.01\pm \sqrt{{\left(0.01\right)}^2+\left(4\times 64\times {10}^{-8}\right)}}{2} \\ \mathrm{S}=\frac{-0.01\pm {10}^{-2}\sqrt{1.256}}{2} \\ \mathrm{S}=\frac{-{10}^{-2}+\left(1.12\times {10}^{-2}\right)}{2} \\ \mathrm{S}=\frac{0.12}{2}\times {10}^{-2}=6\times {10}^{-2} \mathrm{mol} {\mathrm{dm}}^{-3} \end{gathered}$$
 

Answer:

pH of HOCl = 2.85
But, – pH = log [H⁺] 
– 2.85 = log [H⁺]
[H⁺] = 1.413 × 10^–3 
For weak mono basic acid [H⁺] = $\sqrt{K_a\times C}$
$$\begin{gathered} K_a=\frac{{\left[{\mathrm{H}}^{+}\right]}^2}{C}=\frac{{\left(1.413\times {10}^{-3}\right)}^2}{0.08} \\ {K}_a=2.4957\times {10}^{-5} \end{gathered}$$

Answer:

pH of Solution A = 6
Therefore, concentration of [H⁺] ion in solution A = 10^–6 mol L^–1
pH of Solution B = 4
Therefore, concentration of [H⁺] ion concentration of solution B = 10^–4 mol L^–1
On mixing one litre of each solution, total volume = 1L + 1L = 2L
Amount of H⁺ ions in 1L of Solution A= Concentration × volume V = 10^–6 mol L^–1× 1L
Amount of H⁺ ions in 1L of solution B = 10^–4 mol L^–1× 1L
∴ Total amount of H⁺ ions in the solution formed by mixing solutions A and B is (10^–6 mol + 10^–4 mol)
This amount is present in 2L solution.
$$\begin{gathered} \therefore \mathrm{Total} \left[{\mathrm{H}}^{+}\right]=\frac{{10}^{-4}\left(1+0.01\right)}{2}=5\times {10}^{-5} \mathrm{mol} {\mathrm{L}}^{-1} \\ \mathrm{pH}=-\log \left[{\mathrm{H}}^{+}\right] \\ \mathrm{pH}=-\log \left(5\times {10}^{-5}\right) \\ \mathrm{pH}=4.3010\approx 4.3 \end{gathered}$$

Answer:

Let S be the solubility of Al(OH)₃.
Al(OH)₃ â‡Œ Al³⁺(aq) + 3OH^–(aq)
K_sp = [Al³⁺][OH^–]³ = (S)(3S)³ = 27S⁴
$$\begin{gathered} {\mathrm{S}}^4=\frac{{\mathrm{K}}_{\mathrm{sp}}}{27}=\frac{2.7\times {10}^{-11}}{27}=1\times {10}^{-12} \\ \mathrm{S}=1\times {10}^{-3} \mathrm{mol} {\mathrm{L}}^{-1} \end{gathered}$$
Molar mass of Al(OH)₃ is 78 g.
Therefore, Solubility of Al(OH)₃ in g L^–1 = 1 × 10^–3 × 78 g L^–1 = 7.8 × 10^–2 g L^–1
[OH^–] = 3S = 3×1×10^–3 = 3 × 10^–3
pOH = 3 – log 3
pH = 14 – pOH = 11 + log 3
pH = 11.4771 

Answer:

Let S be the solubility of PbCl₂.
PbCl₂(s) â‡Œ Pb²⁺(aq) + 2Cl^–(aq)
K_sp = [Pb²⁺][Cl^–]³ = (S)(2S)² = 4S³
$$\begin{gathered} {\mathrm{S}}^3=\frac{{\mathrm{K}}_{\mathrm{sp}}}{4}=\frac{3.2\times {10}^{-8}}{4} \\ {\mathrm{S}}^3=8\times {10}^{-9} \mathrm{mol} {\mathrm{L}}^{-1} \\ \mathrm{S}=\sqrt[3]{8\times {10}^{-9}} \\ \mathrm{S}=2\times {10}^{-3} \mathrm{mol} {\mathrm{L}}^{-1} \end{gathered}$$
Molar mass of PbCl₂ = 278
∴ Solubility of PbCl₂ in g L^–1 = 2 × 10^–3 × 278 g L^–1 = 0.556 g L^–1
To get saturated solution, 0.556 g of PbCl₂ is dissolved in 1 L water.
0.1 g PbCl₂ is dissolved in $\frac{0.1}{0.556}$L = 0.1798 L water.
To make a saturated solution, dissolution of 0.1 g PbCl₂ in 0.1798 L ≈ 0.2 L of water will be required.

Answer:

The Lewis acid is BF₃ because it is an electron deficient compound and accepts the lone pair of electrons from NH₃ while NH₃ is lewis base because of the presence of lone pair of electrons on nitrogen atom.

Lewis theory of acids and bases explains it.

The hybridisation of B in BF₃ is sp² and hybridisation of N in NH₃ is sp³.

Answer:

∆_r H^⊖ = âˆ†_f H^⊖ [CaO(s)] + âˆ†_f H^⊖ [CO₂(g)] – âˆ†_f H^⊖ [CaCO₃(s)]
∆_r H^⊖ = – 635.1 – 393.5 – (– 1206.9)
∆_r H^⊖ = 178.3 kJ mol^–1
The reaction is endothermic. Hence, according to Le-Chatelier’s principle, reaction will proceed in forward direction on increasing temperature.

Answer:

Answer:

Answer:

Answer:

Answer:

Match the following graphical variation with their description
 

	A		B
(i)		(c)	Variation in reactant concentration with time
(ii)		(a)	Variation in product concentration with time
(iii)		(b)	Reaction at equilibrium

Answer:

(i) → (b) and (c)
(ii) → (d)
(iii) → (a)

Answer:

While comparing acids formed by the elements belonging to the same group of periodic table, H–A bond strength is a more important factor in determining acidity of acid than the polar nature of the bond. The weaker the H–A bond strength, the more acidic is the hydrogen halide. On moving down the halogen group, size of halogen atom increases, H–A bond strength decreases and as a result, acidic character of hydrogen halides increases. Therefore, the increasing order of acidity of hydrogen halides is HF < HCl < HBr < HI.

Hence, the correct answer is option (i).

Answer:

A solution containing a mixture of acetic acid and sodium acetate maintains a constant value of pH on addition of small amounts of acid or alkali because it forms an acidic buffer solution (solution of weak acid and its salt with strong base).

Hence, the correct answer is option (i).

Answer:

The ionisation of hydrogen sulphide, a weak acid in water is low in the presence of hydrochloric acid because of common ion effect. It can be defined as a shift in equilibrium on adding a substance that provides more of an ionic species already present in the dissociation equilibrium. Thus, the addition of H⁺ ions from HCl moves the equilibrium in the direction of undissociated hydrogen sulphide i.e., in a direction of reducing the ionisation and concentration of hydrogen ions, [H⁺].

Hence, the correct answer is option (ii).

Answer:

Equilibrium constant is temperature dependent having one unique value for a particular reaction represented by a balanced equation at a given temperature.

Hence, the correct answer is option (iii).
 

Answer:

The acidic/basic nature of a salt solution of a salt of weak acid and weak base depends on K_a and K_b value of the acid and the base forming it.
pH = 7 + $\frac{1}{2}$[pK_a − pK_b]
The salt ammonium carbonate, (NH₄)₂CO₃ is formed from weak base, NH₄OH (K_b = 1.77 × 10^–5) and weak acid, H₂CO₃ (K_a = 4.5 × 10^–7). Therefore, pH of its aqueous solution can be find out as follows:
pH = 7 + $\frac{1}{2}$[6.35 − 4.75]
pH = 7.8
Since, the pH is greater than 7, the aqueous solution of ammonium carbonate is basic.

Hence, the correct answer is option (i).

Answer:

The solutions which resist change in pH on dilution or with the addition of small amounts of acid or alkali are called Buffer Solutions. An acidic buffer is prepared by mixing a weak acid and its salt formed with strong base while a basic buffer is prepared by mixing â€‹a weak base and its salt formed with strong acid. Ammonium acetate is a salt formed from acetic acid and NH₄OH. Acetic acid is a weak acid and NH₄OH is a weak base. Therefore, an aqueous solution of ammonium acetate cannot act as a buffer.

Hence, the correct answer is option (iii).

Answer:

The addition of an inert gas such as helium at constant volume does not change the partial pressures or the molar concentrations of the substance involved in any reaction and the equilibrium remains undisturbed.

Hence, the correct answer is option (iv).

Answer:

The reaction quotient, Q_c is useful in predicting the direction of reaction by comparing the values of K_c and Q_c. 
(i) If K_c > Q_c, net reaction proceeds in the forward direction.
(ii) If K_c < Q_c, net reaction proceeds in the backward direction.
(iii) If K_c = Q_c, no net reaction occurs.

Answer:

Production of ammonia according to the reaction is an exothermic process. According to Le Chatelier’s principle, raising the temperature shifts the equilibrium to left and decreases the equilibrium concentration of ammonia. In other words, low temperature is favourable for high yield of ammonia, but practically very low temperatures slow down the reaction and thus a catalyst is used.
Since the number of moles formed in the reaction is less than those of reactants, the yield of NH₃ can be improved by increasing the pressure.
The addition of an inert gas such as argon at constant volume does not change the partial pressures or the molar concentrations of the substance involved in the reaction. 

Answer:

${\mathrm{A}}_x^{p+}{\mathrm{B}}_y^{q-}$(s) ⇌ xA^p+(aq) + yB^q–(aq)
(where x × p⁺ = y × q^–)
S is the molar solubility
And its solubility product constant is given by:
K_sp = [A^p+]^x[B^q–]^y = (xS)^x(yS)^y
K_sp = x^x.y^y.S^{(x + y)}
S^{(x + y)} = $\frac{K_{sp}}{x^x.y^y}$
S = ${\left(\frac{K_{sp}}{x^x.y^y}\right)}^{\frac{1}{x+y}}$

Answer:

The relation between ∆G and Q is as follows:

∆G = Free energy change
Q = Reaction quotient
R = Gas constant
T = Absolute temperature

(a) Since âˆ†G^Θ = = – RT lnK
∴ ∆G = – RT lnK + RT lnQ
∆G = RT ln$\frac{Q}{K}$
When Q < K, ∆G will be negative, then the reaction is spontaneous and proceeds in the forward direction.
When Q = K, ∆G = 0, reaction has achieved equilibrium; at this point, there is no longer any free energy left to drive the reaction. Therefore, no net reaction occurs.

(b) For the given reaction: CO(g) + 3H₂(g) ⇌ CH₄(g) + H₂O(g)
$K=\frac{\left[{\mathrm{CH}}_4\right]\left[{\mathrm{H}}_2\mathrm{O}\right]}{\left[\mathrm{CO}\right]{\left[{\mathrm{H}}_2\right]}^3}$
$Q=\frac{\left[{\mathrm{CH}}_4\text{'}\right]\left[{\mathrm{H}}_2\mathrm{O}\text{'}\right]}{\left[\mathrm{CO}\text{'}\right]{\left[{\mathrm{H}}_2\text{'}\right]}^3}$
When the pressure is increased, Q becomes lesser than K and thus, reaction moves in forward direction.