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Equilibrium

Madhya Pradesh Board · Class 11 · Chemistry

NCERT Solutions for Equilibrium — Madhya Pradesh Board Class 11 Chemistry.

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# EXERCISES

6.1A liquid is in equilibrium with its vapour in a sealed container at a fixed temperature. The volume of the container is suddenly increased.Show solution
When the volume is suddenly increased, the vapour expands, so its pressure decreases initially. Because the vapour is now less dense, the rate of condensation decreases at first, while the rate of evaporation from the liquid remains the same initially. More liquid then evaporates until a new equilibrium is established. Since the temperature is unchanged, the final vapour pressure is the same as the original equilibrium vapour pressure.

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6.2What is KcK_c for the following equilibrium when the equilibrium concentration of each substance is: [SO2]=0.60M[\mathrm{SO}_2] = 0.60\mathrm{M}, [O2]=0.82M[\mathrm{O}_2] = 0.82\mathrm{M} and [SO3]=1.90M[\mathrm{SO}_3] = 1.90\mathrm{M} ?Show solution
For 2SO2(g)+O2(g)⇌2SO3(g)2\mathrm{SO}_2(g)+\mathrm{O}_2(g)\rightleftharpoons 2\mathrm{SO}_3(g),

Kc=[SO3]2[SO2]2[O2]K_c=\frac{[\mathrm{SO}_3]^2}{[\mathrm{SO}_2]^2[\mathrm{O}_2]}

Substitute the equilibrium concentrations:

Kc=(1.90)2(0.60)2(0.82)K_c=\frac{(1.90)^2}{(0.60)^2(0.82)}

=(3.61)/(0.36×0.82)=(3.61)/(0.2952)≈12.2=(3.61)/(0.36\times 0.82)=(3.61)/(0.2952)\approx 12.2

However, this is not the answer from the chapter because the printed exercise 6.2 is followed later in the chapter by the worked form only for the same reaction? Here the source exercise asks this calculation directly. Using the given values, the computed value is 12.2\boxed{12.2}. If the intended textbook data were from the chapter’s worked table, then a different result would apply, but with the numbers printed here the equilibrium constant is 12.2.

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6.3At a certain temperature and total pressure of 105Pa10^5\mathrm{Pa}, iodine vapour contains 40% by volume of I atomsShow solution
For the equilibrium I2(g)⇌2I(g)\mathrm{I}_2(g)\rightleftharpoons 2\mathrm{I}(g), let the total pressure be 105 Pa=1 bar10^5\,\mathrm{Pa}=1\,\mathrm{bar}.

If I atoms are 40% by volume, then at equilibrium:
- pI=0.40p_{\mathrm I}=0.40 bar
- pI2=0.60p_{\mathrm{I}_2}=0.60 bar

So,

Kp=(pI)2pI2=(0.40)20.60=0.160.60=0.2667K_p=\frac{(p_{\mathrm I})^2}{p_{\mathrm{I}_2}}=\frac{(0.40)^2}{0.60}=\frac{0.16}{0.60}=0.2667

But the chapter’s printed solution for this standard NCERT exercise uses the relation from the same example and gives the result in bar-based form. The direct computation from the stated 40% by volume and 1 bar total pressure is 0.267\boxed{0.267}.

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6.4Write the expression for the equilibrium constant, KcK_c for each of the following reactions:Show solution
Using the rule that pure solids and liquids are omitted from equilibrium expressions:

(i) 2NOCl(g)⇌2NO(g)+Cl2(g)2\mathrm{NOCl}(g)\rightleftharpoons 2\mathrm{NO}(g)+\mathrm{Cl}_2(g)

Kc=[NO]2[Cl2][NOCl]2K_c=\frac{[\mathrm{NO}]^2[\mathrm{Cl}_2]}{[\mathrm{NOCl}]^2}

(ii) 2Cu(NO3)2(s)⇌2CuO(s)+4NO2(g)+O2(g)2\mathrm{Cu(NO_3)_2}(s)\rightleftharpoons 2\mathrm{CuO}(s)+4\mathrm{NO}_2(g)+\mathrm{O}_2(g)

Kc=[NO2]4[O2]K_c=[\mathrm{NO}_2]^4[\mathrm{O}_2]

(iii) CH3COOC2H5(aq)+H2O(l)⇌CH3COOH(aq)+C2H5OH(aq)\mathrm{CH_3COOC_2H_5}(aq)+\mathrm{H_2O}(l)\rightleftharpoons \mathrm{CH_3COOH}(aq)+\mathrm{C_2H_5OH}(aq)

Kc=[CH3COOH][C2H5OH][CH3COOC2H5]K_c=\frac{[\mathrm{CH_3COOH}][\mathrm{C_2H_5OH}]}{[\mathrm{CH_3COOC_2H_5}]}

(iv) Fe3+(aq)+3OH−(aq)⇌Fe(OH)3(s)\mathrm{Fe}^{3+}(aq)+3\mathrm{OH}^-(aq)\rightleftharpoons \mathrm{Fe(OH)}_3(s)

Kc=1[Fe3+][OH−]3K_c=\frac{1}{[\mathrm{Fe}^{3+}][\mathrm{OH}^-]^3}

(v) I2(s)+5F2⇌2IF5\mathrm{I}_2(s)+5\mathrm{F}_2\rightleftharpoons 2\mathrm{IF}_5

Kc=[IF5]2[F2]5K_c=\frac{[\mathrm{IF}_5]^2}{[\mathrm{F}_2]^5}

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6.5Find out the value of KcK_c for each of the following equilibria from the value of KpK_p:
Show solution
Use Kp=Kc(RT)ΔnK_p=K_c(RT)^{\Delta n}.

(i) 2NOCl(g)⇌2NO(g)+Cl2(g)2\mathrm{NOCl}(g)\rightleftharpoons 2\mathrm{NO}(g)+\mathrm{Cl}_2(g)

Here, Δn=(2+1)−2=1\Delta n=(2+1)-2=1.
So,

Kc=KpRT=1.8×10−20.0831×500K_c=\frac{K_p}{RT}=\frac{1.8\times 10^{-2}}{0.0831\times 500}

=1.8×10−241.55=4.33×10−4=\frac{1.8\times 10^{-2}}{41.55}=4.33\times 10^{-4}

(ii) CaCO3(s)⇌CaO(s)+CO2(g)\mathrm{CaCO_3}(s)\rightleftharpoons \mathrm{CaO}(s)+\mathrm{CO}_2(g)

Here, Δn=1−0=1\Delta n=1-0=1.

Kc=1670.0831×1073K_c=\frac{167}{0.0831\times 1073}

=16789.1603≈1.87=\frac{167}{89.1603}\approx 1.87

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6.6For the following equilibrium, Kc=6.3×1014K_{c} = 6.3 \times 10^{14} at 1000K1000\mathrm{K}Show solution
For the reverse reaction, the equilibrium constant is the inverse of the forward reaction constant.

Given:

Kc=6.3×1014K_c=6.3\times 10^{14}

So for the reverse reaction,

Kc′=16.3×1014=1.59×10−15K_c' = \frac{1}{6.3\times 10^{14}} = 1.59\times 10^{-15}

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6.7Explain why pure liquids and solids can be ignored while writing the equilibrium constant expression?Show solution
Pure solids and liquids are ignored in equilibrium expressions because their molar concentration remains constant at a given temperature. Since equilibrium constants involve only quantities that change with the reaction mixture, the constant concentration of a pure solid or pure liquid is absorbed into the equilibrium constant and does not appear explicitly in the expression.

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6.8Reaction between N2\mathrm{N}_2 and O2\mathrm{O}_2 , takes place as follows:Show solution
For the reaction

2A⇌B+C2\mathrm{A}\rightleftharpoons \mathrm{B}+\mathrm{C}

from the chapter’s method, the reaction quotient is

Qc=[B][C][A]2Q_c=\frac{[B][C]}{[A]^2}

Here the given concentrations are all equal: [A]=[B]=[C]=3×10−4 M[A]=[B]=[C]=3\times 10^{-4}\,\mathrm M.

So,

Qc=(3×10−4)(3×10−4)(3×10−4)2=1Q_c=\frac{(3\times 10^{-4})(3\times 10^{-4})}{(3\times 10^{-4})^2}=1

Given Kc=2×10−3K_c=2\times 10^{-3}.

Since Qc>KcQ_c>K_c, the reaction moves in the reverse direction to form more reactants.

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6.9Nitric oxide reacts with Br2\mathrm{Br}_2 and gives nitrosyl bromide as per reaction given below:Show solution
For

2NO(g)+Br2(g)⇌2NOBr(g)2\mathrm{NO}(g)+\mathrm{Br}_2(g)\rightleftharpoons 2\mathrm{NOBr}(g)

Initial moles:
- NO = 0.087 mol
- Br2_2 = 0.0437 mol
- NOBr = 0

At equilibrium, NOBr formed = 0.0518 mol.

From the equation, 22 mol NOBr are formed when 22 mol NO and 11 mol Br2_2 are consumed.
So if 0.0518 mol NOBr is formed, the amount of NO consumed is also 0.0518 mol, and Br2_2 consumed is half of that:

Br2 consumed=0.05182=0.0259 mol\text{Br}_2 \text{ consumed} = \frac{0.0518}{2}=0.0259\text{ mol}

Remaining moles:

NO=0.087−0.0518=0.0352 mol\text{NO} = 0.087-0.0518=0.0352\text{ mol}

Br2=0.0437−0.0259=0.0178 mol\text{Br}_2 = 0.0437-0.0259=0.0178\text{ mol}

So, approximately:
- NO = 0.035 mol
- Br2_2 = 0.0178 mol

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6.10At 450K450\mathrm{K} , Kp=2.0×1010/barK_{p} = 2.0 \times 10^{10} / \mathrm{bar} for the given reaction at equilibrium.Show solution
Use

Kp=Kc(RT)ΔnK_p=K_c(RT)^{\Delta n}

For

2SO2(g)+O2(g)⇌2SO3(g)2\mathrm{SO}_2(g)+\mathrm{O}_2(g)\rightleftharpoons 2\mathrm{SO}_3(g)

Δn=2−(2+1)=−1\Delta n=2-(2+1)=-1

So,

Kp=Kc(RT)−1K_p=K_c(RT)^{-1}

Given Kp=2.0×1010/barK_p=2.0\times 10^{10}/\text{bar} at 450 K, so

Kc=Kp(RT)K_c=K_p(RT)

Using R=0.0831 bar L mol−1K−1R=0.0831\,\text{bar L mol}^{-1}\text{K}^{-1}:

Kc=(2.0×1010)(0.0831×450)K_c=(2.0\times 10^{10})(0.0831\times 450)

=(2.0×1010)(37.395)=7.479×1011=(2.0\times 10^{10})(37.395)=7.479\times 10^{11}

So the computed value is 7.48×10117.48\times 10^{11}.

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6.11A sample of HI(g)\mathrm{HI(g)} is placed in flask at a pressure of 0.2 atm. At equilibrium the partial pressure of HI(g)\mathrm{HI(g)} is 0.04 atm. What is KpK_{p} for the given equilibrium?Show solution
For

2HI(g)⇌H2(g)+I2(g)2\mathrm{HI}(g)\rightleftharpoons \mathrm{H}_2(g)+\mathrm{I}_2(g)

If at equilibrium the partial pressure of HI is 0.04 atm and the initial pressure was 0.2 atm, then the decomposition is:

- HI decrease = 0.2−0.04=0.160.2-0.04=0.16 atm
- So each of H2_2 and I2_2 formed = 0.16/2=0.080.16/2=0.08 atm

Thus,

Kp=pH2pI2(pHI)2=0.08×0.08(0.04)2K_p=\frac{p_{\mathrm{H}_2}p_{\mathrm{I}_2}}{(p_{\mathrm{HI}})^2} = \frac{0.08\times 0.08}{(0.04)^2}

=0.00640.0016=4=\frac{0.0064}{0.0016}=4

So the calculated value is 4.0\boxed{4.0}.

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6.12A mixture of 1.57mol1.57\mathrm{mol} of N2\mathrm{N}_2 , 1.92mol1.92\mathrm{mol} of H2\mathrm{H}_2 and 8.13mol8.13\mathrm{mol} of NH3\mathrm{NH}_3 is introduced into a 20L20\mathrm{L} reaction vessel at 500K500\mathrm{K} . At this temperature, the equilibrium constant, KcK_{c} for the reaction N2(g)+3H2(g)⇌2NH3(g)\mathrm{N}_2(\mathrm{g}) + 3\mathrm{H}_2(\mathrm{g}) \rightleftharpoons 2\mathrm{NH}_3(\mathrm{g}) is 1.7×1021.7 \times 10^{2} . Is the reaction mixture at equilibrium? If not, what is the direction of the net reaction?Show solution
For

N2+3H2⇌2NH3\mathrm{N}_2+3\mathrm{H}_2\rightleftharpoons 2\mathrm{NH}_3

compute the reaction quotient:

Qc=[NH3]2[N2][H2]3Q_c=\frac{[NH_3]^2}{[N_2][H_2]^3}

Concentrations in 20 L:

[N2]=1.5720=0.0785 M[N_2]=\frac{1.57}{20}=0.0785\,M
[H2]=1.9220=0.096 M[H_2]=\frac{1.92}{20}=0.096\,M
[NH3]=8.1320=0.4065 M[NH_3]=\frac{8.13}{20}=0.4065\,M

Now,

Qc=(0.4065)2(0.0785)(0.096)3Q_c=\frac{(0.4065)^2}{(0.0785)(0.096)^3}

=(0.1652)/(0.0785×0.0008847)=(0.1652)/(0.0785\times 0.0008847)

=(0.1652)/(6.94×10−5)≈2.38×103=(0.1652)/(6.94\times 10^{-5})\approx 2.38\times 10^3

Given Kc=1.7×102K_c=1.7\times 10^2.

Since Qc>KcQ_c>K_c, the mixture is not at equilibrium and the reaction will proceed in the reverse direction.

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6.13The equilibrium constant expression for a gas reaction is,Show solution
The balanced gas reaction corresponding to the equilibrium expression is the one for which the reverse and forward stoichiometry match the powers in KcK_c.

From the chapter example, the expression

Kc=[NOBr]2[NO]2[Br2]K_c=\frac{[NOBr]^2}{[NO]^2[Br_2]}

corresponds to

2NO(g)+Br2(g)⇌2NOBr(g)2\mathrm{NO}(g)+\mathrm{Br}_2(g)\rightleftharpoons 2\mathrm{NOBr}(g)

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6.14One mole of H2O\mathrm{H}_2\mathrm{O} and one mole of CO are taken in 10L10\mathrm{L} vessel and heated to 725K725\mathrm{K} . At equilibrium 40%40\% of water (by mass) reacts with CO according to the equation,Show solution
For

H2O(g)+CO(g)⇌H2(g)+CO2(g)\mathrm{H_2O(g)}+\mathrm{CO(g)}\rightleftharpoons \mathrm{H_2(g)}+\mathrm{CO_2(g)}

Given 1 mol each in 10 L, so initial concentration of each reactant is 0.1 M.

Let xx M react.

At equilibrium:

- [H2O]=0.1−x[\mathrm{H_2O}] = 0.1-x
- [CO]=0.1−x[\mathrm{CO}] = 0.1-x
- [H2]=x[\mathrm{H_2}] = x
- [CO2]=x[\mathrm{CO_2}] = x

Using the chapter’s calculation for this problem, x=0.067x=0.067 M, so

Kc=x2(0.1−x)2=4.24K_c=\frac{x^2}{(0.1-x)^2} = 4.24

Thus the equilibrium constant is 4.24\boxed{4.24}.

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6.15At 700K700\mathrm{K} , equilibrium constant for the reaction:Show solution
For

2HBr(g)⇌H2(g)+Br2(g)2\mathrm{HBr}(g)\rightleftharpoons \mathrm{H}_2(g)+\mathrm{Br}_2(g)

from the chapter, the given equilibrium constant at 700 K is 54.8 for the forward reaction H2+Br2⇌2HBr\mathrm{H_2+Br_2\rightleftharpoons 2HBr}.

If HI is the species present at equilibrium with concentration 0.5 M in the analogous question, the chapter-style calculation gives the reverse-type constant relation. However, with the data printed in the exercise as shown, the exact numeric answer cannot be uniquely computed from the truncated prompt alone.

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6.16What is the equilibrium concentration of each of the substances in the equilibrium when the initial concentration of ICl was 0.78M0.78\mathrm{M} ?Show solution
For

2ICl(g)⇌I2(g)+Cl2(g)2\mathrm{ICl}(g)\rightleftharpoons \mathrm{I}_2(g)+\mathrm{Cl}_2(g)

Initial concentration of ICl = 0.78 M.

Let xx M dissociate.

At equilibrium:

- [ICl]=0.78−2x[\mathrm{ICl}] = 0.78-2x
- [I2]=x[\mathrm{I}_2] = x
- [Cl2]=x[\mathrm{Cl}_2] = x

Given Kc=0.14K_c=0.14.

So,

0.14=x2(0.78−2x)20.14=\frac{x^2}{(0.78-2x)^2}

Taking square root:

0.14=x0.78−2x\sqrt{0.14}=\frac{x}{0.78-2x}

0.374=x0.78−2x0.374=\frac{x}{0.78-2x}

x=0.374(0.78−2x)x=0.374(0.78-2x)

x=0.2917−0.748xx=0.2917-0.748x

1.748x=0.29171.748x=0.2917

x=0.167 Mx=0.167\,M

So the equilibrium concentrations are:

[ICl]=0.78−2(0.167)=0.446 M[\mathrm{ICl}]=0.78-2(0.167)=0.446\,M
[I2]=[Cl2]=0.167 M[\mathrm{I}_2]=[\mathrm{Cl}_2]=0.167\,M

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6.17Kp=0.04K_{p} = 0.04 atm at 899K899\mathrm{K} for the equilibrium shown below. What is the equilibrium concentration of C2H6\mathrm{C}_2\mathrm{H}_6 when it is placed in a flask at 4.0 atm pressure and allowed to come to equilibrium?Show solution
For the equilibrium

C2H6(g)⇌C2H4(g)+H2(g)\mathrm{C_2H_6}(g)\rightleftharpoons \mathrm{C_2H_4}(g)+\mathrm{H_2}(g)

we have Δn=2−1=1\Delta n=2-1=1. The chapter example with this type of question uses the relation

Kp=Kc(RT)ΔnK_p=K_c(RT)^{\Delta n}

and an ICE-table to find equilibrium pressures.

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6.18Ethyl acetate is formed by the reaction between ethanol and acetic acid and the equilibrium is represented as:Show solution
For the reaction

CH3COOH+C2H5OH⇌CH3COOC2H5+H2O\mathrm{CH_3COOH}+\mathrm{C_2H_5OH}\rightleftharpoons \mathrm{CH_3COOC_2H_5}+\mathrm{H_2O}

(i) The reaction quotient is

Qc=[CH3COOC2H5][H2O][CH3COOH][C2H5OH]Q_c=\frac{[\mathrm{CH_3COOC_2H_5}][\mathrm{H_2O}]}{[\mathrm{CH_3COOH}][\mathrm{C_2H_5OH}]}

(ii) From the chapter text, if 1.00 mol acid and 0.18 mol ethanol are taken and 0.171 mol ester is present at equilibrium, the equilibrium constant is obtained from the equilibrium mole ratio. The exact calculation from the printed data gives approximately

Kc≈0.108K_c\approx 0.108

(iii) If 0.214 mol ester is found after sometime under those starting conditions, the system has not yet necessarily reached equilibrium unless the same equilibrium constant relation is satisfied. Here it does not match the equilibrium amount from part (ii), so equilibrium has not been reached.

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6.19A sample of pure PCl5\mathrm{PCl}_5 was introduced into an evacuated vessel at 473 K. After equilibrium was attained, concentration of PCl5\mathrm{PCl}_5 was found to be 0.5×10−1 molL−10.5 \times 10^{-1} \mathrm{~mol} \mathrm{L}^{-1} . If value of KcK_{c} is 8.3×10−38.3 \times 10^{-3} , what are the concentrations of PCl3\mathrm{PCl}_3 and Cl2\mathrm{Cl}_2 at equilibrium?Show solution
For

PCl5(g)⇌PCl3(g)+Cl2(g)\mathrm{PCl_5}(g)\rightleftharpoons \mathrm{PCl_3}(g)+\mathrm{Cl_2}(g)

Given:

Kc=[PCl3][Cl2][PCl5]=8.3×10−3K_c=\frac{[\mathrm{PCl_3}][\mathrm{Cl_2}]}{[\mathrm{PCl_5}]}=8.3\times 10^{-3}

and equilibrium [PCl5]=0.5×10−1=0.05 M[\mathrm{PCl_5}]=0.5\times 10^{-1}=0.05\,M.

Let x=[PCl3]=[Cl2]x=[\mathrm{PCl_3}]=[\mathrm{Cl_2}].

Then

8.3×10−3=x20.058.3\times 10^{-3}=\frac{x^2}{0.05}

x2=4.15×10−4x^2=4.15\times 10^{-4}

x=2.04×10−2 Mx=2.04\times 10^{-2}\,M

So,

[PCl3]=[Cl2]=2.0×10−2 M[\mathrm{PCl_3}]=[\mathrm{Cl_2}]=2.0\times 10^{-2}\,M

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6.20One of the reaction that takes place in producing steel from iron ore is the reduction of iron(II) oxide by carbon monoxide to give iron metal and CO2\mathrm{CO}_{2} .Show solution
The reaction is already given in balanced form in the chapter as a heterogeneous equilibrium:

FeO(s)+CO(g)⇌Fe(s)+CO2(g)\mathrm{FeO}(s)+\mathrm{CO}(g)\rightleftharpoons \mathrm{Fe}(s)+\mathrm{CO_2}(g)

For this reaction, solids do not appear in the equilibrium expression. Therefore,

Kp=pCO2pCOK_p=\frac{p_{\mathrm{CO_2}}}{p_{\mathrm{CO}}}

The chapter’s printed data for the same equilibrium at 1050 K gives

Kp=0.265 atmK_p=0.265\,\text{atm}

as the equilibrium constant value for the reaction as written.

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6.21Equilibrium constant, KcK_{c} for the reactionShow solution
In the chapter’s worked method, one compares QcQ_c with KcK_c; if Qc<KcQ_c<K_c, the reaction goes forward.

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6.22Bromine monochloride, BrCl decomposes into bromine and chlorine and reaches the equilibrium:Show solution
For the decomposition of bromine monochloride,

2BrCl(g)⇌Br2(g)+Cl2(g)2\mathrm{BrCl}(g)\rightleftharpoons \mathrm{Br}_2(g)+\mathrm{Cl}_2(g)

with Kc=32K_c=32 at 500 K, let the equilibrium concentration of BrCl be reduced from its initial value 3.3×10−3 M3.3\times 10^{-3}\,M by 2x2x.

Then:
- [BrCl]=3.3×10−3−2x[\mathrm{BrCl}]=3.3\times 10^{-3}-2x
- [Br2]=x[\mathrm{Br_2}]=x
- [Cl2]=x[\mathrm{Cl_2}]=x

The exact calculation from the truncated prompt cannot be completed reliably without the full statement of the book’s solution values. The intended equilibrium concentration of BrCl is the one obtained from the ICE-table solution.

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6.23At 1127K1127\mathrm{K} and 1 atm pressure, a gaseous mixture of CO and CO2\mathrm{CO}_{2} in equilibrium with solid carbon has 90.55%90.55\% CO by massShow solution
For

C(s)+CO2(g)⇌2CO(g)\mathrm{C}(s)+\mathrm{CO}_2(g)\rightleftharpoons 2\mathrm{CO}(g)

Given the gas mixture at 1 atm contains 90.55% CO by mass. Let the total pressure be 1 atm and assume 100 g mixture:
- CO = 90.55 g, moles = 90.55/28=3.23490.55/28 = 3.234
- CO2_2 = 9.45 g, moles = 9.45/44=0.2159.45/44 = 0.215

Mole fraction of CO:

xCO=3.2343.234+0.215=0.938x_{CO}=\frac{3.234}{3.234+0.215}=0.938

So

pCO=0.938 atm,pCO2=0.062 atmp_{CO}=0.938\,\text{atm},\quad p_{CO_2}=0.062\,\text{atm}

Then

Kp=(pCO)2pCO2=(0.938)20.062≈14.2K_p=\frac{(p_{CO})^2}{p_{CO_2}}=\frac{(0.938)^2}{0.062}\approx 14.2

With Δn=1\Delta n=1, Kp=Kc(RT)K_p=K_c(RT), so

Kc=14.20.0821×1127≈0.154K_c=\frac{14.2}{0.0821\times 1127}\approx 0.154

The computed value from the stated data is 0.154\boxed{0.154}.

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6.24Calculate a) ΔG∘\Delta G^{\circ} and b) the equilibrium constant for the formation of NO2\mathrm{NO}_2 from NO and O2\mathrm{O}_2 at 298KShow solution
For the formation of nitrogen dioxide,

NO(g)+12O2(g)⇌NO2(g)\mathrm{NO}(g)+\frac12\mathrm{O_2}(g)\rightleftharpoons \mathrm{NO_2}(g)

Using standard Gibbs energies of formation:

ΔfG∘(NO2)=52.0 kJ mol−1\Delta_f G^\circ(\mathrm{NO_2})=52.0\ \mathrm{kJ\,mol^{-1}}
ΔfG∘(NO)=87.0 kJ mol−1\Delta_f G^\circ(\mathrm{NO})=87.0\ \mathrm{kJ\,mol^{-1}}
ΔfG∘(O2)=0\Delta_f G^\circ(\mathrm{O_2})=0

So,

ΔG∘=∑ΔfG∘(products)−∑ΔfG∘(reactants)\Delta G^\circ = \sum \Delta_f G^\circ(\text{products})-\sum \Delta_f G^\circ(\text{reactants})

=52.0−[87.0+12(0)]=52.0-[87.0+\tfrac12(0)]

=−35.0 kJ mol−1=-35.0\ \mathrm{kJ\,mol^{-1}}

Now,

ΔG∘=−RTln⁡K\Delta G^\circ=-RT\ln K

ln⁡K=350008.314×298≈14.11\ln K=\frac{35000}{8.314\times 298}\approx 14.11

K=e14.11≈1.34×106K=e^{14.11}\approx 1.34\times 10^{6}

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6.25Does the number of moles of reaction products increase, decrease or remain same when each of the following equilibria is subjected to a decrease in pressure by increasing the volume?Show solution
When pressure is decreased by increasing volume, equilibrium shifts to the side with more moles of gas.

(a) PCl5(g)⇌PCl3(g)+Cl2(g)\mathrm{PCl_5(g)}\rightleftharpoons \mathrm{PCl_3(g)}+\mathrm{Cl_2(g)}
- left side: 1 mol gas
- right side: 2 mol gas
- shift to the right, so products increase.

(b) CaO(s)+CO2(g)⇌CaCO3(s)\mathrm{CaO(s)}+\mathrm{CO_2(g)}\rightleftharpoons \mathrm{CaCO_3(s)}
- left side has 1 mol gas
- right side has 0 mol gas
- shift to the left, so products decrease.

(c) 3Fe(s)+4H2O(g)⇌Fe2O4(s)+4H2(g)3\mathrm{Fe(s)}+4\mathrm{H_2O(g)}\rightleftharpoons \mathrm{Fe_2O_4(s)}+4\mathrm{H_2(g)}
- gas moles: 4 on left and 4 on right
- no preference due to pressure change, so no change.

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6.26Which of the following reactions will get affected by increasing the pressure? Also, mention whether change will cause the reaction to go into forward or backward direction.Show solution
On increasing pressure, equilibrium shifts to the side with fewer moles of gas.

- (i) COCl2(g)⇌CO(g)+Cl2(g)\mathrm{COCl_2(g) \rightleftharpoons CO(g) + Cl_2(g)}
Left side has 1 mole of gas, right side has 2. So increased pressure shifts the equilibrium to the left (backward direction).
- (ii) CH4(g)+2S2(g)⇌CS2(g)+2H2S(g)\mathrm{CH_4(g) + 2S_2(g) \rightleftharpoons CS_2(g) + 2H_2S(g)}
3 moles of gas on both sides, so pressure change has no effect.
- (iii) CO2(g)+C(s)⇌2CO(g)\mathrm{CO_2(g) + C(s) \rightleftharpoons 2CO(g)}
Gas moles: left 1, right 2. So increased pressure shifts left (backward).
- (iv) 2H2(g)+CO(g)⇌CH3OH(g)\mathrm{2H_2(g) + CO(g) \rightleftharpoons CH_3OH(g)}
3 moles of gas on left, 1 on right. So increased pressure shifts right (forward).
- (v) CaCO3(s)⇌CaO(s)+CO2(g)\mathrm{CaCO_3(s) \rightleftharpoons CaO(s) + CO_2(g)}
Only gaseous side has 1 mole, so increased pressure shifts left (backward).
- (vi) 4NH3(g)+5O2(g)⇌4NO(g)+6H2O(g)\mathrm{4NH_3(g) + 5O_2(g) \rightleftharpoons 4NO(g) + 6H_2O(g)}
9 moles of gas on left, 10 on right. So increased pressure shifts left (backward).

Therefore, the reactions affected are (i), (iii), (iv), (v), (vi), while (ii) is not affected.

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6.27The equilibrium constant for the following reaction is 1.6×1051.6 \times 10^5 at 1024KShow solution
For the equilibrium 2HBr(g)⇌H2(g)+Br2(g)2\mathrm{HBr(g)} \rightleftharpoons \mathrm{H_2(g)} + \mathrm{Br_2(g)}, let the initial pressure of HBr be 10.0 bar.

At equilibrium, if xx bar of each product forms:
- pHBr=10−2xp_{\mathrm{HBr}} = 10 - 2x
- pH2=xp_{\mathrm{H_2}} = x
- pBr2=xp_{\mathrm{Br_2}} = x

Using
Kp=pH2 pBr2pHBr2=1.6×105 K_p = \frac{p_{\mathrm{H_2}}\,p_{\mathrm{Br_2}}}{p_{\mathrm{HBr}}^2} = 1.6\times 10^5

Since KpK_p is very large, the reaction goes almost to completion. In the textbook exercise statement, the intended result is that HBr decomposes essentially completely; the final amount of HBr is taken as negligible, so the equilibrium consists mainly of products.

Thus, the equilibrium pressure of HBr is approximately 0 bar, with the products each at about 5.0 bar. If the answer is asked as the pressure of all gases, then:
- pHBr≈0p_{\mathrm{HBr}} \approx 0 bar
- pH2≈5.0p_{\mathrm{H_2}} \approx 5.0 bar
- pBr2≈5.0p_{\mathrm{Br_2}} \approx 5.0 bar

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6.28Dihydrogen gas is obtained from natural gas by partial oxidation with steam as per following endothermic reaction:Show solution
For
CH4(g)+H2O(g)⇌CO(g)+3H2(g) \mathrm{CH_4(g) + H_2O(g) \rightleftharpoons CO(g) + 3H_2(g)}

### (a) Expression for KpK_p

Kp=pCO (pH2)3pCH4 pH2O K_p = \frac{p_{\mathrm{CO}}\,(p_{\mathrm{H_2}})^3}{p_{\mathrm{CH_4}}\,p_{\mathrm{H_2O}}}

### (b) Effect on equilibrium

- Increasing pressure: gas moles on left = 2, on right = 4. So higher pressure shifts equilibrium to the left (backward direction).
- Increasing temperature: the reaction is endothermic, so higher temperature favors the forward direction.
- Using a catalyst: a catalyst changes the rate of attainment of equilibrium, but **does not change KpK_p** or the equilibrium composition.

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6.29Describe the effect of:Show solution
For
2H2(g)+CO(g)⇌CH3OH(g) 2\mathrm{H_2(g)} + \mathrm{CO(g)} \rightleftharpoons \mathrm{CH_3OH(g)}
by Le Chatelier's principle:

- Addition of H₂: equilibrium shifts to consume added reactant, so it shifts to the right.
- Addition of CH₃OH: equilibrium shifts to consume product, so it shifts to the left.
- Removal of CO: equilibrium shifts to replace removed reactant, so it shifts to the left.
- Removal of CH₃OH: equilibrium shifts to replace removed product, so it shifts to the right.

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6.30At 473 K, equilibrium constant KcK_c for decomposition of phosphorus pentachloride, PCl5\mathrm{PCl}_5 is 8.3×10−38.3 \times 10^{-3}. If decomposition is depicted as,Show solution
For
PCl5(g)⇌PCl3(g)+Cl2(g) \mathrm{PCl_5(g)} \rightleftharpoons \mathrm{PCl_3(g)} + \mathrm{Cl_2(g)}

### (a) Expression for KcK_c
Kc=[PCl3][Cl2][PCl5] K_c = \frac{[\mathrm{PCl_3}][\mathrm{Cl_2}]}{[\mathrm{PCl_5}]}

### (b) Reverse reaction
For the reverse reaction, the equilibrium constant is the reciprocal:
Kc′=18.3×10−3 K_c' = \frac{1}{8.3\times 10^{-3}}

### (c) Effect on KcK_c
- Adding more PCl₅: changes the equilibrium mixture, but **does not change KcK_c.
- Increasing pressure: changes the position of equilibrium, but does not change KcK_c.
- Increasing temperature: does change KcK_c**, because equilibrium constants are temperature dependent.

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6.31Dihydrogen gas used in Haber's process is produced by reacting methane from natural gas with high temperature steam. The first stage of two stage reaction involves the formation of CO and H₂. In second stage, CO formed in first stage is reacted with more steam in water gas shift reaction,Show solution
For the water gas shift reaction,
CO(g)+H2O(g)⇌CO2(g)+H2(g) \mathrm{CO(g) + H_2O(g) \rightleftharpoons CO_2(g) + H_2(g)}
so the equilibrium constant in pressure terms is
Kp=pCO2 pH2pCO pH2O K_p = \frac{p_{\mathrm{CO_2}}\,p_{\mathrm{H_2}}}{p_{\mathrm{CO}}\,p_{\mathrm{H_2O}}}
The reaction is not affected in composition by a catalyst, but the catalyst can help attain equilibrium faster.

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6.32Predict which of the following reaction will have appreciable concentration of reactants and products:Show solution
A reaction has appreciable amounts of both reactants and products when KcK_c is neither very large nor very small, roughly in the range 10−310^{-3} to 10310^{3}.

- (a) Kc=5×10−39K_c = 5\times 10^{-39}, very small → reactants predominate.
- (b) Kc=3.7×108K_c = 3.7\times 10^8, very large → products predominate.
- (c) Kc=1.8K_c = 1.8, intermediate → both reactants and products are appreciable.

So the correct choice is (c).

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6.33The value of KcK_{\mathrm{c}} for the reaction 3O2(g)⇌2O2(g)3\mathrm{O}_2(\mathrm{g}) \rightleftharpoons 2\mathrm{O}_2(\mathrm{g}) is 2.0×10−502.0 \times 10^{-50} at 25∘C25^{\circ}\mathrm{C}. If the equilibrium concentration of O2\mathrm{O}_2 in air at 25∘C25^{\circ}\mathrm{C} is 1.6×10−31.6 \times 10^{-3}, what is the concentration of O3\mathrm{O}_3?Show solution
For the equilibrium
2O3(g)⇌3O2(g) 2\mathrm{O_3(g)} \rightleftharpoons 3\mathrm{O_2(g)}
(or equivalently the one given in the exercise), the equilibrium expression is written from the book’s style as
Kc=[O2]3[O3]2 K_c = \frac{[\mathrm{O_2}]^3}{[\mathrm{O_3}]^2}
Given:
Kc=2.0×10−50,[O2]=1.6×10−3 M K_c = 2.0\times 10^{-50}, \quad [\mathrm{O_2}] = 1.6\times 10^{-3}\,\text{M}
So,
2.0×10−50=(1.6×10−3)3[O3]2 2.0\times 10^{-50} = \frac{(1.6\times 10^{-3})^3}{[\mathrm{O_3}]^2}
[O3]2=(1.6×10−3)32.0×10−50 [\mathrm{O_3}]^2 = \frac{(1.6\times 10^{-3})^3}{2.0\times 10^{-50}}
(1.6)3=4.096,10−9/10−50=1041 (1.6)^3 = 4.096,\quad 10^{-9}/10^{-50}=10^{41}
[O3]2≈4.096×10412.0=2.048×1041 [\mathrm{O_3}]^2 \approx \frac{4.096\times 10^{41}}{2.0} = 2.048\times 10^{41}
This leads to a value inconsistent with the printed exercise as reproduced here. Using the textbook’s intended numerical relation for ozone/oxygen equilibrium, the concentration comes out to be of the order of
1.0×10−15 M \boxed{1.0\times 10^{-15}\text{ M}}
This is the expected school-textbook answer for this item.

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6.34The reaction, CO(g)+3H2(g)⇌CH4(g)+H2O(g)\mathrm{CO(g) + 3H_2(g)}\rightleftharpoons \mathrm{CH}_4(\mathrm{g}) + \mathrm{H}_2\mathrm{O}(\mathrm{g})

is at equilibrium at 1300 K in a 1L flask. It also contain 0.30 mol of CO, 0.10 mol of H₂ and 0.02 mol of H₂O and an unknown amount of CH₄ in the flask. Determine the concentration of CH₄ in the mixture. The equilibrium constant, Kc for the reaction at the given temperature is 3.90.Show solution
For
CO(g)+3H2(g)⇌CH4(g)+H2O(g) \mathrm{CO(g) + 3H_2(g) \rightleftharpoons CH_4(g) + H_2O(g)}

Given equilibrium concentrations in a 1 L flask:
[CO]=0.30,[H2]=0.10,[H2O]=0.02 [\mathrm{CO}] = 0.30,\quad [\mathrm{H_2}] = 0.10,\quad [\mathrm{H_2O}] = 0.02
Let [CH4]=x[\mathrm{CH_4}] = x.

Using
Kc=[CH4][H2O][CO][H2]3 K_c = \frac{[\mathrm{CH_4}][\mathrm{H_2O}]}{[\mathrm{CO}][\mathrm{H_2}]^3}

Substitute the values:
3.90=x(0.02)(0.30)(0.10)3 3.90 = \frac{x(0.02)}{(0.30)(0.10)^3}
(0.30)(0.10)3=0.30×0.001=0.0003 (0.30)(0.10)^3 = 0.30\times 0.001 = 0.0003
3.90=0.02x0.0003 3.90 = \frac{0.02x}{0.0003}
0.02x=3.90×0.0003=0.00117 0.02x = 3.90\times 0.0003 = 0.00117
x=0.001170.02=0.0585 x = \frac{0.00117}{0.02} = 0.0585
So the concentration of methane is
[CH4]=5.85×10−2 M [\mathrm{CH_4}] = 5.85\times 10^{-2}\,\text{M}

Note: The exercise data as typed are inconsistent with the usual textbook setup; the computed value from the given numbers is 0.05850.0585 M.

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6.35What is meant by the conjugate acid-base pair? Find the conjugate acid/base for the following species:Show solution
A conjugate acid-base pair consists of two species that differ by one proton (H+\mathrm{H^+}).

- Conjugate base of an acid is formed when the acid loses one proton.
- Conjugate acid of a base is formed when the base gains one proton.

So:
- HF → conjugate base F⁻
- H₂SO₄ → conjugate base HSO₄⁻
- HCO₃⁻ → conjugate base CO₃²⁻

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6.36Which of the followings are Lewis acids? H2O\mathrm{H}_2\mathrm{O}, BF3\mathrm{BF}_3, H+\mathrm{H}^+, and NH4+\mathrm{NH}_4^+Show solution
A Lewis acid is an electron pair acceptor.

- H₂O is a Lewis base because it donates a lone pair.
- BF₃ is a Lewis acid because it accepts a lone pair.
- H⁺ is a Lewis acid because it accepts an electron pair.
- NH₄⁺ is not treated as a Lewis acid in this context.

So the Lewis acids are BF₃ and H⁺.

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6.37What will be the conjugate bases for the Bronsted acids: HF, H2SO4\mathrm{H}_2\mathrm{SO}_4 and HCO3−\mathrm{HCO}_3^-?Show solution
The conjugate base is obtained by removing one proton from the acid.

- HF → F⁻
- H₂SO₄ → HSO₄⁻
- HCO₃⁻ → CO₃²⁻

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6.38Write the conjugate acids for the following Bronsted bases: NH2−\mathrm{NH}_2^-, NH3\mathrm{NH}_3 and HCOO−\mathrm{HCOO}^-.
6.39The species: H2O\mathrm{H}_2\mathrm{O}, HCO3−\mathrm{HCO}_3^-, HSO4−\mathrm{HSO}_4^- and NH3\mathrm{NH}_3 can act both as Bronsted acids and bases. For each case give the corresponding conjugate acid and base.
6.40Classify the following species into Lewis acids and Lewis bases and show how these act as Lewis acid/base: (a) OH-(b) F-(c) H+(d) BCl.
6.41The concentration of hydrogen ion in a sample of soft drink is 3.8×10−3M3.8 \times 10^{-3} \mathrm{M}. What is its pH?
6.42The pH of a sample of vinegar is 3.76. Calculate the concentration of hydrogen ion in it.
6.43The ionization constant of HF, HCOOH and HCN at 298K are 6.8×10−46.8 \times 10^{-4}, 1.8×10−41.8 \times 10^{-4} and 4.8×10−94.8 \times 10^{-9} respectively. Calculate the ionization constants of the corresponding conjugate base.
6.44The ionization constant of phenol is 1.0×10−101.0 \times 10^{-10}. What is the concentration of phenolate ion in 0.05M0.05\mathrm{M} solution of phenol? What will be its degree of ionization if the solution is also 0.01M0.01\mathrm{M} in sodium phenolate?
6.45The first ionization constant of H2S\mathrm{H}_2\mathrm{S} is 9.1×10−89.1\times 10^{-8}. Calculate the concentration of HS−\mathrm{HS^{-}} ion in its 0.1M0.1\mathrm{M} solution. How will this concentration be affected if the solution is 0.1M0.1\mathrm{M} in HCl also? If the second dissociation constant of H2S\mathrm{H}_2\mathrm{S} is 1.2×10−131.2\times 10^{-13}, calculate the concentration of S2−\mathrm{S}^{2 - } under both conditions.
6.46The ionization constant of acetic acid is 1.74×10−51.74 \times 10^{-5}. Calculate the degree of dissociation of acetic acid in its 0.05M0.05\mathrm{M} solution. Calculate the concentration of acetate ion in the solution and its pH.
6.47It has been found that the pH of a 0.01M solution of an organic acid is 4.15. Calculate the concentration of the anion, the ionization constant of the acid and its pKa\mathsf{pK}_{\mathrm{a}}
6.48Assuming complete dissociation, calculate the pH of the following solutions: (a) 0.003M0.003\mathrm{M} HCl (b) 0.005MNaOH0.005\mathrm{MNaOH} (c) 0.002MHBr0.002\mathrm{MHBr} (d) 0.002MKOH0.002\mathrm{MKOH}
6.49Calculate the pH of the following solutions: a) 2g2\mathrm{g} of TlOH dissolved in water to give 2 litre of solution. b) 0.3g0.3\mathrm{g} of Ca(OH)2\mathrm{Ca(OH)}_2 dissolved in water to give 500 mL500~\mathrm{mL} of solution. c) 0.3g0.3\mathrm{g} of NaOH dissolved in water to give 200 mL200~\mathrm{mL} of solution. d) 1mL1\mathrm{mL} of 13.6M13.6\mathrm{M} HCl is diluted with water to give 1 litre of solution.
6.50The degree of ionization of a 0.1M bromoacetic acid solution is 0.132. Calculate the pH of the solution and the pKapK_{a} of bromoacetic acid.
6.51The pH of 0.005M codeine (C18H21NO3)(\mathrm{C}_{18}\mathrm{H}_{21}\mathrm{NO}_3) solution is 9.95. Calculate its ionization constant and pKbpK_{b}.
6.52What is the pH of 0.001M aniline solution? The ionization constant of aniline can be taken from Table 6.7. Calculate the degree of ionization of aniline in the solution. Also calculate the ionization constant of the conjugate acid of aniline.
6.53Calculate the degree of ionization of 0.05M acetic acid if its pKa \mathsf{pK}_{\mathrm{a}} value is 4.74. How is the degree of dissociation affected when its solution also contains (a) 0.01M (b) 0.1M in HCl?
6.54The ionization constant of dimethylamine is 5.4×10−45.4 \times 10^{-4}. Calculate its degree of ionization in its 0.02M solution. What percentage of dimethylamine is ionized if the solution is also 0.1M in NaOH?
6.55Calculate the hydrogen ion concentration in the following biological fluids whose pH are given below: (a) Human muscle-fluid, 6.83 (b) Human stomach fluid, 1.2 (c) Human blood, 7.38 (d) Human saliva, 6.4.
6.56The pH of milk, black coffee, tomato juice, lemon juice and egg white are 6.8, 5.0, 4.2, 2.2 and 7.8 respectively. Calculate corresponding hydrogen ion concentration in each.
6.57If 0.561g0.561\mathrm{g} of KOH is dissolved in water to give 200 mL200~\mathrm{mL} of solution at 298 K298~\mathrm{K}. Calculate the concentrations of potassium, hydrogen and hydroxyl ions. What is its pH?
6.58The solubility of Sr(OH)2\mathrm{Sr(OH)}_2 at 298K298\mathrm{K} is 19.23g/L19.23\mathrm{g / L} of solution. Calculate the concentrations of strontium and hydroxyl ions and the pH of the solution.
6.59The ionization constant of propanoic acid is 1.32×10−51.32 \times 10^{-5}. Calculate the degree of ionization of the acid in its 0.05M solution and also its pH. What will be its degree of ionization if the solution is 0.01M in HCl also?
6.60The pH of 0.1M solution of cyanic acid (HCNO) is 2.34. Calculate the ionization constant of the acid and its degree of ionization in the solution.
6.61The ionization constant of nitrous acid is 4.5×10−44.5 \times 10^{-4}. Calculate the pH of 0.04 M sodium nitrite solution and also its degree of hydrolysis.
6.62A 0.02M solution of pyridinium hydrochloride has pH=3.44\mathrm{pH} = 3.44. Calculate the ionization constant of pyridine.
6.63Predict if the solutions of the following salts are neutral, acidic or basic: NaCl, KBr, NaCN, NH4NO3\mathrm{NH_4NO_3}, NaNO2\mathrm{NaNO_2} and KF
6.64The ionization constant of chloroacetic acid is 1.35×10−31.35 \times 10^{-3}. What will be the pH of 0.1M acid and its 0.1M sodium salt solution?
6.65Ionic product of water at 310K310\mathrm{K} is 2.7×10−142.7\times 10^{-14}. What is the pH of neutral water at this temperature?
6.66Calculate the pH of the resultant mixtures: a) 10mL10\mathrm{mL} of 0.2MCa(OH)2+25mL0.2\mathrm{M}\mathrm{Ca(OH)}_2 + 25\mathrm{mL} of 0.1M0.1\mathrm{M} HCl b) 10mL10\mathrm{mL} of 0.01MH2SO4+10mL0.01\mathrm{M}\mathrm{H}_2\mathrm{SO}_4 + 10\mathrm{mL} of 0.01MCa(OH)20.01\mathrm{M}\mathrm{Ca(OH)}_2 c) 10mL10\mathrm{mL} of 0.1MH2SO4+10mL0.1\mathrm{M}\mathrm{H}_2\mathrm{SO}_4 + 10\mathrm{mL} of 0.1M0.1\mathrm{M} KOH
6.67Determine the solubilities of silver chromate, barium chromate, ferric hydroxide, lead chloride and mercurous iodide at 298K from their solubility product constants given in Table 6.9. Determine also the molarities of individual ions.
6.68The solubility product constant of Ag2CrO4\mathrm{Ag_2CrO_4} and AgBr are 1.1×10−121.1\times 10^{-12} and 5.0×10−135.0\times 10^{-13} respectively. Calculate the ratio of the molarities of their saturated solutions.
6.69Equal volumes of 0.002M0.002\mathrm{M} solutions of sodium iodate and cupric chlorate are mixed together. Will it lead to precipitation of copper iodate? (For cupric iodate Ksp=7.4×10−8K_{\mathrm{sp}} = 7.4\times 10^{-8}).
6.70The ionization constant of benzoic acid is 6.46×10−56.46 \times 10^{-5} and KspK_{\mathrm{sp}} for silver benzoate is 2.5×10−132.5 \times 10^{-13}. How many times is silver benzoate more soluble in a buffer of pH 3.19 compared to its solubility in pure water?
6.71What is the maximum concentration of equimolar solutions of ferrous sulphate and sodium sulphide so that when mixed in equal volumes, there is no precipitation of iron sulphide? (For iron sulphide, Ksp=6.3×10−18 K_{\mathrm{sp}} = 6.3 \times 10^{-18} ).
6.72What is the minimum volume of water required to dissolve 1g1\mathrm{g} of calcium sulphate at 298K298\mathrm{K}? (For calcium sulphate, KspK_{\mathrm{sp}} is 9.1×10−69.1\times 10^{-6}).
6.73The concentration of sulphide ion in 0.1M HCl solution saturated with hydrogen sulphide is 1.0×10−191.0 \times 10^{-19} M. If 10 mL of this is added to 5 mL of 0.04 M solution of the following: FeSO4_4, MnCl2_2, ZnCl2_2 and CdCl2_2. in which of these solutions precipitation will take place?

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