Equilibrium Class 11 Chemistry NCERT Solutions

Step-by-step solution

1. 1(a) If the volume of the container is suddenly increased, the vapour pressure decreases initially. The amount of vapour remains the same but the volume increases suddenly, so the same amount of vapour is now distributed in a larger volume.
2. 2(b) The rate of evaporation remains the same initially, because it depends only on temperature and the surface area of the liquid. As volume increases, the density of the vapour phase decreases, so the rate of collisions between vapour particles and the liquid surface falls — hence the rate of condensation decreases initially.
3. 3(c) When equilibrium is restored, the rate of evaporation again becomes equal to the rate of condensation. Only the volume has changed while temperature is constant, and vapour pressure depends only on temperature, not on volume.
4. 4The final vapour pressure is therefore equal to the original vapour pressure of the system.

Step-by-step solution

1. 1Write the equilibrium constant expression for the reaction:
2. 2
   $$K_c=\frac{[\mathrm{S}{\mathrm{O}}_3{]}^2}{[\mathrm{S}{\mathrm{O}}_2{]}^2[{\mathrm{O}}_2]}$$
3. 3Substitute the given equilibrium concentrations:
4. 4
   $$K_c=\frac{(1.90{)}^2{\mathrm{M}}^2}{(0.60{)}^2(0.82){\mathrm{M}}^3}=\frac{3.61}{0.2952}{\mathrm{M}}^{-1}$$
5. 5
   $$K_c=12.23{\mathrm{M}}^{-1}$$

Step-by-step solution

1. 1Volume per cent of a gas equals its mole fraction, so the atom pressure is 40% of the total pressure:
2. 2
   $$p_I=\frac{40}{100}\times 10^5=4\times 10^4\mathrm{Pa}$$
3. 3The remaining 60% is molecular iodine:
4. 4
   $$p_{{\mathrm{I}}_2}=\frac{60}{100}\times 10^5=6\times 10^4\mathrm{Pa}$$
5. 5Write K_p, noting that I₂ is reactant and I is product of the forward reaction I₂ ⇌ 2I:
6. 6
   $$K_p=\frac{(p_I{)}^2}{p_{{\mathrm{I}}_2}}=\frac{(4\times 10^4{)}^2\mathrm{P}{\mathrm{a}}^2}{6\times 10^4\mathrm{Pa}}$$
7. 7
   $$K_p=\frac{16\times 10^8}{6\times 10^4}=2.67\times 10^4\mathrm{Pa}$$

Step-by-step solution

1. 1(i) For 2NOCl(g) ⇌ 2NO(g) + Cl₂(g):
2. 2
   $$K_c=\frac{[\mathrm{NO}{]}^2[\mathrm{C}{\mathrm{l}}_2]}{[\mathrm{NOCl}{]}^2}$$
3. 3(ii) For 2Cu(NO₃)₂(s) ⇌ 2CuO(s) + 4NO₂(g) + O₂(g), pure solids are omitted from the expression:
4. 4
   $$K_c=\frac{[\mathrm{N}{\mathrm{O}}_2{]}^4[{\mathrm{O}}_2]}{[\mathrm{Cu}(\mathrm{N}{\mathrm{O}}_3{)}_2{]}^2}=[\mathrm{N}{\mathrm{O}}_2{]}^4[{\mathrm{O}}_2]$$
5. 5(iii) Pure liquid water, present in excess, is omitted:
6. 6
   $$K_c=\frac{[\mathrm{C}{\mathrm{H}}_3\mathrm{COOH}][{\mathrm{C}}_2{\mathrm{H}}_5\mathrm{OH}]}{[\mathrm{C}{\mathrm{H}}_3\mathrm{COO}{\mathrm{C}}_2{\mathrm{H}}_5][{\mathrm{H}}_2\mathrm{O}]}=\frac{[\mathrm{C}{\mathrm{H}}_3\mathrm{COOH}][{\mathrm{C}}_2{\mathrm{H}}_5\mathrm{OH}]}{[\mathrm{C}{\mathrm{H}}_3\mathrm{COO}{\mathrm{C}}_2{\mathrm{H}}_5]}$$
7. 7(iv) The solid Fe(OH)₃ is omitted:
8. 8
   $$K_c=\frac{[\mathrm{Fe}(\mathrm{OH}{)}_3]}{[\mathrm{F}{\mathrm{e}}^{3+}][\mathrm{O}{\mathrm{H}}^{-}{]}^3}=\frac{1}{[\mathrm{F}{\mathrm{e}}^{3+}][\mathrm{O}{\mathrm{H}}^{-}{]}^3}$$
9. 9(v) Solid I₂ is omitted:
10. 10
    $$K_c=\frac{[\mathrm{I}{\mathrm{F}}_5{]}^2}{[{\mathrm{I}}_2][{\mathrm{F}}_2{]}^5}=\frac{[\mathrm{I}{\mathrm{F}}_5{]}^2}{[{\mathrm{F}}_2{]}^5}$$

Step-by-step solution

1. 1The relation between K_p and K_c is:
2. 2
   $$K_p=K_c(RT{)}^{\Delta n}$$
3. 3(i) For 2NOCl ⇌ 2NO + Cl₂, Δn = 3 − 2 = 1. With R = 0.0831 bar L mol⁻¹ K⁻¹ and T = 500 K:
4. 4
   $$1.8\times 10^{-2}=K_c(0.0831\times 500{)}^1$$
5. 5
   $$K_c=\frac{1.8\times 10^{-2}}{0.0831\times 500}=4.33\times 10^{-4}$$
6. 6(ii) For CaCO₃(s) ⇌ CaO(s) + CO₂(g), only CO₂ is gaseous, so Δn = 1 − 0 = 1. With T = 1073 K:
7. 7
   $$K_c=\frac{167}{(0.0831)(1073{)}^1}=1.87$$

Step-by-step solution

1. 1The equilibrium constant of the reverse reaction is the reciprocal of that of the forward reaction:
2. 2
   $$K_c^{\prime }=\frac{1}{K_c}=\frac{1}{6.3\times 10^{14}}$$
3. 3
   $$K_c^{\prime }=1.59\times 10^{-15}$$

Step-by-step solution

1. 1The molar concentration of a pure substance is its mass per unit volume divided by its molecular mass:
2. 2
   $$[\text{Pure substance}]=\frac{\text{Number of moles}}{\text{Volume}}=\frac{\text{Mass}/\text{molecular mass}}{\text{Volume}}$$
3. 3
   $$=\frac{\text{Mass}}{\text{Volume}\times \text{Molecular mass}}=\frac{\text{Density}}{\text{Molecular mass}}$$
4. 4At a given temperature the density and molecular mass of a pure solid or liquid are fixed constants. Their ratio is therefore constant and is already absorbed into the equilibrium constant.
5. 5Hence the molar concentrations of pure liquids and solids need not (and cannot conveniently) appear in the equilibrium constant expression.

Step-by-step solution

1. 1Let the concentration of N₂O formed at equilibrium be x mol. Using a 10 L vessel the ICE table reads:
2. 2Initial: 0.482 mol N₂, 0.933 mol O₂, 0 mol N₂O.
3. 3At equilibrium: (0.482 − x) mol N₂, (0.933 − x/2) mol O₂, x mol N₂O, giving:
4. 4
   $$[N_2]=\frac{0.482-x}{10},[O_2]=\frac{0.933-x/2}{10},[N_{2}O]=\frac{x}{10}$$
5. 5Since K_c = 2.0 × 10⁻³⁷ is extremely small, x is negligible relative to the initial amounts of N₂ and O₂:
6. 6
   $$[N_2]\approx \frac{0.482}{10}=0.0482\mathrm{mol}{\mathrm{L}}^{-1},[O_2]\approx \frac{0.933}{10}=0.0933\mathrm{mol}{\mathrm{L}}^{-1}$$
7. 7Substitute into the equilibrium constant expression:
8. 8
   $$K_c=\frac{[N_{2}O{]}^2}{[N_2{]}^2[O_2]}\Rightarrow 2.0\times 10^{-37}=\frac{(x/10{)}^2}{(0.0482{)}^2(0.0933)}$$
9. 9
   $$\frac{x^2}{100}=2.0\times 10^{-37}(0.0482{)}^2(0.0933)=4.335\times 10^{-41}$$
10. 10
    $$x^2=4.335\times 10^{-39}\Rightarrow x=6.6\times 10^{-20}\mathrm{mol}$$
11. 11Concentration of N₂O at equilibrium:
12. 12
    $$[N_{2}O]=\frac{6.6\times 10^{-20}}{10}=6.6\times 10^{-21}\mathrm{mol}{\mathrm{L}}^{-1}$$

Step-by-step solution

1. 1From the stoichometry, 2 mol of NO produce 2 mol of NOBr, so 0.0518 mol of NOBr is formed from 0.0518 mol of NO.
2. 22 mol of NOBr require 1 mol of Br₂, so 0.0518 mol of NOBr is formed from 0.0518/2 = 0.0259 mol of Br₂.
3. 3Amount of NO left at equilibrium:
4. 4
   $$[NO]=0.087-0.0518=0.0352\mathrm{mol}$$
5. 5Amount of Br₂ left at equilibrium:
6. 6
   $$[B{r}_2]=0.0437-0.0259=0.0178\mathrm{mol}$$

Step-by-step solution

1. 1For the reaction 2SO₂ + O₂ ⇌ 2SO₃, the change in the number of moles of gases is:
2. 2
   $$\Delta n=2-3=-1$$
3. 3Use the relation K_p = K_c(RT)^Δn, with R = 0.0831 L bar K⁻¹ mol⁻¹ and T = 450 K:
4. 4
   $$2.0\times 10^{10}\mathrm{ba}{\mathrm{r}}^{-1}=K_c(0.0831\times 450{)}^{-1}$$
5. 5
   $$K_c=(2.0\times 10^{10})(0.0831\times 450)=(2.0\times 10^{10})(37.395)$$
6. 6
   $$K_c=7.479\times 10^{11}\mathrm{L}\mathrm{mo}{\mathrm{l}}^{-1}=7.48\times 10^{11}{\mathrm{M}}^{-1}$$

Step-by-step solution

1. 1The drop in the pressure of HI as it dissociates:
2. 2
   $$0.2-0.04=0.16\mathrm{atm}$$
3. 3From the stoichometry, 2 mol HI give 1 mol H₂ and 1 mol I₂, so the pressure lost by HI produces half of itself in H₂ and I₂:
4. 4
   $$p_{H_2}=p_{I_2}=\frac{0.16}{2}=0.08\mathrm{atm}$$
5. 5At equilibrium: p_HI = 0.04 atm, p_H₂ = p_I₂ = 0.08 atm. Then,
6. 6
   $$K_p=\frac{p_{H_2}\times p_{I_2}}{(p_{HI}{)}^2}=\frac{0.08\times 0.08}{(0.04{)}^2}$$
7. 7
   $$K_p=\frac{0.0064}{0.0016}=4.0$$

Step-by-step solution

1. 1The given concentrations in the 20 L vessel are:
2. 2
   $$[N_2]=\frac{1.57}{20}\mathrm{mol}{\mathrm{L}}^{-1},[H_2]=\frac{1.92}{20}\mathrm{mol}{\mathrm{L}}^{-1},[N{H}_3]=\frac{8.13}{20}\mathrm{mol}{\mathrm{L}}^{-1}$$
3. 3Calculate the reaction quotient Q_c:
4. 4
   $$Q_c=\frac{[N{H}_3{]}^2}{[N_2][H_2{]}^3}=\frac{{\left(\frac{8.13}{20}\right)}^2}{\left(\frac{1.57}{20}\right){\left(\frac{1.92}{20}\right)}^3}$$
5. 5
   $$Q_c=2.4\times 10^3$$
6. 6Compare Q_c with K_c = 1.7 × 10². Since Q_c ≠ K_c, the mixture is not at equilibrium.
7. 7Since Q_c > K_c, the reaction must proceed in the reverse direction to reach equilibrium (the denominator [N₂][H₂]³ must increase).

Step-by-step solution

1. 1The numerator lists the products (each raised to its coefficient) and the denominator lists the reactants:
2. 2Products: NH₃ (coefficient 4), O₂ (coefficient 5).
3. 3Reactants: NO (coefficient 4), H₂O (coefficient 6).
4. 4Hence the balanced equation is:
5. 5
   $$4\mathrm{N}{\mathrm{O}}_{(\mathrm{g})}+6{\mathrm{H}}_2{\mathrm{O}}_{(\mathrm{g})}\rightleftharpoons 4\mathrm{N}{\mathrm{H}}_{3(\mathrm{g})}+5{\mathrm{O}}_{2(\mathrm{g})}$$

Step-by-step solution

1. 1Initial concentrations in the 10 L vessel are [H₂O] = [CO] = 1/10 = 0.1 M; [H₂] = [CO₂] = 0.
2. 240% of water reacts, so the amount reacting is 0.4 mol; the ICE table becomes (mol per litre):
3. 3At equilibrium: [H₂O] = (1 − 0.4)/10 = 0.06 M, [CO] = (1 − 0.4)/10 = 0.06 M, [H₂] = 0.4/10 = 0.04 M, [CO₂] = 0.4/10 = 0.04 M.
4. 4Substitute into the equilibrium constant expression:
5. 5
   $$K_c=\frac{[H_2][C{O}_2]}{[H_{2}O][CO]}=\frac{0.04\times 0.04}{0.06\times 0.06}$$
6. 6
   $$K_c=\frac{0.0016}{0.0036}=0.444$$

Step-by-step solution

1. 1For the forward reaction H₂ + I₂ ⇌ 2HI, K_c = 54.8. The reverse reaction 2HI ⇌ H₂ + I₂ therefore has:
2. 2
   $$K_c^{\prime }=\frac{1}{K_c}=\frac{1}{54.8}$$
3. 3Let the equilibrium concentrations of hydrogen and iodine be x mol L⁻¹ each. Given [HI] = 0.5 mol L⁻¹:
4. 4
   $$\frac{[H_2][I_2]}{[HI{]}^2}=K_c^{\prime }\Rightarrow \frac{x\times x}{(0.5{)}^2}=\frac{1}{54.8}$$
5. 5
   $$x^2=\frac{0.25}{54.8}=4.562\times 10^{-3}$$
6. 6
   $$x=0.06754\approx 0.068\mathrm{mol}{\mathrm{L}}^{-1}$$

Step-by-step solution

1. 1Let x be the equilibrium concentration of I₂ (and of Cl₂, by stoichometry). The ICE table reads:
2. 2Initial: [ICl] = 0.78 M, [Cl₂] = [I₂] = 0.
3. 3At equilibrium: [ICl] = (0.78 − 2x) M, [Cl₂] = x M, [I₂] = x M.
4. 4Substitute into the equilibrium constant expression:
5. 5
   $$K_c=\frac{[I_2][C{l}_2]}{[ICl{]}^2}=\frac{x\times x}{(0.78-2x{)}^2}=0.14$$
6. 6
   $$\frac{x}{0.78-2x}=\sqrt{0.14}=0.374$$
7. 7
   $$x=0.374(0.78-2x)\Rightarrow x=0.292-0.748x$$
8. 8
   $$1.748x=0.292\Rightarrow x=0.167\mathrm{M}$$
9. 9Equilibrium concentrations:
10. 10
    $$[C{l}_2]=[I_2]=0.167\mathrm{M},[ICl]=0.78-2(0.167)=0.446\mathrm{M}$$

Step-by-step solution

1. 1Let p be the equilibrium partial pressure of each of ethene and hydrogen gas.
2. 2Initial pressure of C₂H₆ = 4.0 atm, with p_C₂H₄ = p_H₂ = 0. At equilibrium: p_C₂H₆ = (4.0 − p) atm, p_C₂H₄ = p atm, p_H₂ = p atm.
3. 3Substitute into the equilibrium constant expression:
4. 4
   $$K_p=\frac{p_{C_2{H}_4}\times p_{H_2}}{p_{C_2{H}_6}}=\frac{p\times p}{4.0-p}=0.04$$
5. 5
   $$p^2+0.04p-0.16=0$$
6. 6
   $$p=\frac{-0.04\pm \sqrt{(0.04{)}^2+4(0.16)}}{2}=\frac{-0.04\pm 0.80}{2}$$
7. 7Taking the positive root: p = 0.76/2 = 0.38 atm.
8. 8Equilibrium pressure (concentration) of C₂H₆:
9. 9
   $$p_{C_2{H}_6}=4.0-0.38=3.62\mathrm{atm}$$

Step-by-step solution

1. 1(i) Reaction quotient for the reaction:
2. 2
   $$Q_c=\frac{[C{H}_{3}COO{C}_2{H}_5][H_{2}O]}{[C{H}_{3}COOH][C_2{H}_{5}OH]}$$
3. 3(ii) Let the volume of the mixture be V. The ICE table reads (with x = 0.171 mol of product at equilibrium):
4. 4At equilibrium: [CH₃COOH] = (1 − 0.171)/V, [C₂H₅OH] = (0.18 − 0.171)/V = 0.009/V, [CH₃COOC₂H₅] = 0.171/V, [H₂O] = 0.171/V.
5. 5
   $$K_c=\frac{\left(\frac{0.171}{V}\right)\left(\frac{0.171}{V}\right)}{\left(\frac{0.829}{V}\right)\left(\frac{0.009}{V}\right)}=\frac{0.029241}{0.007461}$$
6. 6
   $$K_c=3.919\approx 3.92$$
7. 7(iii) Now 0.214 mol of ethyl acetate has formed from 1.0 mol acid and 0.5 mol ethanol. The reaction quotient is:
8. 8
   $$Q_c=\frac{\left(\frac{0.214}{V}\right)\left(\frac{0.214}{V}\right)}{\left(\frac{0.786}{V}\right)\left(\frac{0.286}{V}\right)}$$
9. 9
   $$Q_c=\frac{0.0458}{0.2248}=0.2037\approx 0.204$$
10. 10Since Q_c (0.204) < K_c (3.92), the reaction has not yet reached equilibrium and more ethyl acetate must still form.

Step-by-step solution

1. 1Let the equilibrium concentrations of PCl₃ and Cl₂ each be x mol L⁻¹. Given [PCl₅] = 0.5 × 10⁻¹ mol L⁻¹.
2. 2The equilibrium constant expression gives:
3. 3
   $$K_c=\frac{[PC{l}_3][C{l}_2]}{[PC{l}_5]}\Rightarrow \frac{x\times x}{0.5\times 10^{-1}}=8.3\times 10^{-3}$$
4. 4
   $$x^2=8.3\times 10^{-3}\times 0.5\times 10^{-1}=4.15\times 10^{-4}$$
5. 5
   $$x=2.04\times 10^{-2}\approx 0.02\mathrm{mol}{\mathrm{L}}^{-1}$$

Step-by-step solution

1. 1First evaluate the reaction quotient from the initial pressures:
2. 2
   $$Q_p=\frac{p_{C{O}_2}}{p_{CO}}=\frac{0.80}{1.4}=0.571$$
3. 3Since Q_p = 0.571 > K_p = 0.265, the reaction proceeds in the backward direction: p_CO increases and p_CO₂ decreases.
4. 4Let p be the increase in CO pressure, which equals the decrease in CO₂ pressure. Then:
5. 5
   $$K_p=\frac{p_{C{O}_2}}{p_{CO}}\Rightarrow 0.265=\frac{0.80-p}{1.4+p}$$
6. 6
   $$0.265(1.4+p)=0.80-p\Rightarrow 0.371+0.265p=0.80-p$$
7. 7
   $$1.265p=0.429\Rightarrow p=0.339\mathrm{atm}$$
8. 8Equilibrium partial pressures:
9. 9
   $$p_{C{O}_2}=0.80-0.339=0.461\mathrm{atm},p_{CO}=1.4+0.339=1.739\mathrm{atm}$$

Step-by-step solution

1. 1Calculate the reaction quotient from the given concentrations:
2. 2
   $$Q_c=\frac{[N{H}_3{]}^2}{[N_2][H_2{]}^3}=\frac{(0.5{)}^2}{(3.0)(2.0{)}^3}$$
3. 3
   $$Q_c=\frac{0.25}{3.0\times 8.0}=\frac{0.25}{24}=0.0104$$
4. 4Compare with K_c = 0.061. Since Q_c ≠ K_c, the reaction is not at equilibrium.
5. 5Since Q_c < K_c, the numerator must increase; the reaction proceeds in the forward direction to reach equilibrium.

Step-by-step solution

1. 1Let the amount of bromine (and chlorine, by stoichometry) formed at equilibrium be x. The equilibrium concentrations are:
2. 2[BrCl] = (3.3 × 10⁻³ − 2x) M, [Br₂] = x M, [Cl₂] = x M.
3. 3Substitute into the equilibrium constant expression:
4. 4
   $$K_c=\frac{[B{r}_2][C{l}_2]}{[BrCl{]}^2}\Rightarrow \frac{x\times x}{(3.3\times 10^{-3}-2x{)}^2}=32$$
5. 5
   $$\frac{x}{3.3\times 10^{-3}-2x}=\sqrt{32}=5.66$$
6. 6
   $$x=5.66(3.3\times 10^{-3}-2x)\Rightarrow x=1.8678\times 10^{-2}-11.32x$$
7. 7
   $$12.32x=18.678\times 10^{-3}\Rightarrow x=1.5\times 10^{-3}$$
8. 8Equilibrium concentration of BrCl:
9. 9
   $$[BrCl]=3.3\times 10^{-3}-2(1.5\times 10^{-3})=3.0\times 10^{-4}\mathrm{mol}{\mathrm{L}}^{-1}$$

Step-by-step solution

1. 1Take 100 g of the gaseous mixture: mass of CO = 90.55 g and mass of CO₂ = 9.45 g.
2. 2Number of moles:
3. 3
   $$n_{CO}=\frac{90.55}{28}=3.234\mathrm{mol},n_{C{O}_2}=\frac{9.45}{44}=0.215\mathrm{mol}$$
4. 4Partial pressures from mole fractions (total pressure = 1 atm):
5. 5
   $$p_{CO}=\frac{3.234}{3.234+0.215}\times 1=0.938\mathrm{atm},p_{C{O}_2}=\frac{0.215}{3.234+0.215}\times 1=0.062\mathrm{atm}$$
6. 6Then,
7. 7
   $$K_p=\frac{(p_{CO}{)}^2}{p_{C{O}_2}}=\frac{(0.938{)}^2}{0.062}=14.19$$
8. 8For this reaction Δn = 2 − 1 = 1. Using K_p = K_c(RT)^Δn with R = 0.082 L atm K⁻¹ mol⁻¹ and T = 1127 K:
9. 9
   $$K_c=\frac{K_p}{RT}=\frac{14.19}{0.082\times 1127}=0.154$$

Step-by-step solution

1. 1(a) Standard Gibbs energy change of the reaction:
2. 2
   $$\Delta G^{\circ }=\Delta G^{\circ }(\text{Products})-\Delta G^{\circ }(\text{Reactants})$$
3. 3
   $$\Delta G^{\circ }=52.0-(87.0+0)=-35.0\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$$
4. 4(b) Relationship between ΔG° and the equilibrium constant:
5. 5
   $$\Delta G^{\circ }=-RT\ln K_c=-2.303RT\log K_c$$
6. 6
   $$\log K_c=\frac{-35.0\times 10^3}{-2.303\times 8.314\times 298}=6.134$$
7. 7
   $$K_c=\text{antilog}(6.134)=1.36\times 10^6$$

Step-by-step solution

1. 1(a) Moles of products increase. Lowering pressure shifts the equilibrium towards the side with more gas molecules; here the products hold more moles of gas (2 mol) than the reactants (1 mol), so the forward reaction proceeds.
2. 2(b) Moles of products decrease. The products side has fewer gas molecules (0) than the reactants (1 mol CO₂); decreased pressure shifts the equilibrium backwards, consuming product.
3. 3(c) Moles of products remain the same. Both sides hold 4 mol of gas, so a pressure change has no effect on the extent of reaction.

Step-by-step solution

1. 1Increasing the pressure affects a reaction only when the number of moles of gas differs between the two sides. Reactions (ii) has 3 mol gas on each side and is unaffected — all the rest are affected.
2. 2(iv) 2H₂ + CO ⇌ CH₃OH has 3 mol gas reactant and 1 mol gas product; pressure increases shift it to the side of fewer gas molecules, i.e. the forward direction.
3. 3(i), (iii), (v) and (vi) each have more moles of gas on the product side than on the reactant side; increased pressure shifts them in the backward direction.

Step-by-step solution

1. 1The forward reaction H₂ + Br₂ ⇌ 2HBr has K_p = 1.6 × 10⁵, so the reverse reaction 2HBr ⇌ H₂ + Br₂ has:
2. 2
   $$K_p^{\prime }=\frac{1}{K_p}=\frac{1}{1.6\times 10^5}=6.25\times 10^{-6}$$
3. 3Let p be the equilibrium pressure of each of H₂ and Br₂. Then p_HBr = (10 − 2p) bar.
4. 4
   $$\frac{p_{H_2}\times p_{B{r}_2}}{(p_{HBr}{)}^2}=K_p^{\prime }\Rightarrow \frac{p\times p}{(10-2p{)}^2}=6.25\times 10^{-6}$$
5. 5
   $$\frac{p}{10-2p}=2.5\times 10^{-3}$$
6. 6
   $$p=2.5\times 10^{-2}-5.0\times 10^{-3}p\Rightarrow 1.005p=2.5\times 10^{-2}$$
7. 7
   $$p=2.49\times 10^{-2}\mathrm{bar}$$
8. 8Equilibrium pressures:
9. 9
   $$p_{H_2}=p_{B{r}_2}=2.49\times 10^{-2}\mathrm{bar},p_{HBr}=10-2(2.49\times 10^{-2})=9.95\mathrm{bar}$$

Step-by-step solution

1. 1(a) The equilibrium constant expression in terms of partial pressures:
2. 2
   $$K_p=\frac{p_{CO}\times p_{H_2}^3}{p_{C{H}_4}\times p_{H_{2}O}}$$
3. 3(b)(i) Increasing the pressure shifts the equilibrium towards the side with fewer moles of gas. The reactants hold 2 mol gas, the products 4 mol; by Le Chatelier's principle the reaction shifts in the backward direction, so K_p itself is unchanged but the composition changes.
4. 4(b)(ii) The reaction is endothermic, so increasing the temperature shifts the equilibrium in the forward direction. K_p increases with temperature.
5. 5(b)(iii) A catalyst increases the rate of both forward and backward reactions equally and does not affect the value of K_p or the composition; equilibrium is simply reached faster.

Step-by-step solution

1. 1(a) Adding H₂ increases a reactant concentration; by Le Chatelier's principle the equilibrium shifts in the forward direction.
2. 2(b) Adding CH₃OH increases a product concentration; the equilibrium shifts in the backward direction.
3. 3(c) Removing CO decreases a reactant concentration; the equilibrium shifts in the backward direction.
4. 4(d) Removing CH₃OH decreases a product concentration; the equilibrium shifts in the forward direction.

Step-by-step solution

1. 1(a) The equilibrium constant expression:
2. 2
   $$K_c=\frac{[PC{l}_3][C{l}_2]}{[PC{l}_5]}$$
3. 3(b) The equilibrium constant of the reverse reaction is the reciprocal:
4. 4
   $$K_c^{\prime }=\frac{1}{K_c}=\frac{1}{8.3\times 10^{-3}}=1.2048\times 10^2=120.48$$
5. 5(c)(i) Adding more PCl₅ leaves K_c unchanged because K_c depends only on temperature.
6. 6(c)(ii) K_c is constant at constant temperature, so increasing the pressure does not change its value.
7. 7(c)(iii) For an endothermic reaction, the value of K_c increases with an increase in temperature.

Step-by-step solution

1. 1Let the equilibrium partial pressure of each of carbon dioxide and hydrogen be p. Then at equilibrium: p_CO = p_H₂O = (4.0 − p) bar and p_CO₂ = p_H₂ = p bar.
2. 2Substitute into the equilibrium constant expression:
3. 3
   $$K_p=\frac{p_{C{O}_2}\times p_{H_2}}{p_{CO}\times p_{H_{2}O}}\Rightarrow \frac{p\times p}{(4.0-p)(4.0-p)}=10.1$$
4. 4
   $$\frac{p}{4.0-p}=\sqrt{10.1}=3.178$$
5. 5
   $$p=12.712-3.178p\Rightarrow 4.178p=12.712$$
6. 6
   $$p=3.04\mathrm{bar}$$

Step-by-step solution

1. 1When K_c lies in the range 10⁻³ to 10³, neither reactants nor products dominate and the reaction has appreciable concentrations of both.
2. 2(a) K_c = 5 × 10⁻³⁹ is extremely small — products are negligible.
3. 3(b) K_c = 3.7 × 10⁸ is extremely large — reactants are negligible.
4. 4(c) K_c = 1.8 lies within 10⁻³ to 10³, so this reaction has appreciable concentrations of reactants and products.

Step-by-step solution

1. 1The equilibrium constant expression is:
2. 2
   $$K_c=\frac{[O_3{]}^2}{[O_2{]}^3}$$
3. 3Substitute K_c = 2.0 × 10⁻⁵⁰ and [O₂] = 1.6 × 10⁻²:
4. 4
   $$2.0\times 10^{-50}=\frac{[O_3{]}^2}{(1.6\times 10^{-2}{)}^3}$$
5. 5
   $$[O_3{]}^2=2.0\times 10^{-50}\times (1.6\times 10^{-2}{)}^3=8.192\times 10^{-56}$$
6. 6
   $$[O_3]=2.86\times 10^{-28}\mathrm{mol}{\mathrm{L}}^{-1}$$

Step-by-step solution

1. 1Let the equilibrium concentration of methane be x. In the 1 L flask the other concentrations are [CO] = 0.30 M, [H₂] = 0.10 M and [H₂O] = 0.02 M.
2. 2Substitute into the equilibrium constant expression:
3. 3
   $$K_c=\frac{[C{H}_4][H_{2}O]}{[CO][H_2{]}^3}\Rightarrow \frac{x\times 0.02}{0.3\times (0.1{)}^3}=3.90$$
4. 4
   $$x=\frac{3.90\times 0.3\times (0.1{)}^3}{0.02}=\frac{0.00117}{0.02}$$
5. 5
   $$x=0.0585\mathrm{M}=5.85\times 10^{-2}\mathrm{M}$$

Step-by-step solution

1. 1A conjugate acid–base pair is a pair of species that differ only by one proton (H⁺).
2. 2When the species gains a proton it acts as a base, giving its conjugate acid; when it loses a proton it acts as an acid, giving its conjugate base.
3. 3The pairs are tabulated below: HNO₂ and NO₂⁻ (base); CN⁻ and HCN (acid); HClO₄ and ClO₄⁻ (base); F⁻ and HF (acid); OH⁻ with H₂O (acid)/O²⁻ (base); CO₃²⁻ and HCO₃⁻ (acid); S²⁻ and HS⁻ (acid).

Step-by-step solution

1. 1Lewis acids are species that can accept a pair of electrons.
2. 2H₂O has lone pairs and acts as a Lewis base; NH₄⁺ has no vacant orbital that accepts an electron pair in this context.
3. 3BF₃ has an incomplete octet and accepts an electron pair; H⁺ is an electron-pair acceptor.
4. 4Hence BF₃ and H⁺ are Lewis acids.

Step-by-step solution

1. 1The conjugate base of an acid is the species formed when the acid loses one proton.
2. 2HF loses H⁺ to give F⁻.
3. 3H₂SO₄ loses H⁺ to give HSO₄⁻.
4. 4HCO₃⁻ loses H⁺ to give CO₃²⁻.

Step-by-step solution

1. 1The conjugate acid of a base is the species formed when the base gains one proton.
2. 2NH₂⁻ gains H⁺ to give NH₃.
3. 3NH₃ gains H⁺ to give NH₄⁺.
4. 4HCOO⁻ gains H⁺ to give HCOOH.

Step-by-step solution

1. 1A species that is amphiprotic gains a proton (acting as a base, giving a conjugate acid) or loses a proton (acting as an acid, giving a conjugate base).
2. 2H₂O: conjugate acid H₃O⁺, conjugate base OH⁻.
3. 3HCO₃⁻: conjugate acid H₂CO₃, conjugate base CO₃²⁻.
4. 4HSO₄⁻: conjugate acid H₂SO₄, conjugate base SO₄²⁻.
5. 5NH₃: conjugate acid NH₄⁺, conjugate base NH₂⁻.

Step-by-step solution

1. 1(a) OH⁻ is a Lewis base because it can donate its lone pair of electrons.
2. 2(b) F⁻ is a Lewis base because it can donate a pair of electrons.
3. 3(c) H⁺ is a Lewis acid because it can accept a pair of electrons.
4. 4(d) BCl₃ is a Lewis acid because boron has an incomplete octet and can accept a pair of electrons.

Step-by-step solution

1. 1Given [H⁺] = 3.8 × 10⁻³ M.
2. 2
   $$\mathrm{pH}=-\log [H^{+}]=-\log (3.8\times 10^{-3})$$
3. 3
   $$=-(\log 3.8+\log 10^{-3})=-0.58+3$$
4. 4
   $$\mathrm{pH}=2.42$$

Step-by-step solution

1. 1Given pH = 3.76.
2. 2
   $$\mathrm{pH}=-\log [H^{+}]\Rightarrow \log [H^{+}]=-\mathrm{pH}$$
3. 3
   $$[H^{+}]=\text{antilog}(-3.76)=1.74\times 10^{-4}\mathrm{mol}{\mathrm{L}}^{-1}$$

Step-by-step solution

1. 1For a conjugate acid–base pair, K_a × K_b = K_w, so:
2. 2
   $$K_b=\frac{K_w}{K_a}=\frac{10^{-14}}{K_a}$$
3. 3For HF (K_a = 6.8 × 10⁻⁴), the conjugate base F⁻ has:
4. 4
   $$K_b=\frac{10^{-14}}{6.8\times 10^{-4}}=1.5\times 10^{-11}$$
5. 5For HCOOH (K_a = 1.8 × 10⁻⁴), the conjugate base HCOO⁻ has:
6. 6
   $$K_b=\frac{10^{-14}}{1.8\times 10^{-4}}=5.6\times 10^{-11}$$
7. 7For HCN (K_a = 4.8 × 10⁻⁹), the conjugate base CN⁻ has:
8. 8
   $$K_b=\frac{10^{-14}}{4.8\times 10^{-9}}=2.08\times 10^{-6}$$

Step-by-step solution

1. 1Ionisation of phenol: C₆H₅OH + H₂O ⇌ C₆H₅O⁻ + H₃O⁺.
2. 2Initial: 0.05 M phenol, 0 of the ions. At equilibrium: (0.05 − x) M phenol, x M C₆H₅O⁻, x M H₃O⁺.
3. 3
   $$K_a=\frac{[C_6{H}_5{O}^{-}][H_3{O}^{+}]}{[C_6{H}_{5}OH]}=\frac{x\cdot x}{0.05-x}$$
4. 4K_a is very small so x is negligible next to 0.05:
5. 5
   $$x=\sqrt{1.0\times 10^{-10}\times 0.05}=\sqrt{5\times 10^{-12}}=2.2\times 10^{-6}\mathrm{M}$$
6. 6Since [H₃O⁺] = [C₆H₅O⁻]:
7. 7
   $$[C_6{H}_5{O}^{-}]=2.2\times 10^{-6}\mathrm{M}$$
8. 8Now with 0.01 M sodium phenolate present, common-ion effect suppresses the ionisation. Writing the equilibrium with α as the degree of ionisation: [C₆H₅OH] ≈ 0.05 M, [C₆H₅O⁻] ≈ 0.01 M, [H₃O⁺] = 0.05α.
9. 9
   $$K_a=\frac{(0.01)(0.05\alpha )}{0.05}\Rightarrow 1.0\times 10^{-10}=0.01\alpha$$
10. 10
    $$\alpha =1\times 10^{-8}$$