Title 7.EQUILIBRIUM - Explanations.pdf ✅

1 (d) 10−2 M H2SO4 = 2 × 10−2 NH2SO4 [H +] = 2 × 10−2N pH = − log[H +] = −log[2 × 10−2 ] = 2 − log 2 = 2 − 0.3010 = 1.6990 2 (d) H2(g) + I2(g) ⇌ 2HI(g) The equilibrium constant (Kp) changes with the change in temperature. Note : Variation of equilibrium constant with temperature can be express as: log K2 K1 = ∆H 2.303R [ T2−T1 T1.T2 ] 3 (c) Kp = pCO 2 pCO2 = 4 × 4 2 = 8 4 (c) Acidic buffer is a mixture of weak acid and its salt with strong base. Similarly, basic buffer is a mixture of weak base and its salt with strong acid. Hence, 50 mL 0.1 M CH3COOH + 100 mL 0.1 M NaOH does not constitute a buffer solution because in this case millimoles of acid are less than that of strong base, which after reaction with strong base gives salt. Now, the solution contains only strong base and salt but no weak acid. Hence, no buffer is formed. 5 (b) Thermal decomposition of potassium chlorate is irreversible. So, law of mass action cannot be applied on this reaction 6 (c) 50 mL of 0.1 M HCl = 0.1×50 1000 = 5 × 10−3 50 mL of 0.2 M NaOH = 0.2×50 1000 = 10 × 10−3 Hence, after neutralisation NaOH is left = 10 × 10−3 − 5 × 10−3 = 5 × 10−3 Total volume= 100 cc The concentration of NaOH = 5 × 10−3 × 1000 100 = 0.05 M [OH −] = 0.05 M = 5 × 10−2M pOH = − log [OH −] = − log [5 × 10−2 ] = 1.3010 pH + pOH = 14 pH = 14 − 1.3010 = 12.699 7 (c) pOH = pKb + log [salt] [base] = 5 + log 0.01 0.1 pOH = 5 − 1 pOH = 4 pH + pOH = 14 pH = 14 − pOH = 14 − 4 pH = 10 8 (d) Aqueous solutions of HCOONa, C6H5NH3Cl, KCN are basic, acidic and basic in nature. HCOONa + H2O ⇌ HCOOH + NaOH Weak acid strong acid (Basic solution) C6H5NH3Cl(aq) ⇌ C6H5NH2 + HCl Weak base strong acid (Acidic solution) KCN + H2O ⇌ KOH + HCN Strong base weak acid (Basic solution) 9 (a) Ag2CrO4 → 2Ag + + CrO4 2− s 2s s Ksp = (2s) 2 s = 4s 3 s = ( Ksp 4 ) 1/3 = ( 32 × 10−12 4 ) 1/3 = 2 × 10−4M 10 (d) Strong conjugate base has a weak conjugate acid. Weakest conjugate acid is CH3COOH. 11 (d) Moles of HCl = 0.365 36.5 = 0.01 Moles of NaOH in 100 cm3 of 0.2 M NaOH 0.2 = n×1000 100 = 0.02 Moles of NaOH left=0.02−0.01=0.01 Moles of NaOH per litre 0.01×1000 100 = 0.1 [H +] = 10−14 0.1 = 10−13; (Kw = [H +].[OH −]) pH = − log[H +] = − log[10−13] pH = 13 12 (a) Mg(OH)2 ⇌ Mg2+ + 2OH− s 2s KspMg(OH)2 = [Mg2+][OH−] 2 ⇒ Ksp Mg(OH)2 = 4S 3 1.96 × 10−11 = 4S 3 or S = [ 1.96×10−11 4 ] 1⁄3 or S = (4.9 × 10−12) 1⁄3 ∴ S = 1.96 × 10−4 So, concentration of [OH −] = 2S
∴ [OH −] = 3.38 × 10−4 ⇒ pOH = − log[OH−] = − log[3.38 × 10−4 ] pOH = 3.471 pH = 14 − pOH = 14 − 3.471 ∴ pH=10.529 13 (c) Stronger the base, higher the tendency to accept protons. Among the given, CH3COOH and H2S both are acids, thus have very low tendency to accept a proton. Between NH3 and H2O,NH3 is a stronger base (due to less electronegativity of N as compared to O) and hence, it has the highest tendency to accept a proton among the given. Moreover, the conjugate base formed by it also stable one NH3 + H + → NH4 + (stable) 14 (b) Kp Kc (RT) ng Where, ng No. of moles of gaseous products – No. of moles of gaseous reactants CO(g) 1 2 O2 (g) CO2(g) ng 1 − 1.5 − 1 2 ∵ Kp Kc (RT) ng Kp Kc (RT) −1⁄2 Kp Kc (RT) −1⁄2 15 (a) N1V1 = N2V2 10−3 × 10 = N2 × 1000 N2 = 10−5 pH = 5 So,pOH = 14 − 5 = 9 17 (b) H2 (g) + Cl2(g) ⇌ 2HCl(g) We know that, Kp = Kc . (RT) ∆ng ∆ng = no. of moles of gaseous products – no. of moles of gaseous reactants =2-2=0 Kp = Kc . (RT) 0 Kp = Kc 18 (d) In the titrationof weak acid with strong base, phenolphthalein is used 19 (a) Only salts of (weak acid+ strong base) and (strong acid + weak base) get hydrolysed (i. e., show alkalinity or acidity in water). KClO4 a salt of strong acid and strong base, therefore, it does not get hydrolysed in water. KClO4 ⇌ K + + ClO4 − 20 (c) Kh = Ch 2 = 0.5 × ( 0.25 100) 2 = 3.125 × 10−6 21 (c) H2 (g) + I2(g) ⇌ 2HI(g) We know that, Kp = Kc (RT) ∆ng Here, ∆ng = 2 − 2 = 0 ∴ Kp = Kc (RT) 0 = Kc 23 (b) kf = 1.1 × 10−2 , kb = 1.5 × 10−3 Kc = kf kb = 1.1 × 10−2 1.5 × 10−3 = 7.33 24 (a) In the reaction, N2 (g) + O2(g) ⇌ 2NO(g), the number of moles of reactants and products are equal, thus it remain unaffected by change in pressure 25 (a) pH = − log[H +] = − log[0.005] = − log[5 × 10−3 ] = 2.3 26 (a) CH4 has almost no acidic nature and thus, CH3 − is the strongest base 27 (d) Kp is a constant and does not change with pressure 29 (a) N2 (g) + 3H2 (g) ⇌ 3NH3 (g) + 92.3 kJ According to Le-Chatelier’s principle, the favourable conditions for the reaction are 1. Low temperature (as the reaction is exothermic) 2. High pressure (volume is decreasing) 3. Constant removal of ammonia gas as it is formed. 30 (d) We know that pV = nRT p becomes 1 2 p and V becomes 2V, so 1 2 p × 2V = pV = nRT
Hence, there is no effect in equation 31 (d) 3O2(g) ⇌ 2O3(g) O2 and O3 are the allotropes of oxygen, i.e., have different composition, so the equilibrium exist between them is considered as chemical equilibrium. 32 (b) Kc = [HI] 2 [H2 ][I2] ∴ 64 = x 2 0.03 × 0.03 ∴ x 2 = 64 × 9 × 10−4 or, x = 8 × 3 × 10−2 x is the amount of HI At equilibrium, amount of I2 will be 0.30 − 0.24 = 0.06 33 (a) Conjugate acid base pair differ by H +. Acid − H + → Conjugate base HSO4 − − H + → SO4 2− 35 (b) The species which can donate a lone pair of electron, are called Lewis base e.g., NH3,H2O, Cl− etc. 36 (c) In qualitative analysis, in order to detect IInd group radicals, H2S gas is passed in the presence of dilute HCl to decrease the dissociation of H2S by common-ion effect so that less S 2−ions are obtained and only IInd group radical could precipitate. 37 (b) [H +] = 10−pH = 10−4 M 38 (c) By using pH=−log √Ka.C 5 = − log √Ka × 1 = − 1 2 log Ka log Ka = −10 Ka = 10−10 Hence, dissociation constant (Ka ) = 10−10 . 39 (b) A + B ⇌ C + D a a 2a 2a at equilibrium ∴ Kc = [C][D] [A][B] = 2a × 2a a × a = 4 40 (b) The compound which is having least solubility will be precipitated first. BaSO4 : Given, Ksp = 10−11 Let the solubility=x mol/L BaSO4 → Ba 2 + SO4 2− x x ∴ Ksp = [Ba 2+][SO4 2−] or Ksp = x × x ∴ Ksp = x 2 or x = √Ksp = √10−11 = 3.16 × 10−6 mol/L CaSO3 : Given, Ksp = 10−6 Let the solubility=x mol/L CaSO4 → [Ca 2+][SO4 2−] ∴ Ksp = [Ca 2+][So4 2−] or Ksp = x × x ∴ Ksp = x 2 or x = √Ksp = √10−6 = 1 × 10−3 mol/L AgSO4 : Given, Ksp = 10−5 Let the solubility=x mol/L Ag2SO4 → 2Ag + + SO4 2− 2x x ∴ Ksp = [Ag +] 2 [SO4 2−] or = (2x) 2 (x) or Ksp = 4x 3 or x = √Ksp 3 4 = √10−5 4 = 10−2 mol/L ∵ BaSO4 has least solubility. ∴ It will precipitated first. 41 (b) The substance which can donate a pair of electrons is called Lewis base. Amines contain lone pair of electron on nitrogen atom, so behave as Lewis base. 42 (a) pH=5.4 ∴ [H +] = 10−5.4 = 10−6 . 100.6 Antilog of 0.6 is ≈ 4 ∴ [H +] = 4 × 10−6M 43 (a) H2 Electric arc → 2000°C 2H − 104.5 kcal hydrogen atomic molecule hydrogen The reaction is endothermic. For endothermic reaction increase in temperature shift the equilibrium in forward direction. To proceed forward the pressure must be low because for the above reaction, increase of pressure will favoured backward reaction. So, for maximum yield the conditions are high temperature and low pressure. 44 (a) Lewis bases are electron pair donor. I + is electron deficient, hence do not act as Lewis base.
45 (c) Given, vol. of HCl = 5 mL, molarity =M/5 Vol. of NaOH = 10 mL, molarity =M/10 Mulliequivalents of HCl = 5 × 1 5 = 1 Mulliequivalents of NaOH = 10 × 1 10 = 1 HCl + NaOH → NaCl + H2O ∵ Mulliequivalents of HCl = Mulliequivalents of NaOH ∴ Solution is neutral and pH=7. 46 (d) Strongest Bronsted base is that which has weakest conjugate acid. Base Conjugate acid (base + H +) ClO− HClO ClO2 − HClO2 ClO3 − HClO3 ClO4 − HClO4 ∵ HClO is weak conjugate acid. ∵ ClO − is strongest Bronsted base. 47 (c) According to Le-Chatelier principle the reactions in which number of moles of reactants is equal to number of moles of products, is not effected by change in pressure. 2NO(g) ⇌ N2 (g) + O2(g) Moles of reactants =2 Moles of products = 2 ∵ There is no change in number of moles of reactants and products. ∵The reaction is not effected by change in pressure. 48 (d) AB is binary electrolyte, s = √Ksp = √1.21 × 10−6 = 1.1 × 10−3 M 49 (b) pH of the solution A = 3 [H +]A = 10−3 M pH of the solution B = 2 [H +]B = 10−2 M [H +] = 10−3 + 10−2 = 11 × 10−3 pH = − log(11 × 10−3) = 3 − log 11 = 3 − 1.04 = 1.9 50 (a) ∆H is positive, so it will shift toward the product by increasing temperature 51 (a) [H +] = Ka. C Given, [H +]HCOOH = [H +]CH3COOH Ka. C = Ka ′C′ ⇒ 1.8 × 10−4 × 0.001 = 1.8 × 10−5 × C′ ∴ C ′ = 0.01M 52 (b) In this reaction, ∆H is negative, so reaction moves forward by decrease in temperature while value of ∆ng = 2 − 3 = −1, ie, negative, so the reaction moves in forward direction by increase in pressure 53 (b) Kp = Kc (RT) ∆ng ∆ng = −1 (For the reaction, 2SO2 + O2 ⇌ 2SO3). Thus, for this reaction, Kp is less than Kc 54 (d) H2 + I2 ⇌ 2HI Initial 0.4 0.4 0 At equilibrium 0.4-0.25 0.4-0.25 0.05 =0.15 =0.15 Kc = [HI] 2 [H2 ][I2] = ( 0.50 2 ) 2 ( 0.15 2 )( 0.15 2 ) = 0.5×0.5 0.15×0.15 = 11.11 55 (c) The solubility of AgI in NaI solution is less than that in pure water because of common ion effect. 56 (b) pV = nRT Volume become 1 2 V then pressure become 2p, So, 2p × 1 2 V = pV = nRT Hence, there is no effect on Kp 57 (b) Given, [H2 ] = 8.0 mol/L [I2 ] = 3.0 mol/L [HI] = 28 mol/L K =? H2 + I2 ⇌ 2HI ∴ K = [HI] 2 [H2 ][I2 ] = (28) 2 (8)×(3) = 28 × 28 24 = 32.66 58 (c) Ksp of BaSO4 = 1.5 × 10−9 :[Ba 2+] = 0.01 M [SO4 2−] = 1.5 × 10−9 0.01 = 1.5 × 10−7 59 (a) Kp = Kc (RT) ∆n ∆n = sum of coefficients of gaseous products – that of gaseous reactants CO(g) + Cl2(g) → COCl2(g) ∴ ∆n = 1 − 2 = −1 ∴ Kp = Kc (RT) −1 ∴ Kp Kc = (RT) −1 = 1 (RT) 60 (b) NaOH = [OH −] = 10−3