Chapter – 6
Equilibrium
Question 1. A liquid is in equilibrium with its vapours in a sealed container at a fixed temperature. The volume of the container is suddenly increased,
(i) What is the initial effect of the change on the vapour pressure?
(ii) How do the rates of evaporation and condensation change initially?
(iii) What happens when equilibrium is restored finally and what will be the final vapour pressure?
Answer:
(i) On increasing the volume of the container, the vapour pressure will initially decrease because the same amount of vapours are now distributed over a larger space.
(ii) On increasing the volume of the container, the rate of evaporation will increase initially because now more space is available. Since the amount of the vapours per unit volume decrease on increasing the volume, therefore, the rate of condensation will decrease initially.
(iii) Finally, equilibrium will be restored when the rates of the forward and backward processes become equal. However, the vapour pressure will remain unchanged because it depends upon the temperature and not upon the volume of the container.Question 2. What is Kc for the following reaction in state of equilibrium
Question 2. In the case of the reaction H2(g) + I2(g) → 2HI (g), the standard free energy is given as ∆G > 0.The equilibrium constant (K ) will be ___.
(i) K = 0
(ii) K > 1
(iii) K = 1
(iv) K < 1
Answer
(iv) is the correct answer.
Question 3. Which of the following is not the general feature of equilibria involving physical processes?
(i) Equilibrium is possible only in the closed system at the given temperature.
(ii) All measurable properties of the given system remain constant.
(iii) All the physical processes stop at equilibrium.
(iv) The opposing process occurs at the same rate, and there is the dynamic, however stable condition.
Answer
(iii) is the correct answer.
Question 4. PCl5, PCl3, and Cl2 are at equilibrium at 500K in the closed container, and their concentrations are given as 0.8 × 10–3mol L–1, 1.2 × 10–3 mol L–1, and 1.2 × 10–3 mol L–1 respectively. The value of Kc for the given reactionPCl5 (g) PCl3 (g) + Cl2 (g) will be,
(i) 1.8 × 103mol L–1
(ii) 1.8 × 10–3
(iii) 1.8 × 10–3 L moL–1
(iv) 0.55 × 104
Answer
(ii) is the correct answer.
Question 5. If hydrochloric acid is added to the cobalt nitrate solution at room temperature, the following reaction will occur, and the reaction mixture will become blue. On cooling the mixture, it becomes pink. Based on the given information, mark the right answers.[Co (H2O)6]3+ (aq) + 4Cl–(aq) ⇌ [CoCl4]2– (aq) + 6H2O (l)(pink) (blue)
(i) ∆H > 0 for the reaction
(ii) ∆H < 0 for the reaction
(iii) ∆H = 0 for the reaction
(iv) The sign of ∆H can’t be predicted based on the given information.
Answer
(i) is the correct answer.
Question 6. The pH of the neutral water at 25°C is 7.0. When the temperature increases, the ionisation of water increases, but the concentration of H+ ions as well as OH– ions are the same. What would be the pH of pure water at 60°C?
(i) Equal to 7.0
(ii) Greater than 7.0
(iii) Less than 7.0
(iv) Equal to zero
Answer
(iii) is the correct answer.
Question 7. Ka, 2Ka, and 3Ka are the respective ionization constants for the following given reactions.H2S ⇌ H+ + HS–HS– ⇌ H+ + S2–H2S ⇌ 2H+ + S2–The correct relationship in between Ka1, Ka2 and Ka3 will be
(i) Ka3 = Ka1 × Ka2
(ii) Ka3 = Ka1 + Ka2
(iii) Ka3 = Ka1 – Ka2
(iv) Ka3 = Ka1 /Ka2
Answer
(i) is the correct answer.
Question 8. The acidity of the compound BF3 could be explained based on which among the following concepts?
(i) Arrhenius’s concept
(ii) Bronsted Lowry’s concept
(iii) Lewis’s concept
(iv) Bronsted Lowry as well as Lewis’s concept.
Answer
(iii) is the correct answer.
Question 9. Which among the following will produce the buffer solution if mixed in equal volumes?
(i) 0.1 mol dm–3 NH4 OH as well as 0.1 mol dm–3 HCl
(ii) 0.05 mol dm–3 NH4OH as well as 0.1 mol dm–3 HCl
(iii) 0.1 mol dm–3 NH4OH as well as 0.05 mol dm–3 HCl
(iv) 0.1 mol dm–3 CH4 COONa as well as 0.1 mol dm–3 NaOH
Answer
(iii) is the correct answer.
Question 10. Which among the following solvents in silver chloride is most soluble?
(i) 0.1 mol dm–3 AgNO3 solution
(ii) 0.1 mol dm–3 HCl solution
(iii) H2O
(iv) Aqueous ammonia
Answer
(iv) is the correct answer.
Question 11. The ionization for hydrochloric in the water has given below:HCl(aq) + H2O (l ) ⇌ H3O+(aq) + Cl–(aq)Label the two conjugate acid-base pairs in this ionization.
Answer :
The two required conjugate acid-base pairs in the ionization of HCl are (HCl–Cl–), where HCl is the conjugate acid as well as Cl– is the conjugate base. In the same way, the second pair is (H2O–, H3O+ ), where H2O is the conjugate base and H3O+ is the conjugate acid.
Question 12. The conjugate acid of the weak base is always stronger. What would be the decreasing order of the basic strength of the following conjugate bases?OH–, RO–, CH3COO–, Cl–
Answer :
The conjugate base of the strong acid is weak thud the decreasing order of the basic strength would be;RO– > OH– > CH3COO– > Cl–
Question 13. The aqueous solution of the sugar does not conduct electricity. But, when sodium chloride is added to water, it conducts electricity. How would you explain the statement based on the ionisation, as well as how is it affected by the concentration of the sodium chloride?
Answer ;
The aqueous solution of sugar does not conduct electricity as they exist as the molecule in the water. They don’t have free ions to conduct electricity; however, in the case of NaCl, free ions of Na+ as well as Cl– are present to conduct the electricity. Conductance depends on the no. of the ions present in the given solution. More will the given no. of ions of NaCl in water more will be the conductivity.
Question 14. The reaction between ammonia and boron trifluoride is given below as follows:NH3 + BF3 → H3 N : BF3 Identify the acid and base in the given reaction. Which theory gives an explanation to it? What is the hybridisation of B as w N in the given reactants?
Answer :
NH3 is the Lewis base, while BF3 is the Lewis acid. Lewis’s electronic theory of acids, as well as bases, explains it.The hybridisation state of the nitrogen in NH3 is sp3 hybridised, as well as Boron in BF3, which is sp 2 hybridised.
Question 15. The compound BF3 does not have a proton; however, it still acts as an acid and reacts with NH3. Why is it so? What type of bond can be formed between the two?
Answer :
According to the Lewis concept, e- deficient species are known as lewis acid. Thus BF3 will act as the lewis acid while NH3 (N=1S2 2S2 2p3) has the lone pair; hence it will act as the lewis base, and it will donate the lone pair to the empty p-orbital of the Boron through the coordinate bond to form an adduct.
Question 16. Based on the given equation pH = – log [H+], the pH of the given 10-8 mol dm-3 solution of HCl shall be 8. But, it is observed to be less than 7.0. Justify the reason.
Answer :
The solution is given very dilute, and we know that HCl reacts with water to give hydronium ions. The decrease in the pH could be observed as the result of the large concentration of H+. Hydronium ion concentration also needs to be considered here.Then, the total pH would be; [H3O+] = 10-8 + 10-7 M = 7Thus, the solution will be acidic.
Question 17. The ionisation constant of the weak base MOH is given by the expression;Kb = [M+][OH–]/[MOH–]Values of the ionisation constant for some weak bases at particular temperatures give below:Base: Di-methylamine, Urea, Pyridine, and AmmoniaKb: 5.4 × 10-4, 1.3 × 10-14, 1.77× 10-9, 1.77 × 10-5 Arrange the following bases in the decreasing order of the extent of their ionisation at equilibrium. Which among the above base is the strongest?
Answer :
The decreasing order of the bases based on the ionisation constant at equilibrium would be;-Di-methylamine > Ammonia > Pyridine > UreaThe strongest base would be Di-methyl amine because its pKb value is 3.29, and we already know that the less the pKb value, the strong is the base.
Question 18. Arrange the following compounds in increasing order for pHKNO3 (aq), CH3COONa (aq), NH4Cl (aq), C6H5COONH4 (aq)
Answer :
The increasing order of the pH would be;CH3COONa< KNO3< C6H5COONH4<. NH4ClCH3COONa is the salt of a weak acid (CH3COOH) and strong base (NaOH)KNO3 is the salt of strong acid (HNO3)-strong base (KOH)C6H5COONH4 is the salt of a weak acid (benzoic acid) and weak base (NH4OH) NH4Cl is the salt of a strong acid (HCl) and weak base (NH4OH)
Question 19. Conjugate acid of the weak base is always stronger. The decreasing order of the basic strength for the following conjugate bases will be?OH−, RO−, CH3COO−, CI−
Answer :
Conjugate acids of the given bases are H2O, ROH, CH3COOH, and HCl.Their acidic strength has the order.HCl > CH3COOH > H2O >ROH.Therefore, their conjugate bases would have strength in the order.RO− > OH− > CH3COO− > CI−
Question 20. The value of Kc for the given reaction 2HI (g) ⇋ H2 (g) + I2 (g) is 1 × 10-4At the given time, the composition of the reaction mixture is given as follows;[HI] =2 × 10-5 mol, [H2] =1 × 10-5 mol as well as [I2] =1 × 10-5 mol. In which direction would the reaction proceed?
Answer :
Given as; Kc = 1×10 -4 Kc = [H2][I2]/[HI]2Qc expresses the relative ratio of the products to reactants at the given instant.Qc = [H2][I2]/[HI]2= (1×10-5)(1×10-5)/(2×10-4)Qc = 1/4 = 0.25Where; Qc >Kc, the reaction would proceed in the reverse direction.
Question 21. When 0.561 g of KOH is dissolved in water to obtain 200 mL of solution at 298 K. Calculate the concentrations for the potassium, hydrogen and hydroxyl ions. What is the pH?
Answer .
[KOH(aq)] = 0.561 / (1/5)g/L= 2.805 g/L= 2.805 x (1/56.11)= 0.05MKOH(aq) → 10-13 (aq) + OH–(aq)[OH–] = 0.05M = [10-13] [H+][OH–] = Kw [H+] = Kw/[OH–] [H+] = 10-14/0.05 [H+] = 2 × 10-13 MpH = -log[H+]pH = -log[2 × 10-13 ]pH = 12.70
Question 22. The pH of the given 0.08 mol dm−3 HOCI solution is 2.85. Calculate the ionisation constant.
Answer :
pH of the given HOCl = 2.85However, – pH = log [H+]So, -2.85 = log [H+]-3.15 = log [H+] [H+] = 1.413 × 10−3 For the given weak monobasic acid [H+] = (Ka X C) 1 / 2Ka = [H+]*2 / CKa = (1.413 × 10−3)*2 / 0.08Ka = 24.957 × 10−6 Ka = 2.4957 × 10−5