Class 11th Chemistry - Equilibrium Case Study Questions and Answers 2022 - 2023
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Equilibrium Case Study Questions With Answer Key
11th Standard CBSE
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Reg.No. :
Chemistry
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Equilibrium can be established for both physical and chemical processes and at equilibrium rate of forward and reverse processes are equal. Equilibrium constant (Kc) is expressed as the concentration of products divided by reactants, each raised to Stoichiometric Coefficient. Kc has fixed value of constant temperature and at this stage concentration, pressure, etc. become constant. Kp is equilibrium constant in terms of partial pressure of gases or vapours. The direction of reaction can be predicted by reaction quotient Qc and Qp Qc = Kc at equilibrium and Qp = Kp Le Chatelier's principle states 'Equilibrium' will shift in the direction so as to counter balance the effect of change in T, P and conc. The equilibrium between ions and unionised molecules in weak electrolyte is called ionic equilibrium.
(a) At a certain temperature Kc = \(\frac{9}{4}\) for the reaction:
CO(g) + H2O(g) \(\rightleftharpoons\) CO2 (g) + H2 (g)
If we take 10 moles of each of four gases in one litre container, what will be concentration of H2 at equilibrium?
(b) In the equilibrium CO(g) + H2O(g) \(\rightleftharpoons\) CO2 (g) + H2 (g) if at a given temperature, the concentration of the reactants are increased, what will be the value of equilibrium constant Kc.
(c) If the value of an equilibrium constant for a particular reaction is 1.6 x 1012, what will be present in the system at equilibrium.
(d) The Ksp of AgCl, AgBr, AgI are 1.8 x 10-10 , 5.0 x 10-13 , 8.3 x 10-17 , which will precipitate last if AgNO3 solution is added to NaCI, NaBr. NaI?
(e) The equilibrium constant for reaction N2 (g) + 3H2 (g) \(\stackrel{K_{1}}{\rightleftharpoons}\) 2NH3(g) is 49, what is equilibrium constantfor NH3(g) \(\stackrel{\mathbf{K}_{2}}{\rightleftharpoons} \frac{1}{2} \) N2(g) + \(\frac{3}{2}\) H2(g)?(a) -
Observe the table of the ionisation constants of some common polyprotic acid at 298 K. Answer the questions based on this table and related studied concepts.
The Ionisatlon Constants of Some Common Polyprotic Acids (298K)Acid Ka1 Ka2 Ka3 Oxalic acid 5.9 x 10-2 6.4 x 10-5 Ascorbic acid 7.4 x 10-4 1.6 x 10-12 Sulphurous acid 1.7 x 10-2 6.4 x 10-11 Sulphuric acid Very large 1.2 x 10-2 Carbonic acid 4.3 x 10-7 5.6 x 10-11 Citric acid 7.4 x 10-4 1.7 x 10-5 4.0 x 10-7 Phosphoric acid 7.5 x 10-3 6.2 x 10-11 4.2 x 10-13 (a) Why is Ka1 greater than Ka2?
(b) Arrange Ka1, Ka2 and Ka3 in phosphoric acid.
(c) Why is Ka1 >>> Ka2 in sulphuric acid?
(d) What is relationship between Ka1, K,a2 and Ka3?
(e) Write expression for Ka1 and Ka2 and Ka of H2CO3
(f) What is basicity of H3 PO4?
(g) Out of oxalic acid and citric acid, which is stronger?(a) -
Arrhenius acids give W ion in aqueous solution where as bases give OIr in aqueous solution. Bronsted acids are proton donor where as Bronsted bases are proton acceptors. Acids, on donating proton form conjugate base where as bases form conjugate acid after accepting proton. Buffer solution is a solution whose pH does not change by adding small amount of H+ or OH-. The decrease in conc. of one of the ion by adding other ions as common ion is called common ion effect. Lewis acids are electron deficient or +vely charged. Lewis bases are electron rich or negatively charged, Ksp (solubility product) is the product of molar concentration of ions raised to power number of ions per formula of the compound in sparingly soluble salt. Precipitation occurs only if ionic product exceeds solubility product. Solubility of salt decreases in presence of common ion. KW, is ionic product of water, 1 x 10-14 at 298 K. KW, increases with increase in temperature. pH is -log [H3O+] where [H3O+] = C\(\alpha\) in monoprotic acid 'C' is molar cone.,'\(\alpha\)' is degree of ionisation. A salt is said to be hydrolysed if pH of solution changes. KH is hydrolytic constant. pH of salts of strong acid and strong base is equal to 7. pH of other salts can be 7. pH of buffer solution can be calculated with the help of Henderson equation.
pH = pKa + \(\log \frac{[\text { Salt }]}{[\text { Acid }]}\)
(a) What win be conjugate base of
(i) H2 SO4
(ii) H\(\mathbf{C}{O}_{3}^{-}\)\(\)?
(b) What are conjugate acids of
(i) \(\mathbf{N H}_{2}^{-}\)
(ii) NH3
(c) The conc. of [H3O+] is 4 x 10-4. What is pH of solution? [log 4 = 0.6021], log 10 = 1.
(d) Kb for NH3 is 1.80 x 10-5 .What will be Ka? [KW = 1 x 10-14]
(e) Ksp of BaSO4 is 1.0 x 10-10. What is its solubility?
(f) ) Calculate pH of buffer solution containing 0.01 M CH3COOH and 0.1 M CH3COONa,pKa = 4.75.
(g) ThepKa of CH3COOH and pKb of NH4 OH are 4.76 and 4.75 respectively. Calculate pH of CH3COONH4 solution. (PKw = 14)(a)
Case Study
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Answers
Equilibrium Case Study Questions With Answer Key Answer Keys
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(a) CO(g) + H2O(g) \(\rightleftharpoons\) CO2 (g) + H2 (g)
Initial cone 10 10 10 10 Final cone 10 - x 10 - x 10+x 10+x \(\mathrm{K}_{c}=\frac{\left[\mathrm{CO}_{2}\right]\left[\mathrm{H}_{2}\right]}{[\mathrm{CO}]\left[\mathrm{H}_{2} \mathrm{O}\right]}=\frac{(10+x)(10+x)}{(10-x)(10-x)}\)
\(\Rightarrow \quad \frac{9}{4}=\frac{(10+x)^{2}}{(10-x)^{2}}\)
\(\Rightarrow \quad \frac{10+x}{10-x}=\frac{3}{2}\) (By taking square root both sides)
\(\Rightarrow\) 20 + 2x = 30 - 3x
5x = 10
\(\Rightarrow\) x=2
Cone, of H2 at equilibrium = 10 + 2 = 12 mol L-1
(b) Kc will remain the same. It changes only with change in temperature.
(c) The system will contain mostly products since value of Kc is high.
(d) AgI will precipitate because its Ksp is low, ionic product will exceed Ksp.
(e) \(\frac{1}{7} \quad \because \mathrm{K}_{2}=\frac{1}{\sqrt{\mathrm{K}_{1}}}\) -
(a) It is because degree of ionisation is more in first step than in second step and it is difficult to take out proton (H+) from its negatively charged ions.
(b) Ka1 > Ka2 > Ka3
(c) It is because H2 SO4 is strong acid and almost completely ionised in 1st step.
(d) Ka = Ka1 x Ka2
(e) H2CO3(aq) \(\rightleftharpoons\) H+ + \(\mathrm{HCO}_{3}^{-}\) \(\mathrm{K}_{a_{1}}=\frac{\left[\mathrm{H}^{+}\right]\left[\mathrm{HCO}_{3}^{\top}\right]}{\left[\mathrm{H}_{2} \mathrm{CO}_{3}\right]}\)
\(\mathrm{HCO}_{3}^{-}(a q) \rightleftharpoons \mathrm{H}^{+}+\mathrm{CO}_{3}^{2-}\) \(\mathrm{K}_{a_{2}}=\frac{\left[\mathrm{H}^{+}\right]\left[\mathrm{CO}_{3}^{2-}\right]}{\left[\mathrm{HCO}_{3}^{-}\right]}\)
HCO3(aq) \(\rightleftharpoons\) 2H+ + \(\mathrm{CO}_{3}^{2-}\) \(\mathrm{K}_{a}=\frac{\left[\mathrm{H}^{+}\right]^{2}\left[\mathrm{CO}_{3}^{2-}\right]}{\left[\mathrm{H}_{2} \mathrm{CO}_{3}\right]}\)
(f) It is tribasic acid.
(g) Oxalic acid is stronger than citric acid because (Ka = Ka1 x Ka2 ) is higher in oxalic acid than citric acid. -
(a) (i) \(\mathrm{HSO}_{4}^{-}\)
(ii) \(\mathrm{CO}_{3}^{2-}\)
(b) (i) NH3
(ii) \(\mathrm{NH}_{4}^{+}\)
(c) pH = -log[H3P+]
= - log 4 x 10-4
= - log 4 + 4.000
= - 0.6021 + 4.000
= 3.3979
(d) Ka x Kb = Kw
\(\Rightarrow \quad K_{a}=\frac{1 \times 10^{-14}}{1.80 \times 10^{-5}}\) = 0.555 x 10-9 = 5.55 x 10-10
(e) \(\mathrm{BaSO}_{4}(s) \rightleftharpoons \mathrm{Ba}^{2+}(a q)+\mathrm{SO}_{4}^{2-}\)
's' 's'
Ksp = [Ba2+][SO\({ }_{4}^{2-}\)] = s x s
\(\Rightarrow\) s2 = 1 x 10-10
\(\Rightarrow\) S = 1 x 10-5 mol L-1
(f) \(\mathrm{pH}=p \mathrm{~K} a+\log \frac{[\mathrm{Salt}]}{[\text { Acid }]}\)
\(\mathrm{pH}=4.75+\log \frac{0.1}{0.01}\)
= 4.75 + log 10= 4.75 -+ 1.00 = 5.75.
(g) \(\mathrm{pH}=\frac{1}{2}\) (Pw + pKa - pKb)
\(\mathrm{pH}=\frac{1}{2}\) (14 + 4.76 - 4.75)
= 7 + 0.005 = 7.005.
Case Study
