Case Study Questions for Equilibrium

 1. Read the passage given below and answer the following questions.

Reactants and products coexist at equilibrium, so that the conversion of reactant to products is always less than 100%. Equilibrium reaction may involve the decomposition of a covalent (nonpolar) reactant or ionization of ionic compound into their ions in polar solvents. Ostwald dilution law is the application of the law of mass action to the weak electrolytes in solution.
A binary electrolyte AB which dissociates into A+ and B– ions i.e.

for every weak electrolyte, Since α <<1 (1 – α) = 1

(i) A monobasic weak acid solution has a molarity of 0.005 M and pH of 5. What is its percentage ionization in this solution?
(a) 2.0
(b) 0.2
(c) 0.5
(d) 0.25

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(ii) Calculate ionisation constant for pyridinium hydrogen chloride. (Given that H+ ion concentration is 3.6 × 10–4 M and its concentration is 0.02 M.)

(a) 6.48 × 10–2
(b) 6 × 10–6
(c) 1.5 × 10–9
(d) 12 × 10–8

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(iii) The hydrogen ion concentration of a 10–8 M HCl aqueous solution at 298 K (Kw = 10–14) is
(a) 9.525 × 10–8 M
(b) 1.0 × 10–8 M
(c) 1.0 × 10–6 M
(d) 1.0525 × 10–7 M

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(iv) Ostwald dilution law is applicable to
(a) weak electrolytes
(b) non-electrolyte
(c) strong electrolyte
(d) all type of electrolyte.

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(v) If a is the fraction of HI dissociated at equilibrium in the reaction: 2HI ⇔ H2 + I2
then starting with 2 mol of HI, the total number of moles of reactants and products at equilibrium are
(a) 1
(b) 2
(c) 1 + α
(d) 2 + 2α

2. Read the paragraph and choose correct answer of following questions

 Predicting the Direction of the Reaction- The equilibrium constant helps in predicting the direction in which a given reaction will proceed at any stage. For this purpose, we calculate the reaction quotient Q. The reaction quotient, Q (Qc  with molar concentrations and QP with partial pressures) is defined in the same way as the equilibrium constant Kc  except that the concentrations in Qc  are not necessarily equilibrium values. For a general reaction:

a A + b B  ⇌ c C + d D

Qc  = [C]c [D]d  / [A]a [B]b

Relationship between equilibrium constant K, reaction quotient Q and gibbs energy G The value of Kc  for a reaction does not depend on the rate of the reaction. However, it is directly related to the thermodynamics of the reaction and in particular, to the change in Gibbs energy, ∆G.

A mathematical expression of this thermodynamic view of equilibrium can be described by the following equation:

∆G = ∆Gø  + RT lnQ

where, Gø  is standard Gibbs energy. At equilibrium, when ∆G = 0 and Q = Kc , the equation  becomes,

∆G = Gø  + RT lnK = 0

∆Gø = – RT lnK

lnK = – ∆Gø  / RT

Taking antilog of both sides, we get,

 K = e–∆G0/RT

i) If … the reaction will proceed in the direction of reactants (reverse reaction).

a) Qc > Kc        b) Qc < Kc

c) Qc = Kc         d) None of above

ii) If … the reaction will proceed in the direction of the products (forward reaction).

a) Qc > Kc             b) Qc < Kc             c) Qc = Kc                    d) None of above

iii) If … the reaction mixture is already at equilibrium. Consider the gaseous reaction.

a) Qc > Kc              b) Qc < Kc               c) Qc = Kc         d) All of above

iv) If ∆G is …. then the reaction is spontaneous and proceeds in the forward direction.

a) zero             b) positive                      c) negative                 d) None of above

v) ∆G is … reaction has achieved equilibrium; at this point, there is no longer any free energy left to drive the reaction.

a) zero              b) positive         c) negative      d) None of above