Equilibrium

Le Chatilier's Principle

Q30 At 473K, equilibrium constant, Kc, for the decomposition of phosphorous pentachloride PCl5, is 8.3x10-3. If decomposition is depicted as: 

PCl5(g) ⇌  PCl3(g) + Cl2(g)    ΔrH0=124.0kJ/mol

a. write an expression for Kc for the reaction?

b. What is the value of Kc for the reverse reaction at the same temperature?

c. What would be the effect on Kc if (i) more PCl5 is added (ii) pressure is increased (iii) temperature is increased?

Ans. 

a. Kc ⇌  [PCl3(g)] [Cl2(g)]/[PCl5(g)]

b. K' ⇌  1/Kc ⇌  1/8.3x10-3 ⇌  120.48

c. (i) No effect as Kc is constant at constant temperature 

(ii) No effect

(iii) As given reaction is endothermic, on increasing the temperature, kf will increase. As Kc = kf/kb, Kc will increase with increase of temperature.

If Qc < Kc , net reaction goes from left to right 

• If Qc > Kc , net reaction goes from right to left.

• If Qc = Kc , no net reaction occurs. 

Problem 7.15 Classify the following species into Lewis acids and Lewis bases and show how these act as such: 

(a) HO – 

(b)F – 

(c) H+ 

(d) BCl3 

Solution (a) Hydroxyl ion is a Lewis base as it can donate an electron lone pair (:OH – ). 

(b) Flouride ion acts as a Lewis base as it can donate any one of its four electron lone pairs. 

(c) A proton is a Lewis acid as it can accept a lone pair of electrons from bases like hydroxyl ion and fluoride ion. 

(d) BCl3 acts as a Lewis acid as it can accept a lone pair of electrons from species like ammonia or amine molecules 

Problem 7.25 The pKa of acetic acid and pKb of ammonium hydroxide are 4.76 and 4.75 respectively. Calculate the pH of ammonium acetate solution. 

Solution pH = 7 + ½ [pKa – pKb ] = 7 + ½ [4.76 – 4.75] = 7 + ½ [0.01] = 7 + 0.005 = 7.005 

Problem 7.16 The concentration of hydrogen ion in a sample of soft drink is 3.8 × 10-3 M. what is its pH ? 

Solution pH = – log[3.8 × 10-3] = – {log[3.8] + log[10-3]} = – {(0.58) + (– 3.0)} = – { – 2.42} = 2.42

Therefore, the pH of the soft drink is 2.42 and it can be inferred that it is acidic. 

Q42 The pH of a sample of vinegar is 3.76, calculate the concentration of hydrogen ion in it.

Ans. pH = -log [H+] or log [H+] = -pH = -3.76 = 4.24

[H+] = Antilog 4.24 = 1.738 x 10-4 M = 1.74 x 10-4 M

Q43 The ionisation constants of HF, HCOOH and HCN at 298 K are is 6.8x10-4, 1.8x10-4 and 4.8x10-9 respectively. Calculate the ionisation constant of the corresponding conjugate base.

Ans. for F- , Kb = Kw/Ka = 10-14/ (6.8x10-4) = 1.47x10-11

for HCOO- Kb=10-14/(1.8x10-4) = 5.6 x 10-11

for CN- , Kb = 10-14/(4.8x10-9)  = 2.08 x 10-6