.

**Q1.**When rain is accompanied by a thunderstorm the collected rain water will have a pH value:

Solution

c) Thunderstorm produces acidic oxides of N,S which produce acidic rain on dissolution in water.

c) Thunderstorm produces acidic oxides of N,S which produce acidic rain on dissolution in water.

**Q2.**What is the pH of a 1M CH

_{3}COONa solution? K

_{a}of acetic acid =1.8×10

^{-5},K

_{w}=10

^{-14}mol

^{2}litre

^{-2}

Solution

d) CH

d) CH

_{3}COO^{-}+H_{2}O⇌CH_{3}COOH+OH^{-}∴[OH^{-}]=c∙h=c√([K_{h}/c] ) =√([K_{w}/K_{a}∙c] ) =√((10^{-14}×1)/(1.8×10^{-5}))=2.35×10^{-5}∴ pOH=4.6289 ∴ pH=9.3710**Q3.**Ammonia under a pressure of 15 atm at 27℃ is heated to 347℃ in a closed vessel in the presence of catalyst. Under the

conditions, NH

_{3}is partially decomposed according to the equation, 2NH

_{3}⇌N

_{2}+3H

_{2}.

The vessel is such that the volume remains effectively constant whereas pressure increases to 50 atm. Calculate the

percentage of NH

_{3}actually decomposed

Solution

a) 2NH

a) 2NH

_{3}⇌N_{2}+3H_{2}Initial moles a 0 0 At equilibrium(a-2x) x 3x Initial pressure of NH_{3}of ‘a’mole = 15 atm at 27℃. The pressure of ’a’ mole of NH_{3}=p atm at 347℃ ∴15/300=p/620 ∴p=31 atm At constant volume and at 347℃, mole ∝ pressure a∝31 (before equilibrium) ∴(a-2x)∝50 (after equilibrium) ∴ ((a-2x))/a=50/31 ∴x=19/62 a ∴% of NH_{3}decomposed =2x/a×100 =(2×19a)/(62×a)×100 =61.33%

**Q4.**In the equilibrium,2SO

_{2}(g)+O

_{2}(g) ⇌2SO

_{3}(g), the partial pressure of SO

_{2},O

_{2}and SO

_{3}are 0.662, 0.101 and 0.331 atm respectively. What should be the partial pressure of oxygen so that the equilibrium concentration of SO

_{2}and SO

_{3}are equal?

Solution

a) K

a) K

_{p}=(p_{SO3})^{2/((p(SO2 ) )2 (p(O2 ) ) )=(0.331)2/((0.662)2 (0.101))=2.5 Now, Kp=(p(SO3 ) )2/((pSO2 )2 PO2 ); If pSO3= p(SO2 ) Then, p(O2 )=1/Kp =1/2.5=0.4 atm }**Q5.**The exothermic formation of ClF

_{3}is represented by the equation

Cl

_{2}(g)+3F

_{2}(g)⇌2ClF

_{3}(g); ∆H=-329 kJ

Which of the following will increase the quantity of ClF

_{3}in an equilibrium mixture of Cl

_{2},F

_{2}and ClF

_{3}?

Solution

a) Reaction is exothermic. By Le-Chatelier’s principle, a reaction is spontaneous in forward side (in the direction of formation of more ClF_3) if F_2 is added, temperature is lowered and ClF_3 is removed.

a) Reaction is exothermic. By Le-Chatelier’s principle, a reaction is spontaneous in forward side (in the direction of formation of more ClF_3) if F_2 is added, temperature is lowered and ClF_3 is removed.

**Q6.**In a lime kiln, to get higher yield of CO

_{2}, the measure that can be taken is

Solution

d) CaCO(s)⇌CaO(s)+CO

d) CaCO(s)⇌CaO(s)+CO

_{2}(g) The equilibrium constant for this reaction is given by K=[CO_{2}](as CaCO_{3}and CaO are solid). Hence, to get more CO_{2}, we need to pump out continuously the CO_{2}gas.**Q7.**The dissociation equilibrium of a gas AB

_{2}can be represented as:

2AB

_{2}(g)⇌2AB(g)+B

_{2}(g)

The degree of dissociation is 'x' and is small compared to 1. The expression

relating the degree of dissociation (x) with equilibrium constant K

_{p}and total pressure p is:

Solution

a)

a)

**Q8.**The solubility of AgCl in water at 10℃ is 6.2×10

^{-6}mol/litre. The K

_{sp}of AgCl is:

Solution

d) K

d) K

_{sp}for AgCl=s^{2}.**Q9.**The correct relationship between Kc and Kp in gaseous equilibrium is :

Solution

b) K

b) K

_{p}=K_{c}x (RT)^{∆n}. Where ∆n= mole of products – mole of reactants**Q10.**In a chemical equilibrium, the rate constant of the backward reaction is 7.5×10

^{-4}and the equilibrium constant is 1.5. So, the rate constant of the forward reaction is

Solution

a)K

a)K

_{c}=k_{f}/k_{b}∴k_{f}=K_{c}×k_{b}=1.5×7.5×10^{-4}=1.125×10^{-3}