CHEMICAL EQUILIBRIUM QUIZ-6

CHEMISTRY CHEMICAL EQUILBRIUM QUIZ-6

Dear Readers,

The one subject in NEET which is candidates who can easily attain good marks is Chemistry. That's the reason, often, one doesn’t pay notice and choose to compromise it. But if one wants to rank above others, the tip is to be thorough with NEET chemistry concepts. The understanding of reactions and definite basic understanding is what requires major attention in Chemistry but once done it only gets simpler from there. The main focus on the to-do list should be on getting a hang of the NCERT syllabus of NEET chemistry.

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

•  Uninfluenced by occurrence of thunderstorm
•  Depending on the amount of dust in air
•  Slightly lower than that of rain water without thunderstorm
•  Slightly higher than that when the thunderstorm is not there

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

Q2.What is the pH of a 1M CH₃COONa solution? K_a of acetic acid =1.8×10^-5,K_w=10^-14 mol² litre^-2

•  2.4
•  3.6
•   4.8
•  9.4

Solution
d) CH₃ COO^-+H₂ O⇌CH₃ 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, NH₃ is partially decomposed according to the equation, 2NH₃⇌N₂+3H₂.
The vessel is such that the volume remains effectively constant whereas pressure increases to 50 atm. Calculate the
percentage of NH₃ actually decomposed

•  61.3%
•  63.5%
•  65.3%
•  66.6%

Solution
a) 2NH₃⇌N₂+3H₂ Initial moles a 0 0 At equilibrium(a-2x) x 3x Initial pressure of NH₃ of ‘a’mole = 15 atm at 27℃. The pressure of ’a’ mole of NH₃=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 NH₃ decomposed =2x/a×100 =(2×19a)/(62×a)×100 =61.33%

Q4. In the equilibrium,2SO₂ (g)+O₂ (g) ⇌2SO₃ (g), the partial pressure of SO₂,O₂ and SO₃ are 0.662, 0.101 and 0.331 atm respectively. What should be the partial pressure of oxygen so that the equilibrium concentration of SO₂ and SO₃ are equal?

•  0.4 atm
•  1.0 atm
•  0.8 atm
•  0.25 atm

Solution
a) K_p=(p_SO3 )^{2/((p_{(SO2 )} )2 (p_{(O2 )} ) )=(0.331)2/((0.662)2 (0.101))=2.5 Now, K_p=(p_{(SO3 )} )2/((p_SO2 )2 P_O2 );
If p_SO3= p_{(SO2 )}
Then, p_{(O2 )=1/Kp =1/2.5=0.4 atm}}

Q5.The exothermic formation of ClF₃ is represented by the equation
Cl₂ (g)+3F₂ (g)⇌2ClF₃ (g); ∆H=-329 kJ
Which of the following will increase the quantity of ClF₃ in an equilibrium mixture of Cl₂,F₂ and ClF₃?

•  Adding F₂
•  Increasing the volume of the container
•  Removing Cl₂
•  Increasing the temperature

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.

Q6.In a lime kiln, to get higher yield of CO₂, the measure that can be taken is

•  To remove CaO
•  To add more CaCO₃
•  To maintain high temperature
•  To pump out CO₂

Solution
d) CaCO(s)⇌CaO(s)+CO₂ (g) The equilibrium constant for this reaction is given by K=[CO₂ ](as CaCO₃ and CaO are solid). Hence, to get more CO₂, we need to pump out continuously the CO₂ gas.

Q7.The dissociation equilibrium of a gas AB₂ can be represented as:
2AB₂ (g)⇌2AB(g)+B₂ (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:

•  (2K_p/ P )^1/3
•   (2K_p/ P )^1/2
•  K_p/P
•  2K_p/P

Solution
a)

Q8.The solubility of AgCl in water at 10℃ is 6.2×10^-6 mol/litre. The K_sp of AgCl is:

•   [6.2×10^-6]^1/2
•  6.2×(10^-6)²
•  (6.2)²×10^-6
•  [6.2×10^-6]²

Solution
d) K_sp for AgCl=s².

Q9.The correct relationship between Kc and Kp in gaseous equilibrium is :

•  Kc= Kp(RT)^∆n
•   Kp=Kc(RT)^∆n
•  K_c/RT = (Kp)^∆n
•  d) K_P/RT=(Kc)^∆n

Solution
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

•   1.125×10^-3
•   2.225×10^-3
•   3.335×10^-5
•  1.125×10^-1

Solution
a)K_c=k_f/k_b ∴k_f=K_c×k_b=1.5×7.5×10^-4=1.125×10^-3

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