CHEMICAL EQUILIBRIUM QUIZ-5

CHEMISTRY CHEMICAL EQUILBRIUM QUIZ-5

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. For a given solution pH=6.9 at 60℃, where K_w=10^-12 The solution is:

•  Acidic
•  Basic
•  Neutral
•  Unpredictable

Solution
b) If K_w=10^-12, then [H⁺ ] for neutral scale =10^-6 or pH=6; thus, pH 6.9 refers for alkaline nature.

Q2.For the reaction H₂ (g)+I₂ (g)⇌2HI(g), the equilibrium constants expressed in terms of concentrations K_c and in terms
of partial pressure K_p, are related as

•  K_p=K_c(RT)²
•  K_p=K_c(RT)^-2
•  K_p=K_c
•  K_c=K_p(RT)

Solution
c) For the reaction. H₂ (g)+I₂ (g)⇌2HI(g) K_p=K_c(RT)^∆n ∆n=2-2=0 so, K_p=K_c (where, K_p and K_c are equilibrium constants in terms of partial pressures and concentrations.)

Q3.  For the reactions, A+B+Q⇌C+D, if the temperature is increased then concentration of the products will

•  Increase
•   Decrease
•   Remains the same
•   Becomes zero

Solution
a) (A+B)+Q⇌(C+D) The reaction is endothermic so, on increase temperature concentration of product will increase

Q4. When a bottle of cold drink is opened, the gas comes out with a fizzle due to:

•  Decrease in pressure suddenly which results in a decrease in solubility of CO₂ gas in water
•  Decrease in temperature
•  Increase in pressure
•  None of the above

Solution
a) The solubility of gas decreases due to decrease in pressure

Q5.The pH of the solution containing 10 mL of a 0.1 N NaOH and 10 mL of 0.05 NH₂SO₄ would be:

•  Zero
•  1
•  >7
•  7

Solution
 c) Meq.of NaOH=10×0.1=1 Meq.of H₂SO₄=10×0.05=0.5 Thus, Meq.of alkali are left and therefore pH>7

Q6.Consider the reversible reaction, HCN(aq)⇌H⁺ (aq)+CN^- (aq) At equilibrium, the addition of CN^- (aq) would:

•  Reduce HCN(aq) concentration
•  Decrease the H⁺(aq) ion concentration
•  Increase the equilibrium constant
•  Decrease the equilibrium constant

Solution
(b) K_a=([H⁺] [CN^-])/([HCN]) ; An increase in [CN^-] will decrease [H⁺] to maintain K_a constant.

Q7.In hydrolysis of a salt of weak acid and strong base A^-+H₂ O⇌HA+OH^-, the hydrolysis constant (K_h) is equal to…

•  K_w/K_a
•  K_w/K_b
•  √(K_a/C)
•  K_w/(K_a×K_b )

Solution
a)

Q8.The Haber's process for the manufacture of ammonia is usually carried out at about 500℃. If a temperature of about
250℃ was used instead of 500℃:

•  Ammonia would not be formed at all
•  The percentage of ammonia in the equilibrium mixture would be too low
•  A catalyst would be of no use at all at this temperature
•  The rate of formation of ammonia would be too slow

Solution
d) The minimum temperature at which the combination of N2 and H2 occurs at measurable rate is 500C in Haber’s process. No doubt if temperature is raised up, above 500C, it will favour backward reaction, because the reaction is exothermic

Q9.Given that the equilibrium constant for the reaction
2SO₂(g))+O₂(g)⇌2SO₃(g) has a value of 278 at a particular temperature.
What is the value of the equilibrium constant for the following reaction at the same temperature?
SO₃(g)⇌SO₂(g)+1/2 O₂(g)

•  1.8 ×10^-3
•  3.6×10^-3
•  6.0×10^-2
•  1.3×10^-5

Solution
c) 2SO₂+O₂ ⇌2SO₃; K=278 SO₃ ⇌ SO₂+1/2 O₂; K^'=1/√K K^'=1/√K=1/√278=6 × 10^-2

Q10. The salt of strong acid and weak base (FeCl₂) is

•   Acidic
•   Basic
•   Neutral
•  None of these

Solution
a)