Equilibrium MCQs for NEET — Chemistry Questions with Answers

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Which one of the following pairs of solution is not an acidic buffer?

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Explanation

Strong acid with its salt cannot form buffer solution. Hence, HClO4 and NaClO4 is not an acidic buffer.

If the equilibrium constant for N2(g)+O2(g)2NO(g) is K, the equilibrium constant for (1/2)N2(g) + (1/2)O2(g)NO(g) will be,

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Which of the following salts will give highest pH in water?

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Explanation

The highest pH refers to the basic solution containing OH- ions. Therefore, the basic salt releasing OH- ions on hydrolysis will give highest pH in water.

Only the salt of strong base and weak acid would release OH- ion on hydrolysis. Among the given salts, Na2CO3 corresponds to the basic salt as it is formed by the neutralisation of NaOH [strong base] and H2CO3 [weak acid].

           CO32- + H2 HCO32- + OH-

For the reversible reaction,

N2(g) + 3H2(g) — 2NH3(g) + Heat

the equilibrium shifts in forward direction

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Explanation

Any change in the concentration, pressure and temperature of the reaction resulls in change in the direction of equilibrium. This change in the direction of equilibrium in governed by Le-Chatelier's principle. According to Le-Chatelier's principle, equilibrium shifts in the opposite direction to undo the change.

N2(g) + 3H2(g) — 2NH3(g)

(a) Increasing the concentration of NH3(g) On increasing (he concentration of NH3(g), the equilibrium shifts in the backward direction where concentration of NH3(g) decreases.

(b) Decreasing the pressure Since, pan (no. of moles), therefore, equilibrium shifts in the backward direction where number of moles are increasing.

(c) Decreasing the concentration of N2(g) and H2(g) Equilibrium shifts in the backward direction when concentration of N2(g) and H2(g) decreases.

(d) Increasing pressure and decreasing temperature On increasing pressure, equilibrium shifts in the forward direction where number of moles decreases while on decreasing temperature, it will move in forward direction where temperature increases.

 

For a given exothermic reaction, Kp and K'p are the equilibrium constants at temperatures T1 and T2 respectively. Assuming that heat of reaction is constant in a temperature range between T1 and T2, it is readily observed that

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Explanation

The equilibrium constant at two different temperatures for a thermodynamic process is given by

 log K2/K1H°2.303R[1/T1 -1/T2]

Here, K1 and K2 are replaced by Kp and K'p.

Therefore, logK'p/KpH°2.303R[1/T1 -1/T2]

For exothermic reaction, T2>T1 and H = -ve

... Kp>K'p

Which is the strongest acid in the following?

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Explanation

(c) The strength of oxyacids can also be decided with the help of the oxidation number central atom. Higher the oxidation number of central atom, more acidic is the oxyacid.

     +6       +5       +7        +4

   H2SO4, HClO3, HClO4, H2SO3

Since, in HClO4, oxidation number of Cl is highest, so HClO4 is the strongest acid among the given.

Which of these is least likely to act as a Lewis base?

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Explanation

(c) Electron rich species are called Lewis base. Among the given, BF3 is an electron deficient species, so have a capacity of electron accepting instead of donating that's why it is least likely act as a Lewis base. It is a Lewis acid.

Equimolar solutions of the following substances were prepared separately. Which one of these will record the highest pH value?

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Explanation

(a) BaCl2 is a salt of strong acid HCl and strong base Ba(OH)2. So, its aqueous solution is neutral with pH 7. All other salts give acidic solution due to cationic hydrolysis, so their pH is less than 7. Thus, pH value is highest for the solution of BaCl2

Which of the following is least likely to behave as Lewis base?

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Explanation

(b) BF3 is an electron deficient species, thus behaves like a Lewis acid.

A buffer solution is prepared in which the concentration of NH3 is 0.30 M and the concentration of NH4+ is 0.20 M. If the equilibrium constant, Kb for NH3 equals 1.8 x 10-5, what is the pH of the solution?

(log 2.7 = 0.43)

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Explanation

pOH = pKb + log[salt]/[Base]

       = -log Kb + log [salt]/[base]

       = -log 1.8 x 10-5 + log(0.20/0.30)

       = 5-0.25 +(-0.176)

       = 4.75-0.176 = 4.57

... pH = 14-4.57

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