Chemistry MCQs for NEET — Practice Questions with Answers

For a reaction,

$$\begin{gathered} \underset{\mathrm{Reactant}}{\mathrm{A}}\rightleftharpoons \underset{\mathrm{Product}}{\mathrm{B}} \\ \mathrm{K}=\frac{{\left[\mathrm{B}\right]}_{\mathrm{eq}}}{{\left[\mathrm{A}\right]}_{\mathrm{eq}}}\Rightarrow 1.6\times {10}^{12}=\frac{{\left[\mathrm{B}\right]}_{\mathrm{eq}}}{{\left[\mathrm{A}\right]}_{\mathrm{eq}}} \\ {\left[\mathrm{B}\right]}_{\mathrm{eq}}>>{\left[\mathrm{A}\right]}_{\mathrm{eq}} \end{gathered}$$

So, mostly the product will be present in te equilibrium mixture.

Mathematical expression of the thermodynamic equilibrium is

ΔG = ΔG° + 2.303 RT logQ

At equilibrium when ΔG = 0 and O = K

then ΔG = ΔG° + 2.303 RT logK = 0

ΔG° = - 2.303 RT log K

(a) Key Concept When equal volumes of acid and base are mixed, then resulting solution become alkaline if concentration of base is taken high.

Let normality of the solution after mixing 0.1 M NaOH and 0.01 M HCl is N.

$$\begin{gathered} \therefore \mathrm{N}1\mathrm{V}1-\mathrm{N}2\mathrm{V}2 = \mathrm{NV} \\ \mathrm{or} 0.1 \times 1 - 0.01 \times 1=\mathrm{N}\times 2 \end{gathered}$$

Since, normality of NaOH is more than that of HCl. Hence, the resulting solution is alkaline.

$$\begin{gathered} \mathrm{or} \left[\mathrm{OH}\right]=\mathrm{N}=\frac{0.09}{2}=0.045\mathrm{N} \\ \mathrm{or} \mathrm{pOH} =-\log (0.045)=1.35 \\ \therefore \mathrm{pH} =14-\mathrm{pOH} \\ =14-1.35=12.65 \end{gathered}$$

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

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].

           ${\mathrm{CO}}_3^{2-}$ + H₂O $\rightleftharpoons {\mathrm{HCO}}_3^{2-} + {\mathrm{OH}}^{-}$

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.

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

 log K₂/K₁ = $\frac{∆\mathrm{H}°}{2.303\mathrm{R}}[1/{\mathrm{T}}_1 -1/{\mathrm{T}}_2]$

Here, K₁ and K₂ are replaced by K_p and K'_p.

Therefore, logK'_p/K_p = $\frac{∆\mathrm{H}°}{2.303\mathrm{R}}[1/{\mathrm{T}}_1 -1/{\mathrm{T}}_2]$

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

.^.. Kp>K'p

(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.