CBSE JEE Main & Advanced Chemistry: Chemical Equilibrium Important Questions & Complete Chapter Notes

Correct Answer: (b)

Explanation: $\mathrm{N}_2 +3 \mathrm{H}_2 2 \rightleftharpoons \mathrm{NH}_3 \rightarrow K_C=64$ $2 \mathrm{NH}_3\rightleftharpoons \mathrm{N}_2 +3 \mathrm{H}_2 \rightarrow K_C=\frac{1}{64}$ $\mathrm{NH}_3 \rightleftharpoons \frac{1}{2}\mathrm{N}_2+\frac{3}{2}\mathrm{H}_2\rightarrow K_C$ $=\left(\frac{1}{64}\right)^{1/2}=\frac{1}{8}$

Correct Answer: (a)

Explanation: NH$_{4}$OH $\rightleftharpoons$ NH$_{4}^+$ + OH$^{-}$ K$_{b}$ = $\frac{[NH_{4}^+][OH^-]}{[NH_4OH]}$ 1.8 $\times$ 10$^{-5}$ = $\frac{[NH_{4}^+][0.01]}{[0.02]}$ $\therefore$ [NH$_{4}^+$] = 3.6 $\times$ 10$^{-5}$ M

Correct Answer: (a)

Explanation: 1 mol $\underset{1-x}{\mathrm{X_2Y(g)}}\;\rightarrow \;\underset{x}{\mathrm{X_2(g)}}+\frac{1}{2}\mathrm{Y_2(g)}$ $\therefore \;\; P_{x_2y} = \frac{1-x}{1+\frac{x}{2}} \times p$ $\; P_{x_2} = \frac{x}{1+\frac{x}{2}} \times p$ $P_{y_2} = \frac{\frac{x}{2}}{1+\frac{x}{2}} \times p$ $\therefore \; K_p = \frac{ \left( \frac{x}{1+\frac{x}{2}} p \right) \left( \frac{x}{2\left(1+\frac{x}{2}\right)} p \right)^{1/2} }{ \left( \frac{1-x}{1+\frac{x}{2}} \right) \times p }$ $\therefore \; K_p = \left( \frac{x}{1-x} \right) \left( \frac{x}{2\left(1+\frac{x}{2}\right)} \right)^{1/2} \times p^{1/2}$ $\therefore \;\; x$ to be very small $\therefore \; K_p = \frac{x^{\frac{3}{2}}}{2^{\frac{1}{2}}} \times p^{\frac{1}{2}}$ $\therefore \; x^{\frac{3}{2}} = \frac{K_p \times 2^{\frac{1}{2}}}{p^{\frac{1}{2}}}$ $x^3 = \frac{K_p^2 \times 2}{p}$ $x = \left( \frac{K_P^2 \times 2}{p} \right)^{\frac{1}{3}}$

Correct Answer: (c)

Explanation: **Given reaction:** $ W + X \rightleftharpoons Y + Z $ **Initial concentrations:** W starts at $2x$, X starts at $x$, and both Y and Z start at 0. **At equilibrium:** Let the amount of Y and Z formed be $y$. So, at equilibrium: W = $2x - y$ X = $x - y$ Y = $y$ Z = $y$ **Relationship between Y and X at equilibrium:** At equilibrium, the amount of Y is four times the amount of X. This gives us: $[Y]_{eq} = 4[X]_{eq}$ So, $ y = 4(x - y) $ Solve for $y$: $ y = 4x - 4y $ $ y + 4y = 4x $ $ 5y = 4x $ $ y = \frac{4}{5}x $ **Write $K_c$ expression and substitute values:** The equilibrium constant is: $ K_c = \frac{[Y][Z]}{[W][X]} $ Plug in the equilibrium values: $ K_c = \frac{\left(\frac{4}{5}x\right)\left(\frac{4}{5}x\right)}{(2x - \frac{4}{5}x)(x - \frac{4}{5}x)} $ **Simplify:** Numerator: $\left(\frac{4}{5}x\right)^2 = \frac{16}{25}x^2$ Denominator: $2x - \frac{4}{5}x = \frac{10x - 4x}{5} = \frac{6x}{5}$ $x - \frac{4}{5}x = \frac{5x - 4x}{5} = \frac{x}{5}$ So denominator: $\frac{6x}{5} \cdot \frac{x}{5} = \frac{6x^2}{25}$ Therefore, $ K_c = \frac{16x^2/25}{6x^2/25} = \frac{16}{6} = 2.666 $

Correct Answer: (c)

Explanation: $K_p = K_c (RT)^{\Delta n}$ Here $\Delta n = 1 - 2 = -1$ $\therefore \;\frac{k_p}{K_c} = \;\frac{1}{R T}$