### Related Formula P = fracnRTV quad textand quad K_p = frac(P_N_2O_4)^2 cdot P_O_2(P_N_2O_5)^2$P = \frac{nRT}{V} \quad \text{and} \quad K_p = \frac{(P_{N_2O_4})^2 \cdot P_{O_2}}{(P_{N_2O_5})^2}$ ### Core Logic First, find the initial moles of N_2O_5$N_2O_5$ using its molar mass (108text g/mol$108\text{ g/mol}$): n_0 = frac37.8108 = 0.35text moles$n_0 = \frac{37.8}{108} = 0.35\text{ moles}$ Using the ideal gas equation, compute the initial pressure (P_i$P_i$): P_i = frac0.35 times 0.082 times 5001 = 14.35text bar$P_i = \frac{0.35 \times 0.082 \times 500}{1} = 14.35\text{ \bar}$ Setting up the equilibrium partial pressures table: beginarraylcccc & 2N_2O_5(g) & rightleftharpoons & 2N_2O_4(g) & + & O_2(g) \\ textInitially: & 14.35 & & 0 & & 0 \\ textAt equilibrium: & 14.35 - 2P & & 2P & & P endarray$\begin{array}{lcccc} & 2N_2O_{5(g)} & \rightleftharpoons & 2N_2O_{4(g)} & + & O_{2(g)} \\ \text{Initially: } & 14.35 & & 0 & & 0 \\ \text{At equilibrium: } & 14.35 - 2P & & 2P & & P \end{array}$ The total pressure at equilibrium is given as: P_texttotal = (14.35 - 2P) + 2P + P = 14.35 + P = 18.65text bar$P_{\text{total}} = (14.35 - 2P) + 2P + P = 14.35 + P = 18.65\text{ \bar}$ P = 18.65 - 14.35 = 4.3text bar$P = 18.65 - 14.35 = 4.3\text{ \bar}$ Now, calculate the equilibrium partial pressures for each component: - P_N_2O_5 = 14.35 - 2(4.3) = 5.75text bar$P_{N_2O_5} = 14.35 - 2(4.3) = 5.75\text{ \bar}$ - P_N_2O_4 = 2(4.3) = 8.6text bar$P_{N_2O_4} = 2(4.3) = 8.6\text{ \bar}$ - P_O_2 = 4.3text bar$P_{O_2} = 4.3\text{ \bar}$ Substitute these partial pressures into the K_p$K_p$ expression: K_p = frac(8.6)^2 times 4.3(5.75)^2 = frac73.96 times 4.333.0625 approx 9.619$K_p = \frac{(8.6)^2 \times 4.3}{(5.75)^2} = \frac{73.96 \times 4.3}{33.0625} \approx 9.619$ Expressing the result in the requested format (x times 10^-2$x \times 10^{-2}$): K_p = 961.9 times 10^-2$K_p = 961.9 \times 10^{-2}$ Rounding to the nearest integer yields **962**. ### Pattern Recognition Always calculate the initial pressure first using the ideal gas law (PV=nRT$PV=nRT$). This provides a clear baseline for tracking equilibrium partial pressures. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Equilibrium

### Related Formula The solubility product constant (K_sp$K_{sp}$) reflects the equilibrium position of a sparingly soluble salt in water. ### Core Logic Based on standard literature K_sp$K_{sp}$ values at 298text K$298\text{ K}$: - HgS$HgS$: approx 4 times 10^-53$\approx 4 \times 10^{-53}$ (extremely insoluble, Group IIB cation analysis) - PbS$PbS$: approx 8 times 10^-28$\approx 8 \times 10^{-28}$ (highly insoluble, Group IIA cation analysis) - AgBr$AgBr$: approx 5 times 10^-13$\approx 5 \times 10^{-13}$ (sparingly soluble halide salt) - Ca(OH)_2$Ca(OH)_2$: approx 5.5 times 10^-6$\approx 5.5 \times 10^{-6}$ (moderately soluble base) ### Step 1: Arrangement Comparing these K_sp$K_{sp}$ orders: 4 times 10^-53 < 8 times 10^-28 < 5 times 10^-13 < 5.5 times 10^-6$4 \times 10^{-53} < 8 \times 10^{-28} < 5 \times 10^{-13} < 5.5 \times 10^{-6}$ Hence, the correct increasing sequence is: HgS < PbS < AgBr < Ca(OH)_2$HgS < PbS < AgBr < Ca(OH)_2$. ### Pattern Recognition Sulphides of heavy transition metals like Hg^2+$Hg^{2+}$ and Pb^2+$Pb^{2+}$ have exceptionally small K_sp$K_{sp}$ values compared to halides or hydroxides. Among sulphides, HgS$HgS$ is famously known to have one of the lowest solubility products found in inorganic qualitative analysis. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Equilibrium

### Related Formula K_p = fracp_mathrmAB cdot p_mathrmB_2^1/2p_mathrmAB_2$K_p = \frac{p_{\mathrm{AB}} \cdot p_{\mathrm{B}_2}^{1/2}}{p_{\mathrm{AB}_2}}$ ### Core Logic Consider the equilibrium reaction setup : beginarrayrcccc & mathrmAB2(g) & ightleftharpoons & mathrmAB(g) & + & frac12mathrmB2(g) \ textInitial moles: & 1 & & 0 & & 0 \ textEquilibrium moles: & 1-x & & x & & fracx2 endarray $\begin{array}{rcccc} & \mathrm{AB}{2(g)} & ightleftharpoons & \mathrm{AB}{(g)} & + & \frac{1}{2}\mathrm{B}{2(g)} \ \text{Initial moles:} & 1 & & 0 & & 0 \ \text{Equilibrium moles:} & 1-x & & x & & \frac{x}{2} \end{array} $ Total equilibrium moles = 1 - x + x + fracx2 = 1 + fracx2$= 1 - x + x + \frac{x}{2} = 1 + \frac{x}{2}$ . Since x ll 1$x \ll 1$, total moles simeq 1$\simeq 1$ and 1-x simeq 1$1-x \simeq 1$ . Partial pressures: pmathrmAB2 simeq p$p{\mathrm{AB}2} \simeq p$ pmathrmAB simeq xp$p{\mathrm{AB}} \simeq xp$ p_mathrmB_2 simeq fracx2p$p_{\mathrm{B}_2} \simeq \frac{x}{2}p$ Substituting into K_p$K_p$ : K_p = frac(xp) cdot left(fracxp2 ight)^1/2p = x cdot left(fracxp2 ight)^1/2 = fracx^3/2 p^1/2sqrt2$K_p = \frac{(xp) \cdot \left(\frac{xp}{2} ight)^{1/2}}{p} = x \cdot \left(\frac{xp}{2} ight)^{1/2} = \frac{x^{3/2} p^{1/2}}{\sqrt{2}}$ Squaring both sides : K_p^2 = fracx^3 p2 implies x^3 = frac2K_p^2p$K_p^2 = \frac{x^3 p}{2} \implies x^3 = \frac{2K_p^2}{p}$ x = sqrt[3]frac2K_p^2p$x = \sqrt[3]{\frac{2K_p^2}{p}}$ ### Pattern Recognition For Delta n_g = 0.5$\Delta n_g = 0.5$ involving degree of dissociation x ll 1$x \ll 1$, tracking total pressure approximations ensures immediate analytical solution without full expansion. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Chemical Equilibrium

### Related Formula For a salt of weak acid and weak base (like ammonium acetate): mathrmpH = frac12 (mathrmpK_w + mathrmpK_a - mathrmpK_b)$\mathrm{pH} = \frac{1}{2} (\mathrm{p}K_w + \mathrm{p}K_a - \mathrm{p}K_b)$ ### Core Logic Ammonium acetate (CH_3COONH_4$CH_3COONH_4$) is a salt formed from a weak acid (CH_3COOH$CH_3COOH$) and a weak base (NH_4OH$NH_4OH$). Given: K_a = 1.8 times 10^-5$K_a = 1.8 \times 10^{-5}$ K_b = 1.8 times 10^-5$K_b = 1.8 \times 10^{-5}$ Since K_a = K_b$K_a = K_b$, taking the negative logarithm gives mathrmpK_a = mathrmpK_b$\mathrm{p}K_a = \mathrm{p}K_b$. ### Step 1: Calculate pH mathrmpH = fracmathrmpK_w + mathrmpK_a - mathrmpK_b2$\mathrm{pH} = \frac{\mathrm{p}K_w + \mathrm{p}K_a - \mathrm{p}K_b}{2}$ Substitute mathrmpK_a = mathrmpK_b$\mathrm{p}K_a = \mathrm{p}K_b$: mathrmpH = fracmathrmpK_w2$\mathrm{pH} = \frac{\mathrm{p}K_w}{2}$ At standard temperature (298 K), mathrmpK_w = 14$\mathrm{p}K_w = 14$. mathrmpH = frac142 = 7$\mathrm{pH} = \frac{14}{2} = 7$ ### Pattern Recognition If K_a = K_b$K_a = K_b$ for a weak acid-weak base salt, the hydrolysis of cation and anion perfectly balance out, making the resulting solution exactly neutral (pH = 7) regardless of the concentration of the salt. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Equilibrium

### Related Formula mathrmN_2(g) + 3mathrmH_2(g) rightleftharpoons 2mathrmNH_3(g)$\mathrm{N}_2(g) + 3mathrm{H}_2(g) \rightleftharpoons 2mathrm{NH}_3(g)$ K_c = frac[mathrmNH_3]^2, [mathrmN_2][mathrmH_2]^3$K_c = \frac{[\mathrm{NH}_3]^2}, {[\mathrm{N}_2][\mathrm{H}_2]^3}$ ### Core Logic Substituting the given equilibrium concentrations into the equilibrium constant expression: K_c = frac(1.5 times 10^-2)^2, (2 times 10^-2) times (3 times 10^-2)^3$K_c = \frac{(1.5 \times 10^{-2})^2}, {(2 \times 10^{-2}) \times (3 \times 10^{-2})^3}$ Evaluating the values step-by-step: K_c = frac2.25 times 10^-4, (2 times 10^-2) times (27 times 10^-6)$K_c = \frac{2.25 \times 10^{-4}}, {(2 \times 10^{-2}) \times (27 \times 10^{-6})}$ ### Step 1: Final Arithmetic Integration K_c = frac2.25 times 10^-4, 54 times 10^-8 = frac2.25, 54 times 10^4 = 0.041666 times 10^4 approx 416.67$K_c = \frac{2.25 \times 10^{-4}}, {54 \times 10^{-8}} = \frac{2.25}, {54} \times 10^4 = 0.041666 \times 10^4 \approx 416.67$ Rounding to the nearest integer yields **417**. ### Pattern Recognition Pay close attention to the cubic exponent in the denominator derived from the hydrogen stoichiometric coefficient (3$3$). Small calculation errors here can significantly alter the result. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Chemical Equilibrium