### Core Logic Let's track the concentration changes for the reversible reaction starting with reactants mathrmH_2$\mathrm{H}_{2}$ and mathrmI_2$\mathrm{I}_{2}$: 1. As the forward reaction proceeds, the concentrations of reactants (mathrmH_2$\mathrm{H}_2$ and mathrmI_2$\mathrm{I}_2$) decrease over time. 2. Simultaneously, the concentration of the product (mathrmHI$\mathrm{HI}$) increases from zero. 3. Once dynamic equilibrium is attained, the rates of the forward and reverse reactions become equal. Consequently, the concentrations of all reactants and products become constant over time, appearing as horizontal lines on a concentration vs. time plot. Graph (2) correctly depicts the concentration of reactants smoothly decreasing and the product concentration increasing until they all plateau horizontally at dynamic equilibrium. ### Pattern Recognition On a concentration vs. time graph, look for lines that become perfectly horizontal after a certain point. This horizontal plateau signifies that equilibrium has been established. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Equilibrium

### 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