### Related Formula K_p = fracp_B cdot p_Cp_A$K_p = \frac{p_B \cdot p_C}{p_A}$ ### Core Logic Let us write down the dissociation dynamics for the reaction starting with a$a$ moles of mathrmA(g)$\mathrm{A(g)}$: beginarrayrcccc & mathrmA(g) & rightleftharpoons & mathrmB(g) & + & mathrmC(g) \\ textInitial (t=0): & a & & 0 & & 0 \\ textEquilibrium (t=eq): & a(1-alpha) & & aalpha & & aalpha endarray$\begin{array}{rcccc} & \mathrm{A(g)} & \rightleftharpoons & \mathrm{B(g)} & + & \mathrm{C(g)} \\ \text{Initial (t=0):} & a & & 0 & & 0 \\ \text{Equilibrium (t=eq):} & a(1-\alpha) & & a\alpha & & a\alpha \end{array}$ textTotal moles at equilibrium = a(1-alpha) + aalpha + aalpha = a(1+alpha)$\text{Total moles at equilibrium} = a(1-\alpha) + a\alpha + a\alpha = a(1+\alpha)$ ### Step 1: Calculate Partial Pressures The mole fractions (X_i$X_i$) are: - X_A = frac1-alpha1+alpha$X_A = \frac{1-\alpha}{1+\alpha}$ - X_B = fracalpha1+alpha$X_B = \frac{\alpha}{1+\alpha}$ - X_C = fracalpha1+alpha$X_C = \frac{\alpha}{1+\alpha}$ If the total pressure of the gas mixture at equilibrium is p$p$, the partial pressures are: - p_A = left( frac1-alpha1+alpha right) p$p_A = \left( \frac{1-\alpha}{1+\alpha} \right) p$ - p_B = left( fracalpha1+alpha right) p$p_B = \left( \frac{\alpha}{1+\alpha} \right) p$ - p_C = left( fracalpha1+alpha right) p$p_C = \left( \frac{\alpha}{1+\alpha} \right) p$ ### Step 2: Relate K_p to alpha and p Using the expression for K_p$K_p$: K_p = fracp_B cdot p_Cp_A = fracleft(fracalpha1+alpharight)p cdot left(fracalpha1+alpharight)pleft(frac1-alpha1+alpharight)p$K_p = \frac{p_B \cdot p_C}{p_A} = \frac{\left(\frac{\alpha}{1+\alpha}\right)p \cdot \left(\frac{\alpha}{1+\alpha}\right)p}{\left(\frac{1-\alpha}{1+\alpha}\right)p}$ K_p = fracalpha^2 p1-alpha^2 implies fracalpha^21-alpha^2 = fracK_pp$K_p = \frac{\alpha^2 p}{1-\alpha^2} \implies \frac{\alpha^2}{1-\alpha^2} = \frac{K_p}{p}$ Since K_p$K_p$ is strictly a function of temperature, it remains constant. Therefore, if total pressure p$p$ **increases**, the term fracK_pp$\frac{K_p}{p}$ decreases, which demands that the term fracalpha^21-alpha^2$\frac{\alpha^2}{1-\alpha^2}$ must decrease. This is only possible if the degree of dissociation **alpha$\alpha$ decreases**. ### Pattern Recognition Le Chatelier's Principle Shortcut: For reactions with Delta n_g > 0$\Delta n_g > 0$, raising the pressure pushes the system in the reverse direction to decrease the gas moles, which logically decreases the degree of dissociation alpha$\alpha$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Equilibrium

### Related Formula Delta G^circ = Delta H^circ - TDelta S^circ$\Delta G^{\circ} = \Delta H^{\circ} - T\Delta S^{\circ}$ ln K_texteq = -fracDelta G^circRT = -fracDelta H^circRT + fracDelta S^circR$\ln K_{\text{eq}} = -\frac{\Delta G^{\circ}}{RT} = -\frac{\Delta H^{\circ}}{RT} + \frac{\Delta S^{\circ}}{R}$ ### Core Logic Let us state Haber's process reaction: mathrmN_2(g) + 3H_2(g) rightarrow 2NH_3(g)$\mathrm{N_2(g) + 3H_2(g) \rightarrow 2NH_3(g)}$ This synthesis reaction is thermodynamically characterized by: 1. **Delta H^circ < 0$\Delta H^{\circ} < 0$** (Exothermic reaction) 2. **Delta S^circ < 0$\Delta S^{\circ} < 0$** (Decrease in the number of gaseous molecules, from 4 moles to 2 moles) Both Delta H^circ$\Delta H^{\circ}$ and Delta S^circ$\Delta S^{\circ}$ remain relatively constant across the temperature range of interest. mathrmN_2(mathrmg) + 3mathrmH_2(mathrmg) rightarrow 2mathrmNH_3$\mathrm{N}_2(\mathrm{g}) + 3\mathrm{H}_2(\mathrm{g}) \rightarrow 2\mathrm{NH}_3$ DeltamathrmH^circ = -textve$\Delta\mathrm{H}^\circ = -\text{ve}$ DeltamathrmS^circ = -textve$\Delta\mathrm{S}^\circ = -\text{ve}$ (As gaseous moles decreases). (1) As temperature increases frac-DeltamathrmH^circ_mathrmRmathrmT$\frac{-\Delta\mathrm{H}^\circ_{\mathrm{R}}}{\mathrm{T}}$ , decreases (2) DeltamathrmG^circ = -mathrmRT ln mathrmK_mathrmeq$\Delta\mathrm{G}^\circ = -\mathrm{RT} \ln \mathrm{K}_{\mathrm{eq}}$ mathrmR ln mathrmK_mathrmeq = -fracDeltamathrmG^circmathrmT$\mathrm{R} \ln \mathrm{K}_{\mathrm{eq}} = -\frac{\Delta\mathrm{G}^\circ}{\mathrm{T}}$ (on increasing temperature in exothermic reaction mathrmK_mathrmeq$\mathrm{K}_{\mathrm{eq}}$ decreases) DeltamathrmH^circ$\Delta\mathrm{H}^\circ$ and DeltamathrmS^circ$\Delta\mathrm{S}^\circ$ are almost constant with temperature. This perfectly matches Graph (1). ### Pattern Recognition Since Haber's process is exothermic, K_texteq$K_{\text{eq}}$ must decrease as temperature increases (according to Le Chatelier's Principle). Because R ln K_texteq = -Delta G^circ/T$R \ln K_{\text{eq}} = -\Delta G^{\circ}/T$, the quantity -Delta G^circ/T$-\Delta G^{\circ}/T$ must also decrease with temperature. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Equilibrium Class 11 Chemistry: Chemical Thermodynamics

### Related Formula Ideal Gas equation layout for aggregate systems: P_texttotal cdot V = n_texttotal cdot RT$P_{\text{total}} \cdot V = n_{\text{total}} \cdot RT$ Partial Pressure expression using mole fractions: p_i = X_i cdot P_texttotal$p_i = X_i \cdot P_{\text{total}}$ ### Core Logic Let's tabulate equilibrium progress row-by-row: * Reaction matrix: beginarraylcccc & mathrmCO(g) & + & mathrm2H_2(g) & rightleftharpoons & mathrmCH_3OH(g) \\ t=0 & 0.1 & & a & & 0 \\ t_texteq & 0.1 - x & & a - 2x & & x endarray$\begin{array}{lcccc} & \mathrm{CO(g)} & + & \mathrm{2H_2(g)} & \rightleftharpoons & \mathrm{CH_3OH(g)} \\ t=0 & 0.1 & & a & & 0 \\ t_{\text{eq}} & 0.1 - x & & a - 2x & & x \end{array}$ * Given x = 0.04mathrm~mol$x = 0.04\mathrm{~mol}$ at equilibrium: - n_mathrmCO = 0.1 - 0.04 = 0.06mathrm~mol$n_{\mathrm{CO}} = 0.1 - 0.04 = 0.06\mathrm{~mol}$ - n_mathrmCH_3OH = 0.04mathrm~mol$n_{\mathrm{CH_3OH}} = 0.04\mathrm{~mol}$ * Determine total moles via system pressure (P = 5mathrm~bar$P = 5\mathrm{~bar}$, V = 2mathrm~L$V = 2\mathrm{~L}$, T = 500mathrmK$T = 500\mathrm{K}$): 5 times 2 = n_texttotal times 0.08 times 500 implies n_texttotal = frac1040 = 0.25mathrm~mol$5 \times 2 = n_{\text{total}} \times 0.08 \times 500 \implies n_{\text{total}} = \frac{10}{40} = 0.25\mathrm{~mol}$ * Find remaining unknown hydrogen moles: n_texttotal = 0.06 + n_mathrmH_2 + 0.04 = 0.25 implies n_mathrmH_2 = 0.15mathrm~mol$n_{\text{total}} = 0.06 + n_{\mathrm{H_2}} + 0.04 = 0.25 \implies n_{\mathrm{H_2}} = 0.15\mathrm{~mol}$ ### Step 1: Calculate Kp Compute partial pressures using fractional allocation fractions (n_texttotal = 0.25$n_{\text{total}} = 0.25$): * p_mathrmCH_3OH = frac0.040.25 times 5 = 0.8mathrm~bar$p_{\mathrm{CH_3OH}} = \frac{0.04}{0.25} \times 5 = 0.8\mathrm{~bar}$ * p_mathrmCO = frac0.060.25 times 5 = 1.2mathrm~bar$p_{\mathrm{CO}} = \frac{0.06}{0.25} \times 5 = 1.2\mathrm{~bar}$ * p_mathrmH_2 = frac0.150.25 times 5 = 3.0mathrm~bar$p_{\mathrm{H_2}} = \frac{0.15}{0.25} \times 5 = 3.0\mathrm{~bar}$ Substitute these pressures into the equilibrium expression: K_p = fracp_mathrmCH_3OHp_mathrmCO cdot (p_mathrmH_2)^2 = frac0.81.2 times 3^2 = frac0.810.8 = 0.07407 = 74.07 times 10^-3$K_p = \frac{p_{\mathrm{CH_3OH}}}{p_{\mathrm{CO}} \cdot (p_{\mathrm{H_2}})^2} = \frac{0.8}{1.2 \times 3^2} = \frac{0.8}{10.8} = 0.07407 = 74.07 \times 10^{-3}$ Rounding to the nearest integer yields 74. ### Pattern Recognition Finding the total moles using the Ideal Gas Law from the final equilibrium pressure and volume cuts down steps, as it avoids explicitly computing the initial hydrogen amount 'a$a$' first. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Equilibrium

### Related Formula Condition for precipitation step to happen dynamically: Q_mathrmsp > K_mathrmsp$Q_{\mathrm{sp}} > K_{\mathrm{sp}}$ where Ionic Product Q_mathrmsp = [mathrmA^2+][mathrmY^-]^2$Q_{\mathrm{sp}} = [\mathrm{A}^{2+}][\mathrm{Y}^-]^2$. ### Core Logic When equal volumes are combined, total fluid volume doubles, meaning individual concentrations are exactly cut in half: [mathrmA^2+] = frac[mathrmAB_2]_02, quad [mathrmY^-] = frac[mathrmXY]_02$[\mathrm{A}^{2+}] = \frac{[\mathrm{AB}_2]_0}{2}, \quad [\mathrm{Y}^-] = \frac{[\mathrm{XY}]_0}{2}$ Let's calculate Q_mathrmsp$Q_{\mathrm{sp}}$ value for item layout (3): * [mathrmA^2+] = frac2.0 times 10^-22 = 10^-2\ mathrmM$[\mathrm{A}^{2+}] = \frac{2.0 \times 10^{-2}}{2} = 10^{-2}\ \mathrm{M}$ * [mathrmY^-] = frac2.0 times 10^-22 = 10^-2\ mathrmM$[\mathrm{Y}^-] = \frac{2.0 \times 10^{-2}}{2} = 10^{-2}\ \mathrm{M}$ * Evaluating total value: Q_mathrmsp = (10^-2) times (10^-2)^2 = 10^-6$Q_{\mathrm{sp}} = (10^{-2}) \times (10^{-2})^2 = 10^{-6}$ * Comparing arrays: 10^-6 > 5.2 times 10^-7$10^{-6} > 5.2 \times 10^{-7}$, confirming precipitation conditions are satisfied. ### Pattern Recognition Do not skip the dilution factor! Halving initial chemical molar values before calculating the reaction quotients prevents incorrect combinations. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Equilibrium

### Related Formula textpH = -log_10[textH^+]$\text{pH} = -\log_{10}[\text{H}^+]$ ### Core Logic For the initial solution: textpH = 1.0 implies [textH^+]_1 = 10^-1 = 0.1 text M$\text{pH} = 1.0 \implies [\text{H}^+]_1 = 10^{-1} = 0.1 \text{ M}$ When we dilute the solution by adding an equal volume of water, the final volume is doubled (V_2 = 2V_1$V_2 = 2V_1$). Thus, the final concentration is halved: [textH^+]_2 = frac[textH^+]_12 = frac0.12 = 0.05 text M$[\text{H}^+]_2 = \frac{[\text{H}^+]_1}{2} = \frac{0.1}{2} = 0.05 \text{ M}$ Now, calculate the new textpH$\text{pH}$: textpH_2 = -log_10(0.05) = -log_10left(frac120right) = log_10 20 = log_10(10 times 2) = 1 + log_10 2$\text{pH}_2 = -\log_{10}(0.05) = -\log_{10}\left(\frac{1}{20}\right) = \log_{10} 20 = \log_{10}(10 \times 2) = 1 + \log_{10} 2$ textpH_2 = 1 + 0.30 = 1.30$\text{pH}_2 = 1 + 0.30 = 1.30$ ### Pattern Recognition Diluting any strong acid by 2$2$ times increases the textpH$\text{pH}$ by exactly log_10 2 approx 0.30$\log_{10} 2 \approx 0.30$. Thus, 1.0 + 0.3 = 1.3$1.0 + 0.3 = 1.3$ immediately. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Ionic Equilibrium