### Related Formula Gas phase dissociation equilibrium setup: textH_2textO(g) rightleftharpoons textH_2text(g) + frac12textO_2text(g)$\text{H}_2\text{O(g)} \rightleftharpoons \text{H}_2\text{(g)} + \frac{1}{2}\text{O}_2\text{(g)}$ Partial pressure equilibrium expression: K_p = fracP_textH_2 cdot (P_textO_2)^1/2P_textH_2textO$K_p = \frac{P_{\text{H}_2} \cdot (P_{\text{O}_2})^{1/2}}{P_{\text{H}_2\text{O}}}$ ### Execution Step 1: Set up the mole distribution table at equilibrium assuming 1 initial mole: * textH_2textO = 1 - alpha$\text{H}_2\text{O} = 1 - \alpha$ * textH_2 = alpha$\text{H}_2 = \alpha$ * textO_2 = fracalpha2$\text{O}_2 = \frac{\alpha}{2}$ Step 2: Calculate the total moles (n_T$n_T$) at equilibrium: n_T = (1 - alpha) + alpha + fracalpha2 = 1 + fracalpha2$n_T = (1 - \alpha) + \alpha + \frac{\alpha}{2} = 1 + \frac{\alpha}{2}$ Given alpha ll 1$\alpha \ll 1$, we can approximate n_T approx 1$n_T \approx 1$. Step 3: Express the partial pressures using total pressure P = 1 text bar$P = 1 \text{ bar}$: P_textH_2textO = frac1-alpha1 cdot P approx 1 cdot 1 = 1$P_{\text{H}_2\text{O}} = \frac{1-\alpha}{1} \cdot P \approx 1 \cdot 1 = 1$ P_textH_2 = alpha cdot P = alpha$P_{\text{H}_2} = \alpha \cdot P = \alpha$ P_textO_2 = fracalpha2 cdot P = fracalpha2$P_{\text{O}_2} = \frac{\alpha}{2} \cdot P = \frac{\alpha}{2}$ Step 4: Substitute these partial pressures into the K_p$K_p$ expression: K_p = fracalpha cdot left(fracalpha2right)^1/21 = fracalpha^3/2sqrt2$K_p = \frac{\alpha \cdot \left(\frac{\alpha}{2}\right)^{1/2}}{1} = \frac{\alpha^{3/2}}{\sqrt{2}}$ Step 5: Equate to the given value of K_p = 8.0 times 10^-3$K_p = 8.0 \times 10^{-3}$ and solve for alpha$\alpha$: 8.0 times 10^-3 = fracalpha^3/2sqrt2 implies alpha^3/2 = 8sqrt2 times 10^-3$8.0 \times 10^{-3} = \frac{\alpha^{3/2}}{\sqrt{2}} \implies \alpha^{3/2} = 8\sqrt{2} \times 10^{-3}$ Cube both sides to clear fractional exponents: alpha^3 = left(8sqrt2 times 10^-3right)^2 = 128 times 10^-6$\alpha^3 = \left(8\sqrt{2} \times 10^{-3}\right)^2 = 128 \times 10^{-6}$ alpha = sqrt[3]128 times 10^-2 approx 5.03 times 10^-2$\alpha = \sqrt[3]{128} \times 10^{-2} \approx 5.03 \times 10^{-2}$ Matching the target template alpha = 5.03 times 10^-2$\alpha = 5.03 \times 10^{-2}$, the integer value is **5**. ### Pattern Recognition When alpha ll 1$\alpha \ll 1$, the total mole expression simplifies to 1, and the denominator (1-alpha)$(1-\alpha)$ drops out. This simplifies the expression to K_p propto alpha^1 + Delta n_g$K_p \propto \alpha^{1 + \Delta n_g}$, allowing you to quickly isolate alpha$\alpha$ via standard powers. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Chemical Equilibrium Class 11 Chemistry: Chemical Thermodynamics

### Core Logic Statement (A) is correct: Increasing pressure by compressing the volume increases active mass/concentration (c = n/V$c = n/V$) for both reactants and products instantly. Statement (B) is correct: The reaction has Delta n_g = 2 - 4 = -2$Delta n_g = 2 - 4 = -2$. Increasing pressure shifts equilibrium towards the direction of fewer gaseous moles, which is the forward path. Statement (C) is incorrect: Equilibrium constant (K$K$) is exclusively temperature-dependent and does not alter with pressure changes. Statement (D) is correct: Confirms that equilibrium constant remains unchanged. ### Pattern Recognition Always remember: pressure changes shift positions but NEVER alter the value of the equilibrium constant K_c$K_c$ or K_p$K_p$. Only temperature changes can change K$K$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Equilibrium

### Related Formula For a weak acid solution: mathrmHA rightleftharpoons mathrmH^+ + mathrmA^-$\mathrm{HA} \rightleftharpoons \mathrm{H}^+ + \mathrm{A}^-$ K_a = frac[mathrmH^+][mathrmA^-][mathrmHA]$K_a = \frac{[\mathrm{H}^+][\mathrm{A}^-]}{[\mathrm{HA}]}$ ### Step 1: Expressing Concentration Let the initial concentration be a$a$. At equilibrium: [mathrmHA] = a(1-x), quad [mathrmH^+] = ax, quad [mathrmA^-] = ax$[\mathrm{HA}] = a(1-x), \quad [\mathrm{H}^+] = ax, \quad [\mathrm{A}^-] = ax$ Substituting into the equilibrium expression: K_a = frac(ax)(x)1-x = [mathrmH^+] left(fracx1-xright)$K_a = \frac{(ax)(x)}{1-x} = [\mathrm{H}^+] \left(\frac{x}{1-x}\right)$ ### Step 2: Logarithmic Rearrangement Taking negative logarithms on both sides: -log K_a = -log [mathrmH^+] - logleft(fracx1-xright)$-\log K_a = -\log [\mathrm{H}^+] - \log\left(\frac{x}{1-x}\right)$ mathrmpK_a = mathrmpH - logleft(fracx1-xright)$\mathrm{pK}_a = \mathrm{pH} - \log\left(\frac{x}{1-x}\right)$ mathrmpH - mathrmpK_a = logleft(fracx1-xright)$\mathrm{pH} - \mathrm{pK}_a = \log\left(\frac{x}{1-x}\right)$ ### Pattern Recognition Sees: mathrmpH - mathrmpK_a$\mathrm{pH} - \mathrm{pK}_a$ for weak acid equilibrium. Shortcut: This is equivalent to the Henderson-Hasselbalch framework: mathrmpH = mathrmpK_a + logfrac[textSalt][textAcid] = mathrmpK_a + logfracx1-x$\mathrm{pH} = \mathrm{pK}_a + \log\frac{[\text{Salt}]}{[\text{Acid}]} = \mathrm{pK}_a + \log\frac{x}{1-x}$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Ionic Equilibrium

### Core Logic A catalyst provides an alternative pathway with a lower activation energy for both forward and backward reactions. As a result, it increases the rate of both reactions to the same extent. Therefore: 1. It does not alter the equilibrium constant K_c$K_{c}$, which depends solely on temperature. 2. It does not shift the position of equilibrium or change the final equilibrium composition of the system; it merely helps the system reach equilibrium faster. ### Step 1: Statement Assessment Statement I is true because K_c$K_{c}$ remains completely unaltered by a catalyst when temperature is constant. Statement II is false because no catalyst (homogeneous or heterogeneous) can alter the net composition of a system at equilibrium. ### Pattern Recognition Shortcut: Equilibrium composition and equilibrium constant can only be altered by changing thermodynamic state parameters like temperature, not by kinetic helpers like a catalyst. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Chemical Equilibrium

### Core Logic When an inert gas like Xenon is added at constant temperature and pressure, the total volume of the container must expand significantly to maintain constant pressure. According to Le Chatelier's principle, the equilibrium shifts towards the side with a higher number of gas moles (the forward direction here, since Delta n_g = 1 > 0$\Delta n_g = 1 > 0$). This increases the absolute number of moles of textPCl_3$\text{PCl}_3$ and textCl_2$\text{Cl}_2$. ### Step 1: Concentration Analysis Concentration is calculated as moles divided by total volume ([textX] = fracnV$[\text{X}] = \frac{n}{V}$). Even though the forward shift generates a few more moles of products, the fractional volume expansion factor is larger. Thus, the actual molar concentrations of **all species** (textPCl_5$\text{PCl}_5$, textPCl_3$\text{PCl}_3$, and textCl_2$\text{Cl}_2$) ultimately decrease. ### Pattern Recognition Trap Alert: Distinguish clearly between "moles" and "concentration". The forward shift increases product moles, but the volume expansion dilutes all chemical species, dropping total concentration values. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Chemical Equilibrium