### Related Formula pH of a weak acid-weak base salt system: textpH = 7 + frac12(textpK_a - textpK_b)$\text{pH} = 7 + \frac{1}{2}(\text{pK}_a - \text{pK}_b)$ ### Core Logic Ammonium carbonate, (textNH_4)_2textCO_3$(\text{NH}_4)_2\text{CO}_3$, is formed from a weak acid (textH_2textCO_3$\text{H}_2\text{CO}_3$, K_a approx 4.3 times 10^-7$K_a \approx 4.3 \times 10^{-7}$) and weak base (textNH_4textOH$\text{NH}_4\text{OH}$, K_b approx 1.8 times 10^-5$K_b \approx 1.8 \times 10^{-5}$). Since K_b > K_a$K_b > K_a$, the aqueous medium accumulates an excess of hydroxyl particles over hydronium, forming a basic system (textpH > 7$\text{pH} > 7$). Both statements are structurally accurate descriptions. ### Chapter Mix Class 11 Chemistry: Equilibrium

### Related Formula K_p = K_c cdot (RT)^Delta n_g$K_p = K_c \cdot (RT)^{\Delta n_g}$ ### Core Logic For the given gaseous equilibrium reaction: N_2O_4(g) rightleftharpoons 2NO_2(g)$N_2O_4(g) \rightleftharpoons 2NO_2(g)$ First, find the change in the number of moles of gas (Delta n_g$\Delta n_g$): Delta n_g = n_p - n_r = 2 - 1 = 1$\Delta n_g = n_p - n_r = 2 - 1 = 1$ ### Step 1: Calculation Substitute the given values into the K_p - K_c$K_p - K_c$ relationship: K_p = 0.492$K_p = 0.492$ R = 0.082$R = 0.082$ T = 300text K$T = 300\text{ K}$ 0.492 = K_c cdot (0.082 times 300)^1$0.492 = K_c \cdot (0.082 \times 300)^1$ K_c = frac0.4920.082 times 300$K_c = \frac{0.492}{0.082 \times 300}$ K_c = frac0.49224.6$K_c = \frac{0.492}{24.6}$ K_c = 0.02$K_c = 0.02$ Converting to the requested format (x times 10^-2$x \times 10^{-2}$): K_c = 2 times 10^-2$K_c = 2 \times 10^{-2}$ So, the value is 2. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Equilibrium

### Related Formula Henderson-Hasselbalch Equation for Acidic Buffers: mathrmpH = pK_a + log left( frac[textSalt][textAcid] right)$\mathrm{pH} = pK_a + \log \left( \frac{[\text{Salt}]}{[\text{Acid}]} \right)$ ### Core Logic Let the volume of 1M Benzoic acid be V_a$V_a$ mL and the volume of 1M Sodium benzoate be V_s$V_s$ mL. Total volume = V_s + V_a = 300\,textmL$V_s + V_a = 300\,\text{mL}$. Millimoles of acid = 1 times V_a = V_a$1 \times V_a = V_a$ Millimoles of salt = 1 times V_s = V_s$1 \times V_s = V_s$ Applying Henderson's Equation: 4.5 = 4.2 + log left(fracV_sV_aright)$4.5 = 4.2 + \log \left(\frac{V_s}{V_a}\right)$ ### Step 1: Calculate Volume Ratio log left(fracV_sV_aright) = 4.5 - 4.2 = 0.3$\log \left(\frac{V_s}{V_a}\right) = 4.5 - 4.2 = 0.3$ Since log 2 approx 0.3$\log 2 \approx 0.3$, we have: fracV_sV_a = 2$\frac{V_s}{V_a} = 2$ V_s = 2 V_a$V_s = 2 V_a$ ### Step 2: Substitute and Solve We know V_s + V_a = 300$V_s + V_a = 300$ Substituting V_s = 2 V_a$V_s = 2 V_a$: 2 V_a + V_a = 300$2 V_a + V_a = 300$ 3 V_a = 300$3 V_a = 300$ V_a = 100 \, textmL$V_a = 100 \, \text{mL}$ ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Equilibrium

### Related Formula K_P = fracP_B cdot (P_C)^frac12P_A$K_P = \frac{P_B \cdot (P_C)^{\frac{1}{2}}}{P_A}$ where P_i$P_i$ is the partial pressure of component i$i$. ### Step 1: Setting up the ICE Table For the reaction A_(g) rightleftharpoons B_(g) + frac12 C_(g)$A_{(g)} \rightleftharpoons B_{(g)} + \frac{1}{2} C_{(g)}$ Let initial moles of A = 1$A = 1$. At equilibrium: Moles of A = 1 - alpha$A = 1 - \alpha$ Moles of B = alpha$B = \alpha$ Moles of C = fracalpha2$C = \frac{\alpha}{2}$ Total moles at equilibrium = (1 - alpha) + alpha + fracalpha2 = 1 + fracalpha2 = frac2 + alpha2$(1 - \alpha) + \alpha + \frac{\alpha}{2} = 1 + \frac{\alpha}{2} = \frac{2 + \alpha}{2}$ ### Step 2: Calculating Partial Pressures Using mole fraction times$\times$ Total Pressure (P$P$): P_A = frac1 - alpha1 + fracalpha2 cdot P$P_A = \frac{1 - \alpha}{1 + \frac{\alpha}{2}} \cdot P$ P_B = fracalpha1 + fracalpha2 cdot P$P_B = \frac{\alpha}{1 + \frac{\alpha}{2}} \cdot P$ P_C = fracfracalpha21 + fracalpha2 cdot P$P_C = \frac{\frac{\alpha}{2}}{1 + \frac{\alpha}{2}} \cdot P$ ### Step 3: Calculating Kp K_P = fracP_B cdot (P_C)^frac12P_A$K_P = \frac{P_B \cdot (P_C)^{\frac{1}{2}}}{P_A}$ K_P = fracleft( fracalpha1 + alpha/2 P right) cdot left( fracalpha/21 + alpha/2 P right)^1/2frac1 - alpha1 + alpha/2 P$K_P = \frac{\left( \frac{\alpha}{1 + \alpha/2} P \right) \cdot \left( \frac{\alpha/2}{1 + \alpha/2} P \right)^{1/2}}{\frac{1 - \alpha}{1 + \alpha/2} P}$ K_P = fracalpha cdot (alpha/2)^1/2 cdot P^3/2(1 + alpha/2)^3/2 cdot frac1 + alpha/2(1 - alpha) P$K_P = \frac{\alpha \cdot (\alpha/2)^{1/2} \cdot P^{3/2}}{(1 + \alpha/2)^{3/2}} \cdot \frac{1 + \alpha/2}{(1 - \alpha) P}$ K_P = fracalpha^3/2 cdot P^1/2sqrt2 cdot (1 + alpha/2)^1/2 cdot (1 - alpha)$K_P = \frac{\alpha^{3/2} \cdot P^{1/2}}{\sqrt{2} \cdot (1 + \alpha/2)^{1/2} \cdot (1 - \alpha)}$ Since 1 + alpha/2 = frac2+alpha2$1 + \alpha/2 = \frac{2+\alpha}{2}$, the sqrt2$\sqrt{2}$ in denominator cancels out perfectly leaving: K_P = fracalpha^frac32 P^frac12(2 + alpha)^frac12(1 - alpha)$K_P = \frac{\alpha^{\frac{3}{2}} P^{\frac{1}{2}}}{(2 + \alpha)^{\frac{1}{2}}(1 - \alpha)}$ ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Equilibrium