Consider the general reaction given below at 400 K xA(g) rightleftharpoons yB(g)$xA(g) \rightleftharpoons yB(g)$ The values of K_p$K_{p}$ and K_c$K_{c}$ are studied under the same condition of temperature but variation in x and y. (i) K_p = 85.87$K_{p} = 85.87$ and K_c = 2.586$K_{c} = 2.586$ appropriate units (ii) K_p = 0.862$K_{p} = 0.862$ and K_c = 28.62$K_{c} = 28.62$ appropriate units The value of x and y in (i) and (ii) respectively are:

### Core Logic For a weak diprotic acid textH_2textX$\text{H}_2\text{X}$, the concentration of textX^2-$\text{X}^{2-}$ in a solution of concentration C$C$ is approximately equal to the second dissociation constant K_a_2$K_{a_2}$ under standard weak acid approximation conditions ([textX^2-] approx K_a_2 = 1.0 times 10^-13 text M$[\text{X}^{2-}] \approx K_{a_2} = 1.0 \times 10^{-13} \text{ M}$). Converting to the required format: 1.0 times 10^-13 = 100 times 10^-15 text M$1.0 \times 10^{-13} = 100 \times 10^{-15} \text{ M}$. ### Step 1: Final Calculation Thus, the integer value is 100. ### Pattern Recognition Sees: polyprotic acid dissociation equilibrium calculation for secondary anion concentration. Trap: Assuming complete dissociation for the second step. ### Chapter Mix Class 11 Chemistry: Equilibrium

### Related Formula textpOH = 14 - textpH = 14 - 9.25 = 4.75$$\text{pOH} = 14 - \text{pH} = 14 - 9.25 = 4.75$$ textpOH = pK_b + log frac[textSalt][textBase]$$\text{pOH} = pK_b + \log \frac{[\text{Salt}]}{[\text{Base}]}$$ ### Core Logic Step 1: Determine condition for textpOH = 4.75$\text{pOH} = 4.75$: 4.75 = 4.75 + log frac[textSalt][textBase] implies log frac[textSalt][textBase] = 0 implies [textSalt] = [textBase]$$4.75 = 4.75 + \log \frac{[\text{Salt}]}{[\text{Base}]} \implies \log \frac{[\text{Salt}]}{[\text{Base}]} = 0 \implies [\text{Salt}] = [\text{Base}]$$ Step 2: Test Option (2): textInitial millimoles of textNH_4textOH = 0.2 times 500 = 100text mmol$$\text{Initial millimoles of } \text{NH}_4\text{OH} = 0.2 \times 500 = 100\text{ mmol}$$ textInitial millimoles of textHCl = 0.1 times 500 = 50text mmol$$\text{Initial millimoles of } \text{HCl} = 0.1 \times 500 = 50\text{ mmol}$$ textMillimoles of textNH_4textCl formed = 50text mmol$$\text{Millimoles of } \text{NH}_4\text{Cl formed} = 50\text{ mmol}$$ textMillimoles of textNH_4textOH remaining = 100 - 50 = 50text mmol$$\text{Millimoles of } \text{NH}_4\text{OH remaining} = 100 - 50 = 50\text{ mmol}$$ Since [textSalt] = [textBase]$[\text{Salt}] = [\text{Base}]$, textpOH = pK_b = 4.75$\text{pOH} = pK_b = 4.75$ and textpH = 9.25$\text{pH} = 9.25$. ### Pattern Recognition Sees: Basic buffer when textpH = 14 - pK_b$\text{pH} = 14 - pK_b$. Shortcut: Requires millimoles of remaining weak base to equal millimoles of formed conjugate salt (i.e. acid amount must be exactly half of base amount). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Equilibrium

### Related Formula K_p = frac(P_N_2)^1/2 (P_H_2)^3/2(P_NH_3)$$K_p = \frac{(P_{N_2})^{1/2} (P_{H_2})^{3/2}}{(P_{NH_3})}$$ ### Core Logic Let the degree of dissociation of NH_3$NH_3$ be alpha$\alpha$. Initial moles: NH_3 = 1$NH_3 = 1$ Equilibrium moles: NH_3 = 1 - alpha$NH_3 = 1 - \alpha$, N_2 = fracalpha2$N_2 = \frac{\alpha}{2}$, H_2 = frac3alpha2$H_2 = \frac{3\alpha}{2}$ Total moles = 1 - alpha + fracalpha2 + frac3alpha2 = 1 + alpha$1 - \alpha + \frac{\alpha}{2} + \frac{3\alpha}{2} = 1 + \alpha$ ### Step 1: Setting up Partial Pressures Using mole fractions and total pressure P_T = sqrt3$P_T = \sqrt{3}$: P_NH_3 = frac1-alpha1+alpha P_T$P_{NH_3} = \frac{1-\alpha}{1+\alpha} P_T$ P_N_2 = fracalpha/21+alpha P_T$P_{N_2} = \frac{\alpha/2}{1+\alpha} P_T$ P_H_2 = frac3alpha/21+alpha P_T$P_{H_2} = \frac{3\alpha/2}{1+\alpha} P_T$ ### Step 2: Solving for alpha K_p = frac (fracalpha/21+alpha P_T)^1/2 cdot (frac3alpha/21+alpha P_T)^3/2 frac1-alpha1+alpha P_T $$K_p = \frac{ (\frac{\alpha/2}{1+\alpha} P_T)^{1/2} \cdot (\frac{3\alpha/2}{1+\alpha} P_T)^{3/2} }{ \frac{1-\alpha}{1+\alpha} P_T }$$ K_p = frac (alpha/2)^1/2 (3alpha/2)^3/2 1-alpha cdot fracP_T^2P_T (1+alpha) cdot (1+alpha)$$K_p = \frac{ (\alpha/2)^{1/2} (3\alpha/2)^{3/2} }{ 1-\alpha } \cdot \frac{P_T^2}{P_T (1+\alpha)} \cdot (1+\alpha)$$ Wait, simplify the P_T$P_T$ term: K_p = frac (alpha/2)^1/2 (3alpha/2)^3/2 1-alpha cdot left(fracP_T1+alpharight)$$K_p = \frac{ (\alpha/2)^{1/2} (3\alpha/2)^{3/2} }{ 1-\alpha } \cdot \left(\frac{P_T}{1+\alpha}\right)$$ Substitute K_p = 9$K_p = 9$ and P_T = sqrt3$P_T = \sqrt{3}$: 9 = frac (alpha/2)^1/2 (3alpha/2)^3/2 (1-alpha) cdot fracsqrt31+alpha$$9 = \frac{ (\alpha/2)^{1/2} (3\alpha/2)^{3/2} }{ (1-\alpha) } \cdot \frac{\sqrt{3}}{1+\alpha}$$ 9 = frac sqrt27 cdot (alpha/2)^2 (1-alpha)(1+alpha) cdot sqrt3$$9 = \frac{ \sqrt{27} \cdot (\alpha/2)^2 }{ (1-\alpha)(1+\alpha) } \cdot \sqrt{3}$$ 9 = frac 9 cdot (alpha^2/4) 1 - alpha^2 $$9 = \frac{ 9 \cdot (\alpha^2/4) }{ 1 - \alpha^2 }$$ 1 = frac alpha^2/4 1 - alpha^2 $$1 = \frac{ \alpha^2/4 }{ 1 - \alpha^2 }$$ 1 - alpha^2 = fracalpha^24$$1 - \alpha^2 = \frac{\alpha^2}{4}$$ 1 = frac5alpha^24$$1 = \frac{5\alpha^2}{4}$$ alpha^2 = frac45 = 0.8$$\alpha^2 = \frac{4}{5} = 0.8$$ alpha = (0.8)^1/2$$\alpha = (0.8)^{1/2}$$ ### Step 3: Matching the given format We are given alpha = (x times 10^-2)^-1/2$\alpha = (x \times 10^{-2})^{-1/2}$. Equating both forms: (0.8)^1/2 = (x times 10^-2)^-1/2$$(0.8)^{1/2} = (x \times 10^{-2})^{-1/2}$$ Invert both sides: x times 10^-2 = frac10.8$$x \times 10^{-2} = \frac{1}{0.8}$$ x times 10^-2 = 1.25$$x \times 10^{-2} = 1.25$$ x = 125$x = 125$ ### Pattern Recognition When doing fractional stoichiometric power K_p$K_p$ calculations, grouping the alpha$\alpha$ terms early simplifies algebra massively. 3^3/2 times 3^1/2 = 9$3^{3/2} \times 3^{1/2} = 9$, which perfectly cancels the K_p$K_p$ value. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Equilibrium

### Related Formula K_c = frac[Z]^2[X_2][Y_2]$$K_c = \frac{[Z]^2}{[X_2][Y_2]}$$ ### Core Logic For the reaction X_2(g) + Y_2(g) rightleftharpoons 2Z(g)$X_{2}(g) + Y_{2}(g) \rightleftharpoons 2Z(g)$: First, calculate the equilibrium constant K_c$K_c$ using the initial equilibrium concentrations (moles in 1 L volume). [X_2] = 3 text M$[X_2] = 3 \text{ M}$ [Y_2] = 3 text M$[Y_2] = 3 \text{ M}$ [Z] = 9 text M$[Z] = 9 \text{ M}$ K_c = frac(9)^2(3)(3) = frac819 = 9$$K_c = \frac{(9)^2}{(3)(3)} = \frac{81}{9} = 9$$ ### Step 1: Shift due to addition of Z Now, 10 moles of Z(g) are added. The new initial moles are: X_2 = 3$X_2 = 3$ Y_2 = 3$Y_2 = 3$ Z = 9 + 10 = 19$Z = 9 + 10 = 19$ The reaction quotient Q_c > K_c$Q_c > K_c$, so the reaction shifts backward. Let the change in X_2$X_2$ and Y_2$Y_2$ be +x$+x$ moles, then Z changes by -2x$-2x$ moles. At new equilibrium: X_2 = 3 + x$X_2 = 3 + x$ Y_2 = 3 + x$Y_2 = 3 + x$ Z = 19 - 2x$Z = 19 - 2x$ ### Step 2: Calculate New Equilibrium K_c = frac(19 - 2x)^2(3 + x)(3 + x) = 9$$K_c = \frac{(19 - 2x)^2}{(3 + x)(3 + x)} = 9$$ Take the square root of both sides (concentrations must be positive): frac19 - 2x3 + x = 3$$\frac{19 - 2x}{3 + x} = 3$$ 19 - 2x = 3(3 + x)$$19 - 2x = 3(3 + x)$$ 19 - 2x = 9 + 3x$$19 - 2x = 9 + 3x$$ 10 = 5x implies x = 2$$10 = 5x \implies x = 2$$ The number of moles of Z(g) at the new equilibrium is: textMoles of Z = 19 - 2x = 19 - 2(2) = 19 - 4 = 15$$\text{Moles of Z} = 19 - 2x = 19 - 2(2) = 19 - 4 = 15$$ ### Pattern Recognition If a reaction's K_c$K_c$ evaluates to a perfect square and Delta n_g = 0$\Delta n_g = 0$, adding products sets up a neat quadratic that can be directly resolved by taking the root of both sides, heavily speeding up the algebraic work. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Equilibrium