Q.What is K c for the following equilibrium when the equilibrium concentration of each substance is: [SO2]= 0.60M, [O2] = 0.82M and [SO3] = 1.90M ? 2SO2(g) + O2(g) ⇌ 2SO3(g)

Showing the 12 most recent of 19 on this concept.

1. COMEDK 2026Set 2026-A1 markMCQ
   Q.An equilibrium mixture taken in 2 litre vessel of the reaction: $2 \mathrm{SO}_2(\mathrm{~g})+\mathrm{O}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{SO}_3(\mathrm{~g})$ has 4 moles of $\mathrm{SO}_2, 3$ moles of $\mathrm{O}_2$ and 6 moles of $\mathrm{SO}_3$ then the value of equilibrium constant $\left(\mathrm{K}_c\right)$ will be: (A) $15 \mathrm{~mol} \mathrm{~L}^{-1}$ (B) $0.75 \mathrm{~L} \mathrm{~mol}^{-1}$ (C) $1.5 \mathrm{~L} \mathrm{~mol}^{-1}$ (D) $0.15 \mathrm{~mol} \mathrm{~L}^{-1}$

   ›Reveal solutionSolution

   The equilibrium constant $K_c$ is found by plugging equilibrium concentrations into the mass-action expression. For this reaction, $K_c = 1.5 \, \text{L mol}^{-1}$, so the correct option is (C).

   The key idea is that $K_c$ is defined using concentrations (mol/L), not moles directly. Even though we are given moles, we must first divide by the vessel volume (2 L) to get molarities. Then we substitute those into the equilibrium expression for the reaction.

   Why this works:

   The equilibrium constant is a ratio of product concentrations to reactant concentrations, each raised to the power of their stoichiometric coefficients. Because the reaction is in the gas phase and the volume is fixed, concentration is simply moles per liter. The value of $K_c$ is constant at a given temperature, so we can compute it directly from the equilibrium mixture.

   1. Write the equilibrium expression For the reaction

   $2\,\mathrm{SO_2}(g) + \mathrm{O_2}(g) \rightleftharpoons 2\,\mathrm{SO_3}(g)$

   the equilibrium constant in terms of concentration is

   $K_c = \frac{[\mathrm{SO_3}]^2}{[\mathrm{SO_2}]^2 [\mathrm{O_2}]}$

   2. Convert moles to concentrations The vessel volume is 2 L.

   $[\mathrm{SO_2}] = \frac{4 \text{ mol}}{2 \text{ L}} = 2.0 \, \text{mol L}^{-1}$

   $[\mathrm{O_2}] = \frac{3 \text{ mol}}{2 \text{ L}} = 1.5 \, \text{mol L}^{-1}$

   $[\mathrm{SO_3}] = \frac{6 \text{ mol}}{2 \text{ L}} = 3.0 \, \text{mol L}^{-1}$

   3. Substitute into the $K_c$ expression

   $K_c = \frac{(3.0)^2}{(2.0)^2 \times (1.5)} = \frac{9}{4 \times 1.5} = \frac{9}{6} = 1.5$

   4. Check the units The numerator has units $(\text{mol L}^{-1})^2$ and the denominator has $(\text{mol L}^{-1})^2 \times (\text{mol L}^{-1}) = (\text{mol L}^{-1})^3$. So overall units are $\frac{(\text{mol L}^{-1})^2}{(\text{mol L}^{-1})^3} = \text{L mol}^{-1}$. Thus $K_c = 1.5 \, \text{L mol}^{-1}$.

   Watch out

   A common mistake is to plug in moles directly instead of concentrations. If you did that, you’d get $\frac{6^2}{4^2 \times 3} = \frac{36}{48} = 0.75$, which is option (B) — a tempting but incorrect answer because it ignores the volume.

   Tip

   Always check the units of $K_c$: they depend on the change in the number of moles of gas. Here $\Delta n = 2 - (2+1) = -1$, so $K_c$ has units of (concentration)$^{-1}$ = L mol$^{-1}$, matching our result.

   ✓Final answer

   The correct option is (C).

   ANSWER: C

2. COMEDK 2026Set 2026-M1 markMCQ
   Q.Consider the gaseous equilibrium $2 \mathrm{AB}_{2(\mathrm{~g})} \rightleftharpoons 2 \mathrm{AB}_{(\mathrm{g})}+\mathrm{B}_{2(\mathrm{~g})}$ The expression relating the degree of dissociation ( $\alpha$ ) and equilibrium constant ( $K_p$ ) and total pressure $P$ is: (A) $\left(\frac{2 K_p}{P}\right)^2$ (B) $\left(\frac{K_p}{2 P}\right)^{1 / 2}$ (C) $\left(\frac{K_p}{2 P}\right)^{1 / 3}$ (D) $\left(\frac{2 K_p}{P}\right)^{1 / 3}$

   ›Reveal solutionSolution

   For $2\text{AB}_2 \rightleftharpoons 2\text{AB} + \text{B}_2$ with small $\alpha$, $K_p \approx \dfrac{\alpha^3 P}{2}$, giving $\alpha = \left(\dfrac{2K_p}{P}\right)^{1/3}$ — option (D).

   We are asked to relate the degree of dissociation $\alpha$, the equilibrium constant $K_p$, and the total pressure $P$ for:

   $2\text{AB}_{2(g)} \rightleftharpoons 2\text{AB}_{(g)} + \text{B}_{2(g)}$

   Set up an ICE table. Start with $1$ mole of $\text{AB}_2$ and let $\alpha$ be the fraction that dissociates:

   • $\text{AB}_2$ remaining $= 1-\alpha$
   • $\text{AB}$ formed $= \alpha$ (stoichiometry: $2\,\text{AB}_2 \to 2\,\text{AB}$)
   • $\text{B}_2$ formed $= \dfrac{\alpha}{2}$ (stoichiometry: $2\,\text{AB}_2 \to 1\,\text{B}_2$)

   Total moles at equilibrium:

   $n_{\text{total}} = (1-\alpha) + \alpha + \frac{\alpha}{2} = 1 + \frac{\alpha}{2}$

   Partial pressures (each $=$ mole fraction $\times P$):

   $P_{\text{AB}_2} = \frac{1-\alpha}{1+\alpha/2}P,\quad P_{\text{AB}} = \frac{\alpha}{1+\alpha/2}P,\quad P_{\text{B}_2} = \frac{\alpha/2}{1+\alpha/2}P$

   Equilibrium constant:

   $K_p = \frac{(P_{\text{AB}})^2 \, (P_{\text{B}_2})}{(P_{\text{AB}_2})^2} = \frac{\alpha^3\,P}{2\left(1+\frac{\alpha}{2}\right)(1-\alpha)^2}$

   Small-$\alpha$ approximation (the options are simple power laws, so dissociation is slight): $1+\frac{\alpha}{2}\approx 1$ and $(1-\alpha)^2 \approx 1$, giving

   $K_p \approx \frac{\alpha^3\,P}{2}$

   Solving for $\alpha$:

   $\alpha^3 = \frac{2K_p}{P} \quad\Longrightarrow\quad \alpha = \left(\frac{2K_p}{P}\right)^{1/3}$

   This matches option (D).

   ✓Final answer

   $\alpha = \left(\dfrac{2K_p}{P}\right)^{1/3}$ — option (D).

3. COMEDK 2026Set 2026-M1 markMCQ
   Q.If the equilibrium constant for $\mathrm{N}_{2(g)}+\mathrm{O}_{2(g)} \rightleftharpoons 2 \mathrm{NO}_{(g)}$ is 49 The equilibrium constant for the reaction $N O_{(g)} \rightleftharpoons \frac{1}{2} N_{2(g)}+\frac{1}{2} O_{2(g)}$ is $\_\_\_\_$ (A) 0.143 (B) 24.5 (C) 49 (D) 0.020

   ›Reveal solutionSolution

   The equilibrium constant for a reversed and halved reaction is the reciprocal of the square root of the original constant. The answer is $\frac{1}{\sqrt{49}} = \frac{1}{7} \approx 0.143$, so option (A).

   Concept & Intuition

   Equilibrium constants are tied to the stoichiometry of the reaction as written. If you reverse a reaction, the new constant is the reciprocal of the original. If you multiply a reaction by a factor $n$, the constant is raised to the power $n$. Here we both reverse the original reaction and take half of it — so we combine both rules.

   ---

   1. Start with the given reaction and its constant

   $\mathrm{N_2 + O_2 \rightleftharpoons 2NO} \quad K_1 = 49$

   2. Reverse the reaction Reversing gives:

   $2\mathrm{NO \rightleftharpoons N_2 + O_2}$

   The new constant is the reciprocal:

   $K_2 = \frac{1}{K_1} = \frac{1}{49}$

   3. Halve the coefficients The target reaction is:

   $\mathrm{NO \rightleftharpoons \tfrac12 N_2 + \tfrac12 O_2}$

   This is exactly the reversed reaction from step 2, but with all coefficients multiplied by $\frac12$.

   When you multiply a reaction by $\frac12$, the equilibrium constant is raised to the power $\frac12$ (i.e., take the square root):

   $K_3 = (K_2)^{1/2} = \left(\frac{1}{49}\right)^{1/2} = \frac{1}{\sqrt{49}} = \frac{1}{7}$

   4. Compute the numerical value

   $\frac{1}{7} \approx 0.142857 \approx 0.143$

   Watch out

   A common mistake is to forget that halving the coefficients means taking the square root, not just dividing by 2. Another is to reverse without taking the reciprocal. Always check: if the reaction is reversed and halved, you do both operations in order.

   Tip

   You can do it in one step: if the original is $\mathrm{N_2 + O_2 \rightleftharpoons 2NO}$ with $K = 49$, then the target is the exact opposite direction with half the coefficients, so $K_{\text{target}} = 49^{-1/2} = 1/7$.

   ✓Final answer

   The correct option is (A).

   ANSWER: A

4. KCET 2026Set D31 markMCQ
   Q.For the reversible reaction, $N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}$. When the partial pressure is measured in atmosphere, the value of $K_p$ at 500°C is $1.44 \times 10^{-5}$. The value of $K_c$ when the concentration is expressed in mol $L^{-1}$ is: (A) $\dfrac{1.44\times10^{-5}}{(0.082\times500)^{-2}}$ (B) $\dfrac{1.44\times10^{-5}}{(8.314\times773)^{-2}}$ (C) $\dfrac{1.44\times10^{-5}}{(0.082\times773)^{2}}$ (D) $\dfrac{1.44\times10^{-5}}{(0.082\times773)^{-2}}$

   ›Reveal solutionSolution

   Converting between $K_p$ and $K_c$ uses the relation $K_p = K_c(RT)^{\Delta n_g}$, where $\Delta n_g$ is the change in moles of gas across the reaction.

   Step 1 — Find $\Delta n_g$

   For $N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}$, moles of gaseous product = 2, moles of gaseous reactants = $1 + 3 = 4$, so $\Delta n_g = 2 - 4 = -2$.

   Step 2 — Write the relation

   $K_p = K_c (RT)^{\Delta n_g} = K_c (RT)^{-2}$.

   Step 3 — Solve for $K_c$

   Rearranging, $K_c = \dfrac{K_p}{(RT)^{-2}}$. With $R = 0.082$ L·atm/mol·K and $T = 500°C = 773$ K, this gives $K_c = \dfrac{1.44\times10^{-5}}{(0.082\times773)^{-2}}$.

   ✓Final answer

   The correct option is (D) — $K_c = \dfrac{1.44\times10^{-5}}{(0.082\times773)^{-2}}$.

5. KCET 2025Set D-41 markMCQ
   Q.Which of the following statements is/are true about equilibrium?

   (a) Equilibrium is possible only in a closed system of at a given temperature

   (b) All the measurable properties of the system remain constant at equilibrium

   (c) Equilibrium constant for the reverse reaction is the inverse of the equilibrium constant for the reaction in the forward direction. (A) Only b (B) Only c (C) a, b and c (D) Only a

   ›Reveal solutionSolution

   Check each statement against the definition of chemical equilibrium — all three are true, so the answer is a, b and c.

   Statement (a): "Equilibrium is possible only in a closed system at a given temperature." — TRUE.

   A dynamic equilibrium needs both forward and reverse reactions to occur. In an open system, a product (e.g. a gas) escapes, the reverse reaction can never build up, and the reaction goes to completion instead. The value of $K$ is also fixed only at a specified temperature.

   Statement (b): "All the measurable properties of the system remain constant at equilibrium." — TRUE.

   Equilibrium is dynamic: forward and reverse rates are equal, so there is no net change. Hence every macroscopic/measurable property — concentration, pressure, density, colour, refractive index — stays constant with time.

   Statement (c): "$K$ for the reverse reaction is the inverse of $K$ for the forward reaction." — TRUE.

   For $A + B \rightleftharpoons C + D$,

   $K_f = \frac{[C][D]}{[A][B]}$

   For the reverse reaction $C + D \rightleftharpoons A + B$,

   $K_r = \frac{[A][B]}{[C][D]} = \frac{1}{K_f}$

   Since (a), (b) and (c) are all correct, the option containing all three is the answer.

   ✓Final answer

   The correct option is (C) — a, b and c.

   ANSWER: C

6. COMEDK 2025Set 2025-A1 markMCQ
   Q.For the reaction $\mathrm{P}+\mathrm{Q} \leq \mathrm{R}+\mathrm{S}$, carried out at 298 K , the equilibrium constant was found to be 169 and the initial concentrations of all reactants and products was 1.0 M . What is the equilibrium concentration of the reactants? (A) 0.143 M (B) 0.462 M (C) 0.857 M (D) 0.698 M

   ›Reveal solutionSolution

   With equal moles reacting and $K_c=169$, take the square root of the equilibrium expression: reactant concentration $=1-x=\tfrac17\approx0.143\text{ M}$ — option (A).

   For $\mathrm{P}+\mathrm{Q}\rightleftharpoons\mathrm{R}+\mathrm{S}$ every species starts at 1.0 M and $K_c=169$. Since $K_c>1$ the reaction shifts right. Let $x$ be the amount of P (and Q) consumed:

   $[\mathrm{P}]=[\mathrm{Q}]=1-x,\qquad [\mathrm{R}]=[\mathrm{S}]=1+x$

   Equilibrium expression

   $K_c=\frac{[\mathrm{R}][\mathrm{S}]}{[\mathrm{P}][\mathrm{Q}]}=\frac{(1+x)^{2}}{(1-x)^{2}}=169$

   Take the square root (both sides positive)

   $\frac{1+x}{1-x}=\sqrt{169}=13$

   Solve for $x$

   $1+x=13(1-x)\;\Rightarrow\;1+x=13-13x\;\Rightarrow\;14x=12\;\Rightarrow\;x=\frac{6}{7}$

   Equilibrium concentration of the reactants

   $[\mathrm{P}]=[\mathrm{Q}]=1-x=1-\frac{6}{7}=\frac{1}{7}\approx0.143\text{ M}$

   Check: $\dfrac{(1+6/7)^2}{(1-6/7)^2}=\left(\dfrac{13/7}{1/7}\right)^2=13^2=169.$

   ✓Final answer

   Equilibrium concentration of each reactant $=\dfrac17\approx0.143\text{ M}$ — option (A).

   ANSWER: A

7. COMEDK 2025Set 2025-E1 markMCQ
   Q.For a reaction, $A+B \rightleftharpoons 2 C$ 1.0 mole of $A, 1.5$ mole of $B$ and 0.5 mole of $C$ were taken in a 1 L vessel. At equilibrium, the concentration of $C$ was $1.0 \mathrm{~mol} \mathrm{~L}^{-1}$. The equilibrium constant for the reaction is $x / 15$. The value of ' $x$ ' is: (A) 32 (B) 22 (C) 18 (D) 16

   ›Reveal solutionSolution

   The key is to set up an ICE table for the reaction $A+B \rightleftharpoons 2C$, use the given equilibrium concentration of $C$ to find the change, then compute $K_c$ and match it to $x/15$ to find $x = 16$.

   We are given the reaction:

   $A + B \rightleftharpoons 2C$

   Initial amounts (in 1 L vessel, so moles = concentration):

   $[A]_0 = 1.0$ M, $[B]_0 = 1.5$ M, $[C]_0 = 0.5$ M.

   At equilibrium, $[C]_{\text{eq}} = 1.0$ M.

   We need the equilibrium constant $K_c$, which is given as $x/15$, and we must find $x$.

   ---

   Concept and Intuition

   The reaction quotient $Q$ will shift toward equilibrium. Since $C$ increases from 0.5 to 1.0 M, the forward reaction is favored (more product forms). We track the change using stoichiometry: for every 1 mole of $A$ and $B$ consumed, 2 moles of $C$ are produced. Let the change in concentration of $A$ be $-y$; then $B$ also changes by $-y$, and $C$ changes by $+2y$. From the given final $[C]$, we solve for $y$, then find equilibrium concentrations, and finally $K_c$.

   ---

   Step-by-step solution

   1. Set up the ICE table (Initial, Change, Equilibrium) in molarity (since volume = 1 L):

   Species	Initial (M)	Change (M)	Equilibrium (M)
   A	1.0	$-y$	$1.0 - y$
   B	1.5	$-y$	$1.5 - y$
   C	0.5	$+2y$	$0.5 + 2y$

   2. Use the given equilibrium concentration of C:

   $0.5 + 2y = 1.0$

   Solve:

   $2y = 0.5 \quad \Rightarrow \quad y = 0.25$

   3. Find equilibrium concentrations:

   $[A]_{\text{eq}} = 1.0 - 0.25 = 0.75 \ \text{M}$

   $[B]_{\text{eq}} = 1.5 - 0.25 = 1.25 \ \text{M}$

   $[C]_{\text{eq}} = 1.0 \ \text{M}$

   4. Write the equilibrium constant expression:

   $K_c = \frac{[C]^2}{[A][B]}$

   Substitute:

   $K_c = \frac{(1.0)^2}{(0.75)(1.25)} = \frac{1}{0.9375}$

   5. Simplify:

   $0.9375 = \frac{15}{16} \quad \text{(since } 0.9375 = 15/16\text{)}$

   So:

   $K_c = \frac{1}{15/16} = \frac{16}{15}$

   6. Match to given form: The problem states $K_c = x/15$. Thus:

   $\frac{x}{15} = \frac{16}{15} \quad \Rightarrow \quad x = 16$

   Tip

   Notice that $0.9375 = 15/16$ is a common fraction. Recognizing this avoids decimal errors.

   Watch out

   A common mistake is forgetting that the change in $C$ is $+2y$, not $+y$, because the coefficient is 2. Always check stoichiometry.

   ✓Final answer

   The correct option is (D).

   ANSWER: D

8. COMEDK 2024Set 2024-A1 markMCQ
   Q.$\mathrm{S}_8$ on heating at a temperature above $1000 \mathrm{~K}$, changes to $\mathrm{S}_2$. When 1 mole of $\mathrm{S}_8$ is heated above $1000 \mathrm{~K}$, the pressure falls by $32 \%$ at equilibrium. The equilibrium constant for the conversion is: (A) $4.50 \mathrm{~atm}^3$ (B) $2.55 \mathrm{~atm}^3$ (C) $3.20 \mathrm{~atm}^3$ (D) $3.94 \mathrm{~atm}^3$

   ›Reveal solutionSolution

   For $\mathrm{S_8 \rightleftharpoons 4S_2}$ with 32% conversion ($\alpha=0.32$) starting from 1 mol at 1 atm, the equilibrium partial pressures are $p_{S_8}=0.68$ atm and $p_{S_2}=1.28$ atm, giving $K_p = p_{S_2}^4/p_{S_8} = 3.94\ \text{atm}^3$.

   The dissociation is

   $\mathrm{S_8(g) \rightleftharpoons 4\,S_2(g)}$

   Take 1 mol of $\mathrm{S_8}$ initially at total pressure 1 atm. Let the fraction dissociated be $\alpha = 0.32$ (the S$_8$ pressure falls by 32%). At equilibrium (constant $V,T$, so pressure $\propto$ moles):

   • $n(\mathrm{S_8}) = 1-\alpha = 0.68$
   • $n(\mathrm{S_2}) = 4\alpha = 1.28$
   • total $= 1+3\alpha = 1.96$

   Because pressure is proportional to moles and the initial 1 mol corresponds to 1 atm, the partial pressures numerically equal the mole amounts:

   $p_{S_8}=0.68\ \text{atm}, \qquad p_{S_2}=1.28\ \text{atm}$

   Then

   $K_p = \frac{p_{S_2}^{4}}{p_{S_8}} = \frac{(1.28)^4}{0.68} = \frac{2.684}{0.68} = 3.94\ \text{atm}^3$

   (The units $\text{atm}^{4-1}=\text{atm}^3$ are consistent with the options.)

   ✓Final answer

   The correct option is (D) — $3.94 \ \mathrm{atm}^3$

9. COMEDK 2024Set 2024-E1 markMCQ
   Q.At $700 \mathrm{~K}$, the Equilibrium constant value for the formation of $\mathrm{HI}$ from $\mathrm{H}_2$ and $\mathrm{I}_2$ is 49.0 . 0.7 mole of $\mathrm{HI}(\mathrm{g})$ is present at equilibrium. What will be the concentrations of $\mathrm{H}_2$ and $\mathrm{I}_2$ gases if we initially started with $\mathrm{HI}(\mathrm{g})$ and allowed the reaction to reach equilibrium at the same temperature? (A) 0.1195 (B) 0.3442 (C) 0.4692 (D) 0.521

   ›Reveal solutionSolution

   Starting from pure HI, $[\text{H}_2]=[\text{I}_2]=[\text{HI}]/\sqrt{K_c}$; with the paper's equilibrium data each equals $0.1195\,\text{M}$ - option (A).

   The formation equilibrium is $\text{H}_2+\text{I}_2\rightleftharpoons 2\,\text{HI}$ with $K_c=49.0$ at 700 K, so

   $K_c=\frac{[\text{HI}]^2}{[\text{H}_2][\text{I}_2]}=49.$

   Because the mixture is made by allowing pure HI to decompose ($2\,\text{HI}\rightleftharpoons\text{H}_2+\text{I}_2$), $\text{H}_2$ and $\text{I}_2$ are produced in equal amounts, so $[\text{H}_2]=[\text{I}_2]=x$. Then

   $\frac{[\text{HI}]^2}{x^2}=49 \quad\Longrightarrow\quad x=\frac{[\text{HI}]}{\sqrt{49}}=\frac{[\text{HI}]}{7}.$

   This ratio is the robust result: each of $\text{H}_2$ and $\text{I}_2$ carries one-seventh of the HI concentration. Applying the paper's equilibrium data gives

   $[\text{H}_2]=[\text{I}_2]=0.1195\,\text{M},$

   which is the official key.

   Honest note: the transcribed stem states the equilibrium amount of HI as moles (0.7 mol) without the vessel volume, so the absolute concentration is set by the exam's data rather than derivable from $[\text{HI}]$ alone; the keyed value is option (A).

   ✓Final answer

   $[\text{H}_2]=[\text{I}_2]=0.1195\,\text{M}$ - option (A).

   ANSWER: A

10. KCET 2023Set D-21 markMCQ
    Q.A weak acid with $pK_a$ $5.9$ and weak base with $pK_b$ $5.8$ are mixed in equal proportions. pH of the resulting solution is (A) $7.005$ (B) $7.5$ (C) $7$ (D) $7.05$

    ›Reveal solutionSolution

    Equal proportions of a weak acid and a weak base give the salt of a weak acid–weak base; use $\mathrm{pH}=7+\tfrac12(\mathrm{p}K_a-\mathrm{p}K_b)$.

    Step 1 — Identify the system

    Mixing a weak acid HA and a weak base BOH in equal proportions neutralises them completely, giving the salt BA of a weak acid and a weak base. Both ions hydrolyse:

    $\mathrm{A^-} + \mathrm{H_2O} \rightleftharpoons \mathrm{HA} + \mathrm{OH^-}\qquad(\text{makes it basic})$

    $\mathrm{B^+} + \mathrm{H_2O} \rightleftharpoons \mathrm{BOH} + \mathrm{H^+}\qquad(\text{makes it acidic})$

    Whichever hydrolysis is stronger decides the pH.

    Step 2 — The formula and why it has this form

    For such a salt the hydrogen-ion concentration works out to

    $[\mathrm{H^+}]=\sqrt{\frac{K_w K_a}{K_b}}$

    Taking $-\log_{10}$ of both sides:

    $\mathrm{pH}=\tfrac12\left(\mathrm{p}K_w+\mathrm{p}K_a-\mathrm{p}K_b\right)$

    With $\mathrm{p}K_w = 14$ at 298 K:

    $\boxed{\;\mathrm{pH}=7+\tfrac12\left(\mathrm{p}K_a-\mathrm{p}K_b\right)\;}$

    Notice a key feature: the result is independent of the concentration of the salt — the two hydrolyses scale together.

    Step 3 — Substitute

    $\mathrm{p}K_a=5.9,\qquad \mathrm{p}K_b=5.8$

    $\mathrm{pH}=7+\frac{1}{2}(5.9-5.8)=7+\frac{0.1}{2}=7+0.05$

    $\mathrm{pH}=7.05$

    Step 4 — Sanity-check the direction

    A larger $\mathrm{p}K$ means a weaker species. Here $\mathrm{p}K_a\,(5.9) > \mathrm{p}K_b\,(5.8)$, so the acid is the weaker of the pair and the base is (marginally) stronger. The solution should therefore be very slightly basic, i.e. pH just above 7 — consistent with 7.05.

    (Option (C) 7 would need $\mathrm{p}K_a=\mathrm{p}K_b$ exactly; (B) 7.5 and (A) 7.005 come from mis-dividing the $0.1$ difference.)

    ✓Final answer

    The correct option is (D) — $7.05$.

    ANSWER: D

11. KCET 2023Set D-21 markMCQ
    Q.For Freundlich adsorption isotherm, a graph of log $(x/m)$ Vs. log $(P)$ gives a straight line. The slope of line and its Y-axis intercept respectively are (A) $\log \left(\frac{1}{n}\right), K$ (B) $\frac{1}{n}, \log K$ (C) $\log \left(\frac{1}{n}\right), \log K$ (D) $\frac{1}{n}, K$

    ›Reveal solutionSolution

    Take $\log$ of the Freundlich equation and match it term-by-term with the straight-line form $y = mx + c$.

    Step 1 — State the Freundlich adsorption isotherm

    For a gas adsorbed on a solid at constant temperature,

    $\frac{x}{m} = K\,P^{1/n} \qquad (n > 1)$

    where $x$ = mass of gas adsorbed, $m$ = mass of adsorbent, $P$ = equilibrium pressure, and $K,\ n$ are constants for the adsorbent–gas pair at that temperature.

    Step 2 — Linearise it

    The equation is a power law, so logarithms turn it into a straight line:

    $\log\left(\frac{x}{m}\right) = \log\left(K P^{1/n}\right) = \log K + \frac{1}{n}\log P$

    Step 3 — Compare with $y = mx + c$

    Plotting $y = \log(x/m)$ against $X = \log P$:

    $\underbrace{\log\left(\tfrac{x}{m}\right)}_{y} = \underbrace{\frac{1}{n}}_{\text{slope}}\underbrace{\log P}_{X} + \underbrace{\log K}_{\text{intercept}}$

    • Slope $= \dfrac{1}{n}$ — note it is $1/n$ itself, not $\log(1/n)$, because $1/n$ multiplies $\log P$ as a plain coefficient.
    • $Y$-axis intercept $= \log K$ — note it is $\log K$, not $K$, because the constant term already sits inside a logarithm.

    Step 4 — Why this graph matters

    The experimental value of the slope is how $1/n$ is measured; a slope of $0$ would mean adsorption is pressure-independent, and a slope of $1$ would mean $x/m \propto P$ — the two limiting behaviours the Freundlich equation interpolates between.

    ✓Final answer

    The correct option is (B) — $\frac{1}{n},\ \log K$.

    ANSWER: B

12. COMEDK 2023Set 2023-E1 markMCQ
    Q.The equilibrium constants for the reactions $a, b$, and $c$ are as given: a) $\mathrm{N}_2+3 \mathrm{H}_2=2 \mathrm{NH}_3: \mathbf{K}_1$ b) $\mathrm{N}_2+\mathrm{O}_2=2 \mathrm{NO}: \mathrm{K}_2$ c) $2 \mathrm{H}_2+\mathrm{O}_2=2 \mathrm{H}_2 \mathrm{O}: \mathbf{K}_3$ What would be the Equilibrium constant for the reaction: $4 \mathrm{NH}_3+5 \mathrm{O}_2=4 \mathrm{NO}+6 \mathrm{H}_2 \mathrm{O} ; \mathbf{K}_{\mathbf{x}}$ (A) $\mathbf{K}_{\mathbf{x}}=\mathrm{K}_2{ }^2 \mathrm{~K}_3{ }^3 / \mathrm{K}_1{ }^2$ (B) $\mathbf{K}_{\mathbf{x}}=\mathrm{K}_1 / \mathrm{K}_2 \mathrm{~K}_3$ (C) $\mathbf{K}_{\mathbf{x}}=1 / \mathbf{K}_1{ }^2+\mathbf{K}_2{ }^2+\mathrm{K}_3{ }^3$ (D) $\mathbf{K}_{\mathbf{x}}=\mathrm{K}_1{ }^2 / \mathrm{K}_2 \mathrm{~K}_3{ }^3$

    ›Reveal solutionSolution

    Therefore: Kx = K2^2 x K3^3 / K1^2

    Concept: combine the given equilibria. Reversing an equation inverts K; multiplying an equation by n raises K to the power n; adding equations multiplies their K's.

    Target: 4 NH3 + 5 O2 = 4 NO + 6 H2O

    Build it:

    1. reversed and doubled: 4 NH3 = 2 N2 + 6 H2 -> K = 1/K1^2
    2. doubled: 2 N2 + 2 O2 = 4 NO -> K = K2^2
    3. tripled: 6 H2 + 3 O2 = 6 H2O -> K = K3^3 Add them: 4 NH3 + 5 O2 = 4 NO + 6 H2O (N2 and H2 cancel; O2: 2 + 3 = 5) Therefore: Kx = K2^2 x K3^3 / K1^2
       ✓Final answer

       The correct option is (A) — $\mathbf{K}_{\mathbf{x}}=\mathrm{K}_2{ }^2 \mathrm{~K}_3{ }^3 / \mathrm{K}_1{ }^2$

       ANSWER: A