Q.Write the expression for the equilibrium constant, Kc for each of the following reactions:

Showing the 12 most recent of 69 on this concept.

1. AP EAPCET 2026Set eng-2026-05-12-FN1 markMCQ
   Q.The equilibrium constants $K_{P_1}$ and $K_{P_2}$ for the gaseous reactions, $X \rightleftharpoons 2Y$ and $Z \rightleftharpoons P + Q$ respectively are in the ratio of $1:9$. If the degree of dissociation of X and Z are equal, then the ratio of their total pressures at these equilibria is (A) $1:36$ (B) $1:3$ (C) $9:1$ (D) $4:1$

   ›Reveal solutionSolution

   Both reactions dissociate 1 mole into 2 moles of gas, but one produces 2 identical molecules (so the mole-fraction factor is $4\alpha^2$) and the other produces 2 different molecules (factor $\alpha^2$) — that factor-of-4 difference is what converts a $1:9$ ratio of $K_P$ into a $1:36$ ratio of total pressures.

   Concept and Intuition

   For a dissociation of the type $1\ \text{mol} \rightarrow 2\ \text{mol}$ starting from 1 mole of pure reactant with degree of dissociation $\alpha$ and total equilibrium pressure $P$, the partial pressures scale with mole fractions times $P$, and total moles at equilibrium are $1+\alpha$. Whether the two product molecules are identical (as in $X\to 2Y$) or different (as in $Z\to P+Q$) changes the numerical prefactor in $K_P$ even though the mole-count bookkeeping is otherwise identical.

   Step-by-Step Solution

   1. Reaction 1: $X \rightleftharpoons 2Y$. Start 1 mol $X$; at equilibrium: $(1-\alpha)$ mol $X$, $2\alpha$ mol $Y$, total $=1+\alpha$.

   $p_X = \frac{(1-\alpha)}{1+\alpha}P_1,\qquad p_Y = \frac{2\alpha}{1+\alpha}P_1$

   $K_{P_1} = \frac{p_Y^2}{p_X} = \frac{4\alpha^2P_1^2/(1+\alpha)^2}{(1-\alpha)P_1/(1+\alpha)} = \frac{4\alpha^2 P_1}{(1-\alpha)(1+\alpha)} = \frac{4\alpha^2P_1}{1-\alpha^2}$

   2. Reaction 2: $Z \rightleftharpoons P+Q$. Start 1 mol $Z$; at equilibrium: $(1-\alpha)$ mol $Z$, $\alpha$ mol $P$, $\alpha$ mol $Q$, total $=1+\alpha$.

   $p_Z=\frac{1-\alpha}{1+\alpha}P_2,\quad p_P=p_Q=\frac{\alpha}{1+\alpha}P_2$

   $K_{P_2} = \frac{p_Pp_Q}{p_Z} = \frac{\alpha^2P_2^2/(1+\alpha)^2}{(1-\alpha)P_2/(1+\alpha)} = \frac{\alpha^2P_2}{1-\alpha^2}$ …

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2. AP EAPCET 2026Set eng-2026-05-12-AN1 markMCQ
   Q.At 540 K, 0.1 mol of $PCl_5$ is heated in a 8.0 L flask. The total pressure of the equilibrium mixture is 1 atm. The $K_p$ (in atm) for the reaction is ($R = 0.082\ L\,atm\,K^{-1}\,mol^{-1}$) (A) 1.77 (B) 0.77 (C) 2.77 (D) 3.77

   ›Reveal solutionSolution

   This finds the degree of dissociation of $PCl_5$ from the observed total pressure (via total moles), then uses it to compute $K_p$; the answer is $\approx1.77$ atm.

   Concept and Intuition

   When $PCl_5$ partially dissociates into $PCl_3$ and $Cl_2$ (1 mole of gas becoming 2 moles), the total moles — and hence the total pressure at fixed $T,V$ — increase above what pure undissociated $PCl_5$ would give. Comparing the observed total pressure to the pressure expected with zero dissociation lets us extract the degree of dissociation $\alpha$, and from there the mole fractions (and $K_p$) follow directly.

   Step-by-Step Solution

   1. Reaction: $PCl_5(g)\rightleftharpoons PCl_3(g)+Cl_2(g)$. Starting with $n_0=0.1$ mol $PCl_5$, at degree of dissociation $\alpha$: moles are $PCl_5=n_0(1-\alpha)$, $PCl_3=n_0\alpha$, $Cl_2=n_0\alpha$; total $=n_0(1+\alpha)$.
   2. Total moles at equilibrium from the ideal gas law: $n_{eq}=\dfrac{P_{total}V}{RT}=\dfrac{(1)(8.0)}{(0.082)(540)}=\dfrac{8.0}{44.28}\approx0.181$ mol.
   3. So $n_0(1+\alpha)=0.181 \Rightarrow (1+\alpha)=\dfrac{0.181}{0.1}\approx1.8 \Rightarrow \alpha\approx0.8$.
   4. Mole fractions at equilibrium: $x_{PCl_5}=\dfrac{1-\alpha}{1+\alpha}$, $x_{PCl_3}=x_{Cl_2}=\dfrac{\alpha}{1+\alpha}$. …

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3. AP EAPCET 2026Set eng-2026-05-13-FN1 markMCQ
   Q.For the reaction $2HI(g) \rightleftharpoons H_2(g) + I_2(g)$, the correct expression relating the degree of dissociation ($\alpha$) of HI and equilibrium constant $K_p$ is (A) $\alpha = \dfrac{1+2\sqrt{K_p}}{2}$ (B) $\alpha = \sqrt{1+\dfrac{2K_p}{2}}$ (C) $\alpha = \dfrac{2\sqrt{K_p}}{1+2\sqrt{K_p}}$ (D) $\alpha = \sqrt{\dfrac{2K_p}{1+2K_p}}$

   ›Reveal solutionSolution

   Setting up the ICE table for $2\mathrm{HI}\rightleftharpoons\mathrm{H_2}+\mathrm{I_2}$ in terms of degree of dissociation $\alpha$, using the fact that total moles don't change, and simplifying $K_p$ in terms of mole fractions gives $\alpha=\dfrac{2\sqrt{K_p}}{1+2\sqrt{K_p}}$.

   Concept and Intuition

   Because 2 moles of HI decompose into exactly 1+1=2 moles of products, the total number of moles (and hence total pressure, at fixed volume/temperature) never changes during this reaction — a special simplification that lets $K_p$ be written purely in terms of mole fractions (the total-pressure factors cancel since the numerator and denominator have the same overall power of 2).

   Step-by-Step Solution

   1. Start with $a$ mol of HI, degree of dissociation $\alpha$. Since 2 mol HI $\to$ 1 mol H$_2$ + 1 mol I$_2$:
      • HI remaining: $a(1-\alpha)$
      • H$_2$ formed: $a\alpha/2$
      • I$_2$ formed: $a\alpha/2$
      • Total moles $=a(1-\alpha)+a\alpha/2+a\alpha/2 = a$ (unchanged).
   2. Mole fractions: $x_{\mathrm{HI}}=(1-\alpha)$, $x_{\mathrm{H_2}}=x_{\mathrm{I_2}}=\alpha/2$.
   3. Since total moles (and hence total pressure at fixed $T,V$) is unchanged, $K_p$ reduces to a pure mole-fraction expression (the $P_{\text{total}}$ factors cancel, as numerator and denominator both carry overall power 2): $K_p = \frac{x_{\mathrm{H_2}}\,x_{\mathrm{I_2}}}{x_{\mathrm{HI}}^2} = \frac{(\alpha/2)(\alpha/2)}{(1-\alpha)^2} = \frac{\alpha^2}{4(1-\alpha)^2}$ …

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4. AP EAPCET 2026Set eng-2026-05-14-FN1 markMCQ
   Q.Observe the following reaction $A(g) + B(g) \rightleftharpoons_{T(K)} C(g) + D(g)$ In a 1L closed flask, 2 moles of A(g) and 1 mole of B(g) were taken and heated to T(K). At equilibrium, the concentration of C(g) is thrice the concentration of B(g). What is the value of $K_c$? (A) 7.2 (B) 3.6 (C) 0.9 (D) 1.8

   ›Reveal solutionSolution

   A standard ICE-table equilibrium problem: use the given ratio between two equilibrium concentrations to pin down the extent of reaction $x$, then plug into the $K_c$ expression. Answer: $K_c = 1.8$.

   Concept and Intuition

   For a reaction run in a fixed 1 L vessel, moles and molar concentrations are numerically identical, which simplifies the ICE (Initial–Change–Equilibrium) table. The extra piece of information — that $[C]$ at equilibrium is exactly three times $[B]$ at equilibrium — is what lets us solve for the single unknown $x$, the amount of A and B consumed (equal to the amount of C and D formed, since all stoichiometric coefficients are 1).

   Step-by-Step Solution

   1. Write the ICE table (1 L flask, so [X] = moles of X):

   $A(g) + B(g) \rightleftharpoons C(g) + D(g)$

   Initial: $A=2,\ B=1,\ C=0,\ D=0$

   Change: $-x,\ -x,\ +x,\ +x$

   Equilibrium: $A=2-x,\ B=1-x,\ C=x,\ D=x$

   2. Apply the given condition $[C] = 3[B]$:

   $x = 3(1-x)$

   $x = 3 - 3x \Rightarrow 4x = 3 \Rightarrow x = 0.75$

   3. So at equilibrium: $[A]=2-0.75=1.25$, $[B]=1-0.75=0.25$, $[C]=[D]=0.75$.
   4. Compute $K_c$: …

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5. AP EAPCET 2026Set eng-2026-05-14-AN1 markMCQ
   Q.At T(K), 15 moles of $\mathrm{H_2}$ reacts with 5.2 moles of $\mathrm{I_2}$ and forms 10 moles of $\mathrm{HI}$. The equilibrium constant for the reaction $2HI(g) \rightleftharpoons H_2(g) + I_2(g)$ is (A) 50 (B) 100 (C) $2 \times 10^{-2}$ (D) $1 \times 10^{-2}$

   ›Reveal solutionSolution

   Compute equilibrium moles from the stoichiometry of $\mathrm{H_2+I_2\to2HI}$, get $K$ for the forward reaction, then invert it since the question asks for $K$ of the reverse reaction $2HI\rightleftharpoons H_2+I_2$.

   Concept and Intuition

   When a reaction is written in reverse, its equilibrium constant is the reciprocal of the forward one: $K_{reverse} = 1/K_{forward}$. Here the physical process described (15 mol $\mathrm{H_2}$ + 5.2 mol $\mathrm{I_2}$ forming 10 mol HI) is the FORWARD synthesis, but the constant asked for is for the reverse decomposition.

   Step-by-Step Solution

   1. Reaction (as it physically occurred): $\mathrm{H_2 + I_2 \to 2HI}$.
   2. To form 10 mol HI, by 1:1:2 stoichiometry, $5$ mol $\mathrm{H_2}$ and $5$ mol $\mathrm{I_2}$ must have reacted.
   3. Equilibrium amounts (in the same fixed volume, so ratios of moles work like concentrations): $[\mathrm{H_2}] \propto 15-5=10$, $[\mathrm{I_2}]\propto 5.2-5=0.2$, $[\mathrm{HI}]\propto 10$.
   4. $K_{forward} = \dfrac{[\mathrm{HI}]^2}{[\mathrm{H_2}][\mathrm{I_2}]} = \dfrac{10^2}{10\times0.2} = \dfrac{100}{2} = 50$. …

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6. AP EAPCET 2026Set eng-2026-05-13-AN1 markMCQ
   Q.At T (K) in a closed 2.0 L vessel, 1 mole of $H_2$ and 2 moles of $I_2$ are taken initially and allowed to react till equilibrium is established. At equilibrium, the number of moles of $H_2$ is found to be 0.2. The equilibrium constant for the dissociation of $HI$ is (A) $1.066\times10^1$ (B) $93.75\times10^{-2}$ (C) $2.132\times10^2$ (D) $9.375\times10^{-2}$

   ›Reveal solutionSolution

   This tests ICE-table equilibrium computation and the reciprocal relationship between $K_c$ of forward and reverse reactions.

   Concept and Intuition

   For $H_2(g) + I_2(g) \rightleftharpoons 2HI(g)$, the equilibrium constant for the reverse reaction (dissociation of HI, $2HI \rightleftharpoons H_2 + I_2$) is simply the reciprocal of the forward $K_c$, since reversing a reaction inverts its equilibrium expression.

   Step-by-Step Solution

   1. Initial moles in 2.0 L: $H_2 = 1$, $I_2 = 2$.
   2. At equilibrium, $H_2 = 0.2$ mol, so $1 - 0.2 = 0.8$ mol of $H_2$ reacted.
   3. By stoichiometry (1:1:2), $I_2$ reacted $= 0.8$ mol, remaining $I_2 = 2 - 0.8 = 1.2$ mol; $HI$ formed $= 2 \times 0.8 = 1.6$ mol.
   4. Concentrations (divide by 2 L): $[H_2] = 0.1\ M$, $[I_2] = 0.6\ M$, $[HI] = 0.8\ M$.
   5. $K_c$ for formation ($H_2+I_2 \to 2HI$): $K_c = \dfrac{[HI]^2}{[H_2][I_2]} = \dfrac{0.64}{0.06} = 10.667$. …

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7. AP EAPCET 2026Set eng-2026-05-15-AN1 markMCQ
   Q.At $T(K)$ in a 10.0 $L$ flask, 2 moles of $N_2O_4(g)$ was heated. At equilibrium 20% of $N_2O_4(g)$ was dissociated into $NO_2(g)$. What is the $K_c$ for this reaction? (A) $2 \times 10^{-2}$ (B) $4 \times 10^{-2}$ (C) $3 \times 10^{-2}$ (D) $6 \times 10^{-2}$

   ›Reveal solutionSolution

   This tests setting up an ICE table for the dissociation equilibrium $N_2O_4 \rightleftharpoons 2NO_2$ using a given degree of dissociation, then computing $K_c$. The value is $4\times10^{-2}$.

   Concept and Intuition

   Degree of dissociation $\alpha$ tells us what fraction of the initial moles broke apart at equilibrium. For $N_2O_4 \rightleftharpoons 2NO_2$, each mole of $N_2O_4$ that dissociates produces 2 moles of $NO_2$, so the ICE table must track this 1:2 stoichiometry carefully before computing concentrations and $K_c$.

   Step-by-Step Solution

   1. Initial concentration: $[N_2O_4]_0 = \dfrac{2\ mol}{10\ L} = 0.2\ M$.
   2. Let $\alpha = 0.20$ (20% dissociated). At equilibrium:
      • $[N_2O_4] = 0.2(1-\alpha) = 0.2\times0.8 = 0.16\ M$
      • $[NO_2] = 2\times0.2\times\alpha = 2\times0.2\times0.2 = 0.08\ M$ …

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8. AP EAPCET 2026Set eng-2026-05-15-FN1 markMCQ
   Q.Observe the following reaction $A(g) + B(g) \underset{}{\overset{T(K)}{\rightleftharpoons}} C(g)$ In a 1L closed flask, 2 moles of A(g) and 1 mole of B(g) were taken and heated to T(K). At equilibrium, the concentration of B(g) is equal to twice the equilibrium concentration of C(g). What is the value of $K_c$? (A) 0.3 (B) 0.6 (C) 1.2 (D) 1.5

   ›Reveal solutionSolution

   Setting up an ICE table with the given constraint $[B]=2[C]$ solves for the extent of reaction, giving $K_c = 0.3$.

   Concept and Intuition

   For $A(g)+B(g)\rightleftharpoons C(g)$ in a 1 L flask, all concentrations equal mole numbers. Using one algebraic constraint from the problem (here, $[B]=2[C]$ at equilibrium) lets us solve for the single unknown extent of reaction $x$, then substitute into the $K_c$ expression.

   Step-by-Step Solution

   1. Initial: $[A]_0=2$, $[B]_0=1$, $[C]_0=0$ (mol/L, since V=1L).
   2. Let $x$ mol/L of A and B react to form $x$ mol/L of C (1:1:1 stoichiometry).
   3. At equilibrium: $[A]=2-x$, $[B]=1-x$, $[C]=x$.
   4. Given constraint: $[B] = 2[C] \Rightarrow 1-x = 2x \Rightarrow 1 = 3x \Rightarrow x = \dfrac{1}{3}$.
   5. So $[A]=2-\dfrac13=\dfrac53$, $[B]=1-\dfrac13=\dfrac23$, $[C]=\dfrac13$. …

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9. AP EAPCET 2026Set eng-2026-05-18-FN1 markMCQ
   Q.At 450 K, for the reaction $2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g)$, the value of $K_p$ is $2.0 \times 10^{10}$. What is the value of $K_c$ for the decomposition of sulphur trioxide at the same temperature? [$R=0.083\ L\ bar\ mol^{-1}K^{-1}$] (A) $7.47 \times 10^{11}$ (B) $1.34 \times 10^{-12}$ (C) $7.47 \times 10^{-12}$ (D) $1.34 \times 10^{12}$

   ›Reveal solutionSolution

   Convert $K_p$ to $K_c$ for the formation reaction using $K_p=K_c(RT)^{\Delta n}$, then invert for the reverse (decomposition) reaction to get $K_c'\approx1.34\times10^{-12}$.

   Concept and Intuition

   $K_p$ and $K_c$ are related through the change in moles of gas, $\Delta n_g$, via $K_p = K_c(RT)^{\Delta n_g}$. The equilibrium constant of a reverse reaction is simply the reciprocal of the forward reaction's constant, since reversing a reaction flips which side is "products."

   Step-by-Step Solution

   1. Forward (formation) reaction: $2SO_2(g)+O_2(g)\rightleftharpoons2SO_3(g)$, $\Delta n_g = 2-(2+1) = -1$.
   2. $K_p = K_c(RT)^{\Delta n_g} = K_c(RT)^{-1} \Rightarrow K_c = K_p \times RT$.
   3. $RT = 0.083 \times 450 = 37.35$ (L bar/mol, matching $K_p$'s bar units).
   4. $K_c(\text{formation}) = 2.0\times10^{10} \times 37.35 = 7.47\times10^{11}$. …

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10. AP EAPCET 2026Set ap-2026-05-19-FN1 markMCQ
    Q.Observe the following reaction $\mathrm{A(g) \underset{}{\overset{T(K)}{\rightleftharpoons}} B(g) + C(g)}$ One mole of A(g) was taken in a 1L closed flask and heated to T(K). At equilibrium, the concentration of A(g) is equal to four times the equilibrium concentration of B(g). What is $K_c$? (A) 0.02 (B) 0.04 (C) 0.05 (D) 0.06

    ›Reveal solutionSolution

    A standard ICE-table equilibrium problem: express everything in terms of one unknown extent of reaction, use the given ratio to pin it down, then plug into $K_c$.

    Concept and Intuition

    Since A dissociates 1:1:1 into B and C, if $x$ is the concentration of B (equal to C, since they're produced in equal amounts) at equilibrium, then the amount of A consumed is also $x$, leaving $[A]=1-x$. The condition "[A] is four times [B]" pins down $x$ directly.

    Step-by-Step Solution

    1. Initial: $[A]_0=1$ mol/L (1 mol in 1 L), $[B]_0=[C]_0=0$.
    2. At equilibrium: $[A]=1-x$, $[B]=[C]=x$.
    3. Given $[A]=4[B]$: $1-x=4x\Rightarrow x=0.2$.
    4. So $[A]=0.8$, $[B]=[C]=0.2$. …

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11. AP EAPCET 2026Set ap-2026-05-19-AN1 markMCQ
    Q.At 300 K, $K_c$ for the reaction $2A(g) \rightleftharpoons B(g) + 2C(g)$ is $10^{-3}\ mol\ L^{-1}$. What is its $K_p$ in bar? ($R = 0.08\ L\ bar\ mol^{-1}K^{-1}$) (A) $10^{-3}$ (B) $2.4 \times 10^{-3}$ (C) $10^{-2}$ (D) $2.4 \times 10^{-2}$

    ›Reveal solutionSolution

    Convert $K_c$ to $K_p$ using $K_p=K_c(RT)^{\Delta n_g}$; here $\Delta n_g=1$, giving $K_p=2.4\times10^{-2}$ bar.

    Concept and Intuition

    $K_p$ and $K_c$ differ because pressure and concentration units scale differently with moles of gas; the relation $K_p = K_c(RT)^{\Delta n_g}$ (with $\Delta n_g$ = moles gaseous products − moles gaseous reactants) converts between them, using $R$ in units matching the desired pressure unit (bar here).

    Step-by-Step Solution

    1. Reaction: $2A(g) \rightleftharpoons B(g) + 2C(g)$.
    2. $\Delta n_g = (1+2) - 2 = 1$.
    3. $K_p = K_c (RT)^{\Delta n_g} = 10^{-3}\ \text{mol L}^{-1} \times (0.08\ \text{L bar mol}^{-1}\text{K}^{-1}\times 300\ \text{K})^{1}$. …

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12. AP EAPCET 2026Set eng-2026-05-18-AN1 markMCQ
    Q.At T(K), in a 10 L flask, the following equilibrium is established $2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g)$ The value of $K_c$ for this reaction is 100. At equilibrium, the number of moles of $SO_3(g)$ is equal to twice the number of moles of $SO_2(g)$. What is the number of moles of $O_2(g)$ at equilibrium? (A) 0.04 (B) 0.4 (C) 0.02 (D) 0.2

    ›Reveal solutionSolution

    Express all equilibrium concentrations in terms of moles of $SO_2$ using the given ratio, plug into $K_c$, and solve for moles of $O_2$; the volume terms conveniently cancel to give $K_c = 40/n(O_2)$.

    Concept and Intuition

    The equilibrium constant $K_c$ relates the concentrations (mol/L) of products and reactants, so we must divide every mole quantity by the flask volume (10 L) before substituting. Using the given relationship between moles of $SO_3$ and $SO_2$ lets us express everything in one unknown.

    Step-by-Step Solution

    1. Reaction: $2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g)$, $K_c = 100$, $V = 10$ L.
    2. Let moles of $SO_2$ at equilibrium $= a$. Given moles of $SO_3 = 2a$. Let moles of $O_2 = b$.
    3. Concentrations: $[SO_2] = a/10$, $[SO_3] = 2a/10 = a/5$, $[O_2] = b/10$. …

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