What is the equilibrium concentration of each of the substances in the equilibrium when the initial concentrat

Exercises · 6.16

Q. What is the equilibrium concentration of each of the substances in the equilibrium when the initial concentration of ICl was 0.78 M ? 2ICl

(g) ⇌ I2

(g) + Cl2 (g); Kc = 0.14

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Concept understanding — Equilibrium Constant Calculation

Equilibrium Constant Calculation: From Intuition to Precision

Imagine you're at a party where people can move between two rooms. Some people prefer the kitchen (more snacks), others prefer the living room (better music). After a while, the number of people in each room stops changing — not because everyone froze, but because the rate of people leaving the kitchen equals the rate of people entering it. The system is in dynamic equilibrium.

Chemical reactions work the same way. A reversible reaction like

$A + B \rightleftharpoons C + D$ doesn't stop when it reaches equilibrium. Instead, the forward reaction (making C and D) and the reverse reaction (making A and B) happen at the same rate. The concentrations of A, B, C, and D become constant — not equal, but constant.

The equilibrium constant $K$ is a number that tells you where this balance lies. It answers the question: At equilibrium, which side of the reaction is favoured?

The Intuitive Idea

Think of a seesaw. If $K$ is very large (say $10^6$ ), the equilibrium sits heavily on the product side — almost all A and B have turned into C and D. If $K$ is very small (say $10^{-6}$ ), the opposite is true: hardly any product forms. If $K$ is around 1, both sides have comparable amounts.

So $K$ is a ratio — a comparison of product concentrations to reactant concentrations at equilibrium, each raised to the power of their stoichiometric coefficients.

The Precise Statement

For a general reversible reaction at a given temperature:

aA + bB \rightleftharpoons cC + dD

the equilibrium constant $K_c$ (for concentrations in mol/L) is:

K_c = \frac{[C]^c [D]^d}{[A]^a [B]^b}

where $[X]$ means the equilibrium concentration of species X in moles per litre.

Important

The exponents come directly from the balanced chemical equation. If the coefficient of A is 2, you square its concentration. This is not optional — it's built into the definition.

What $K$ Actually Depends On

$K$ is constant at a given temperature. Change the temperature, and $K$ changes. But $K$ does not depend on:

Initial concentrations
Presence of a catalyst (catalysts speed up both directions equally)
Pressure or volume changes (for $K_c$ ; $K_p$ for gases has its own rules)

This is a critical exam point: if a problem gives you initial concentrations and asks for $K$ , you must first find equilibrium concentrations — you cannot plug in initial values.

A Worked Example

Problem:

$N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)$

At 500°C, equilibrium concentrations are:

$[N_2] = 0.50\ \text{M}$ , $[H_2] = 1.50\ \text{M}$ , $[NH_3] = 0.20\ \text{M}$

Calculate $K_c$ .

Solution:

Write the expression:

K_c = \frac{[NH_3]^2}{[N_2][H_2]^3}

Substitute:

K_c = \frac{(0.20)^2}{(0.50)(1.50)^3} = \frac{0.04}{0.50 \times 3.375} = \frac{0.04}{1.6875} \approx 0.0237

Note

The units cancel because the numerator and denominator both have units of (mol/L) $^2$ and (mol/L) $^4$ respectively — but by convention, $K_c$ is reported without units. The numerical value is what matters.

Common Pitfalls

Forgetting the exponents: A coefficient of 2 means square the concentration, not double it.
Using initial concentrations: You must use equilibrium concentrations only. …

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For the reaction $2\text{ICl} \rightleftharpoons \text{I}_2 + \text{Cl}_2$ with $K_c = 0.14$ and initial $[\text{ICl}] = 0.78\ \text{M}$ , the equilibrium concentrations are $[\text{ICl}] = 0.48\ \text{M}$ , $[\text{I}_2] = [\text{Cl}_2] = 0.15\ \text{M}$ .

The key idea here is that the equilibrium constant tells us the ratio of products to reactants at equilibrium. Since the reaction starts with only ICl, the products must form in equal amounts (one mole of I₂ and one mole of Cl₂ per two moles of ICl consumed). This symmetry simplifies the algebra.

Let’s walk through it step by step.

Write the balanced equation and the expression for $K_c$ . The reaction is:

2\text{ICl}(g) \rightleftharpoons \text{I}_2(g) + \text{Cl}_2(g)

The equilibrium constant expression is:

K_c = \frac{[\text{I}_2][\text{Cl}_2]}{[\text{ICl}]^2} = 0.14

Set up the change in concentrations using a variable. Let $x$ be the concentration of I₂ that forms at equilibrium. Because the stoichiometry is 1:1 for I₂ and Cl₂, the same $x$ applies to Cl₂:

[\text{I}_2] = x \quad \text{and} \quad [\text{Cl}_2] = x

For every mole of I₂ formed, 2 moles of ICl are consumed. So the change in ICl is $-2x$ , and its equilibrium concentration is:

[\text{ICl}] = 0.78 - 2x

Substitute into the $K_c$ expression.

0.14 = \frac{(x)(x)}{(0.78 - 2x)^2} = \frac{x^2}{(0.78 - 2x)^2}

Solve for $x$ by taking the square root of both sides. This is a neat shortcut — because both numerator and denominator are perfect squares, we avoid solving a quadratic:

\sqrt{0.14} = \frac{x}{0.78 - 2x}

Calculate $\sqrt{0.14}$ :

\sqrt{0.14} \approx 0.3742

So:

0.3742 = \frac{x}{0.78 - 2x}

Solve the linear equation. Multiply both sides by $(0.78 - 2x)$ :

0.3742(0.78 - 2x) = x

0.2919 - 0.7484x = x

0.2919 = x + 0.7484x = 1.7484x

x = \frac{0.2919}{1.7484} \approx 0.167\ \text{M}

Watch out

A common mistake is to forget that the denominator is squared. If you mistakenly write $K_c = \frac{x^2}{0.78 - 2x}$ , you’ll get a different (wrong) answer. Always check the exponent in the $K_c$ expression.

Check if the approximation is valid.

The change $2x = 0.334\ \text{M}$ is less than the initial 0.78 M, so the equilibrium concentration of ICl is positive: $0.78 - 0.334 = 0.446\ \text{M}$ . This is fine — no negative concentrations.
Compute the equilibrium concentrations precisely.

Let’s use more exact arithmetic to avoid rounding errors.

x = \frac{0.78 \cdot \sqrt{0.14}}{1 + 2\sqrt{0.14}}

With $\sqrt{0.14} = \sqrt{\frac{14}{100}} = \frac{\sqrt{14}}{10} \approx \frac{3.7417}{10} = 0.37417$ : …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 69 on this concept.

AP EAPCET 2026Set eng-2026-05-12-FN1 markMCQ
$K_{P_1}$
›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

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

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

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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AP EAPCET 2026Set eng-2026-05-12-AN1 markMCQ
$PCl_5$
›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

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 $=n_0(1+\alpha)$ .

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.

So $n_0(1+\alpha)=0.181 \Rightarrow (1+\alpha)=\dfrac{0.181}{0.1}\approx1.8 \Rightarrow \alpha\approx0.8$ .

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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AP EAPCET 2026Set eng-2026-05-13-FN1 markMCQ
$2HI(g) \rightleftharpoons H_2(g) + I_2(g)$
›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

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).

Mole fractions: $x_{\mathrm{HI}}=(1-\alpha)$ , $x_{\mathrm{H_2}}=x_{\mathrm{I_2}}=\alpha/2$ .

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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AP EAPCET 2026Set eng-2026-05-14-FN1 markMCQ
$A(g) + B(g) \rightleftharpoons_{T(K)} C(g) + D(g)$
›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

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 $[C] = 3[B]$ :
$x = 3(1-x)$ $x = 3 - 3x \Rightarrow 4x = 3 \Rightarrow x = 0.75$

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

Compute $K_c$ : …
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AP EAPCET 2026Set eng-2026-05-14-AN1 markMCQ
$\mathrm{H_2}$
›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

Reaction (as it physically occurred): $\mathrm{H_2 + I_2 \to 2HI}$ .

To form 10 mol HI, by 1:1:2 stoichiometry, $5$ mol $\mathrm{H_2}$ and $5$ mol $\mathrm{I_2}$ must have reacted.

Equilibrium amounts (in the same fixed volume, so ratios of moles work like concentrations): $[\mathrm{H_2}] \propto 15-5=10$ , $[\mathrm{I_2}]\propto 5.2-5=0.2$ , $[\mathrm{HI}]\propto 10$ .

$K_{forward} = \dfrac{[\mathrm{HI}]^2}{[\mathrm{H_2}][\mathrm{I_2}]} = \dfrac{10^2}{10\times0.2} = \dfrac{100}{2} = 50$ . …
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AP EAPCET 2026Set eng-2026-05-13-AN1 markMCQ
$H_2$
›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

Initial moles in 2.0 L: $H_2 = 1$ , $I_2 = 2$ .

At equilibrium, $H_2 = 0.2$ mol, so $1 - 0.2 = 0.8$ mol of $H_2$ reacted.

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.

Concentrations (divide by 2 L): $[H_2] = 0.1\ M$ , $[I_2] = 0.6\ M$ , $[HI] = 0.8\ M$ .

$K_c$ 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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AP EAPCET 2026Set eng-2026-05-15-AN1 markMCQ
$T(K)$
›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

Initial concentration: $[N_2O_4]_0 = \dfrac{2\ mol}{10\ L} = 0.2\ M$ .

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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AP EAPCET 2026Set eng-2026-05-15-FN1 markMCQ
$A(g) + B(g) \underset{}{\overset{T(K)}{\rightleftharpoons}} C(g)$
›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

Initial: $[A]_0=2$ , $[B]_0=1$ , $[C]_0=0$ (mol/L, since V=1L).

Let $x$ mol/L of A and B react to form $x$ mol/L of C (1:1:1 stoichiometry).

At equilibrium: $[A]=2-x$ , $[B]=1-x$ , $[C]=x$ .

Given constraint: $[B] = 2[C] \Rightarrow 1-x = 2x \Rightarrow 1 = 3x \Rightarrow x = \dfrac{1}{3}$ .

So $[A]=2-\dfrac13=\dfrac53$ , $[B]=1-\dfrac13=\dfrac23$ , $[C]=\dfrac13$ . …
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AP EAPCET 2026Set eng-2026-05-18-FN1 markMCQ
$2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g)$
›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

Forward (formation) reaction: $2SO_2(g)+O_2(g)\rightleftharpoons2SO_3(g)$ , $\Delta n_g = 2-(2+1) = -1$ .

$K_p = K_c(RT)^{\Delta n_g} = K_c(RT)^{-1} \Rightarrow K_c = K_p \times RT$ .

$RT = 0.083 \times 450 = 37.35$ (L bar/mol, matching $K_p$ 's bar units).

$K_c(\text{formation}) = 2.0\times10^{10} \times 37.35 = 7.47\times10^{11}$ . …
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AP EAPCET 2026Set ap-2026-05-19-FN1 markMCQ
$\mathrm{A(g) \underset{}{\overset{T(K)}{\rightleftharpoons}} B(g) + C(g)}$
›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

Initial: $[A]_0=1$ mol/L (1 mol in 1 L), $[B]_0=[C]_0=0$ .

At equilibrium: $[A]=1-x$ , $[B]=[C]=x$ .

Given $[A]=4[B]$ : $1-x=4x\Rightarrow x=0.2$ .

So $[A]=0.8$ , $[B]=[C]=0.2$ . …
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AP EAPCET 2026Set ap-2026-05-19-AN1 markMCQ
$K_c$
›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

Reaction: $2A(g) \rightleftharpoons B(g) + 2C(g)$ .

$\Delta n_g = (1+2) - 2 = 1$ .

$K_p = K_c (RT)^{\Delta n_g} = 10^{-3}\ \text{mol L}^{-1} \times (0.08\ \text{L bar mol}^{-1}\text{K}^{-1}\times 300\ \text{K})^{1}$ . …
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AP EAPCET 2026Set eng-2026-05-18-AN1 markMCQ
$2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g)$
›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

Reaction: $2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g)$ , $K_c = 100$ , $V = 10$ L.

Let moles of $SO_2$ at equilibrium $= a$ . Given moles of $SO_3 = 2a$ . Let moles of $O_2 = b$ .

Concentrations: $[SO_2] = a/10$ , $[SO_3] = 2a/10 = a/5$ , $[O_2] = b/10$ . …
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Q.What is the equilibrium concentration of each of the substances in the equilibrium when the initial concentration of ICl was 0.78 M ? 2ICl

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