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Chemical Equilibrium and Kc

What is chemical equilibrium and how do you calculate Kc?

IntermediateEquilibriumLast reviewed 4 October 2026
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What is it?

Many reactions do not go to completion. They are reversible: the products can react to re-form the reactants. In a closed container, a reversible reaction reaches chemical equilibrium:

HX2(g)+IX2(g)⇌2 HI(g)\ce{H2(g) + I2(g) <=> 2HI(g)}

At equilibrium the forward and reverse reactions continue at equal rates, so the concentrations of reactants and products stay constant. This is a dynamic equilibrium: molecules keep reacting in both directions, but there is no overall change.

Key idea

At equilibrium, the concentrations are constant, not equal. The ratio between them is fixed by the equilibrium constant, KcK_\text{c}.

Why does it matter?

  • Yield. Equilibrium sets the maximum amount of product a reversible reaction can give, which matters in industry (ammonia, sulfuric acid, methanol).
  • Acids, bases and solubility. KaK_\text{a} for weak acids, buffers and solubility are all equilibrium constants.
  • The body. Oxygen binding to haemoglobin and the carbonic acid buffer in blood are equilibria.

How does it work?

1. The equilibrium constant, Kc

For a general reaction a A+b B⇌c C+d D\ce{aA + bB <=> cC + dD}, the equilibrium constant expression is:

Kc=[C]c [D]d[A]a [B]bK_\text{c} = \frac{[\ce{C}]^c\,[\ce{D}]^d}{[\ce{A}]^a\,[\ce{B}]^b}
  • Products go on top, reactants on the bottom.
  • Each concentration (in mol/L) is raised to the power of its coefficient in the balanced equation.
  • Pure solids and pure liquids are left out (their “concentration” does not change). For CaCOX3(s)⇌CaO(s)+COX2(g)\ce{CaCO3(s) <=> CaO(s) + CO2(g)}, Kc=[COX2]K_\text{c} = [\ce{CO2}].
  • KcK_\text{c} is constant at a given temperature; it changes only when the temperature changes.

Units of Kc. Strictly, each concentration is divided by the standard concentration c°=1 Mc° = 1\ \text{M}, so the units cancel and KcK_\text{c} has no unit. The worked examples show this step.

2. What the size of K tells you

Value of KcK_\text{c}At equilibrium the mixture contains…
very large (e.g. 101010^{10})mostly products: the reaction goes almost to completion
about 1significant amounts of both
very small (e.g. 10−1010^{-10})mostly reactants: very little reaction
  • Reverse the equation: Kreverse=1KK_\text{reverse} = \dfrac{1}{K}
  • Multiply the equation by nn: Knew=K nK_\text{new} = K^{\,n} (for example, halving the coefficients gives K\sqrt{K})

4. The reaction quotient, Q

QQ has the same form as KK, but uses the concentrations at any moment, not only at equilibrium. Comparing QQ with KK predicts the direction of change:

ComparisonWhat happens
Q<KQ < Ktoo few products: the reaction goes forward (→)
Q=KQ = Kat equilibrium: no net change
Q>KQ > Ktoo many products: the reaction goes in reverse (←)

5. ICE tables

To find equilibrium concentrations, use an ICE table: Initial, Change, Equilibrium. The changes follow the mole ratio of the equation, written in terms of an unknown, xx.

Think of it like this

Picture two shops connected by a door, with customers walking both ways. After a while, as many people walk from shop A to shop B each minute as walk back. The number of people in each shop stays the same, even though people never stop moving. That’s dynamic equilibrium, and the shops need not hold equal numbers.

More precisely

The thermodynamic equilibrium constant is defined with activities, which for dilute solutions are approximately [X]/c°[\ce{X}]/c° and for gases pX/p°p_\text{X}/p°. That is why equilibrium constants are dimensionless. For gases, KpK_\text{p} uses partial pressures instead of concentrations; KpK_\text{p} and KcK_\text{c} are equal only when the number of moles of gas is the same on both sides.

Visualise it

Two graphs for H2 + I2 forming 2HI, starting with 0.100 M H2 and 0.100 M I2 and no HI, with Kc = 49.0. Top: the concentration of H2 (equal to I2) falls and levels off at 0.0222 M, while HI rises and levels off at 0.156 M; the shaded region marks equilibrium, where the concentrations are constant. Bottom: the forward rate starts high and falls, the reverse rate starts at zero and rises, and they become equal at equilibrium.
At equilibrium the concentrations stop changing because the forward and reverse rates are equal. Calculated from a simple rate model.

Worked example

Worked example: Calculating Kc

Question: At a certain temperature, an equilibrium mixture of NX2OX4(g)⇌2 NOX2(g)\ce{N2O4(g) <=> 2NO2(g)} contains [NX2OX4]=0.0450[\ce{N2O4}] = 0.0450 M and [NOX2]=0.0300[\ce{NO2}] = 0.0300 M. Calculate KcK_\text{c}.

  1. Write the expression: Kc=[NOX2]2[NX2OX4]K_\text{c} = \dfrac{[\ce{NO2}]^2}{[\ce{N2O4}]}

  2. Substitute, dividing each concentration by c°=1c° = 1 M:

    Kc=(0.0300 M/1 M)20.0450 M/1 M=(0.0300)20.0450=0.0200\begin{aligned} K_\text{c} &= \frac{(0.0300\ \text{M} / 1\ \text{M})^2}{0.0450\ \text{M} / 1\ \text{M}} \\[4pt] &= \frac{(0.0300)^2}{0.0450} = 0.0200 \end{aligned}

  3. The units cancel: Kc=0.0200K_\text{c} = \textbf{0.0200}.

Worked example: Predicting the direction with Q

Question: For HX2(g)+IX2(g)⇌2 HI(g)\ce{H2(g) + I2(g) <=> 2HI(g)}, Kc=49.0K_\text{c} = 49.0. A mixture contains 0.050 M HX2\ce{H2}, 0.050 M IX2\ce{I2} and 0.20 M HI. Is it at equilibrium? If not, which way will it shift?

  1. Q=(0.20 M)2(0.050 M)(0.050 M)=16Q = \dfrac{(0.20\ \text{M})^2}{(0.050\ \text{M})(0.050\ \text{M})} = 16 (the units of M² cancel)
  2. Q=16<Kc=49.0Q = 16 < K_\text{c} = 49.0, so there is too little HI: the reaction goes forward, making more HI.

Worked example: An ICE table

Question: 0.100 M HX2\ce{H2} and 0.100 M IX2\ce{I2} are mixed with no HI (Kc=49.0K_\text{c} = 49.0). Find the equilibrium concentrations.

  1. ICE table, where xx is the decrease in [HX2][\ce{H2}], in M:

    HX2\ce{H2}IX2\ce{I2}HI\ce{HI}
    Initial0.100 M0.100 M0 M
    Change−x−x+2x
    Equilibrium0.100 M − x0.100 M − x2x

  2. Substitute: 49.0=(2x)2(0.100 M−x)249.0 = \dfrac{(2x)^2}{(0.100\ \text{M} - x)^2}. Both sides are perfect squares, so take the square root: 7.00=2x0.100 M−x7.00 = \dfrac{2x}{0.100\ \text{M} - x}

  3. 0.700 M−7.00x=2x0.700\ \text{M} - 7.00x = 2x, so x=0.700 M9.00=0.0778 Mx = \dfrac{0.700\ \text{M}}{9.00} = 0.0778\ \text{M}

  4. [HI]=2x=[\ce{HI}] = 2x = 0.156 M; [HX2]=[IX2]=0.100 M−0.0778 M=[\ce{H2}] = [\ce{I2}] = 0.100\ \text{M} - 0.0778\ \text{M} = 0.0222 M

  5. Check: (0.156 M)2(0.0222 M)2=49.4≈49.0\dfrac{(0.156\ \text{M})^2}{(0.0222\ \text{M})^2} = 49.4 \approx 49.0 (the small difference comes from rounding).

Common mistake

Common mistake: Thinking equilibrium means equal amounts

At equilibrium the concentrations are constant, not equal. In the example above there is seven times as much HI as HX2\ce{H2}.

Common mistake: Forgetting the coefficients as powers

For HX2+IX2⇌2 HI\ce{H2 + I2 <=> 2HI}, Kc=[HI]2[HX2][IX2]K_\text{c} = \dfrac{[\ce{HI}]^2}{[\ce{H2}][\ce{I2}]}, not 2[HI][HX2][IX2]\dfrac{2[\ce{HI}]}{[\ce{H2}][\ce{I2}]}. Coefficients become exponents.

Common mistake: Including solids, liquids or water

Pure solids and liquids (including water as the solvent) do not appear in KcK_\text{c}. That’s why KaK_\text{a} has no [HX2O][\ce{H2O}] term.

Notation note

  • The double half-arrow ⇌ shows a reversible reaction at equilibrium.
  • Square brackets mean concentration in mol/L: [HI][\ce{HI}] = concentration of HI.
  • “Position of equilibrium” describes whether reactants or products are favoured; it can change, but KK changes only with temperature.

Remember this

Remember this

  • Dynamic equilibrium: forward rate = reverse rate; concentrations constant, not equal.
  • KcK_\text{c} = products over reactants, each to the power of its coefficient; leave out solids and liquids.
  • Q<KQ < K: forward. Q>KQ > K: reverse. Q=KQ = K: at equilibrium.
  • Use an ICE table; KcK_\text{c} depends only on temperature.

Test yourself

Check your understanding before moving on.

Flashcards

10 cards

  1. Question
    What is dynamic equilibrium?
    Answer

    The state in which the forward and reverse reactions continue at equal rates, so concentrations stay constant.

  2. Question
    Are reactant and product concentrations equal at equilibrium?
    Answer

    No. They are constant, not equal.

  3. Question
    Write KcK_\text{c} for NX2+3 HX2⇌2 NHX3\ce{N2 + 3H2 <=> 2NH3}.
    Answer

    Kc=[NHX3]2[NX2][HX2]3K_\text{c} = \dfrac{[\ce{NH3}]^2}{[\ce{N2}][\ce{H2}]^3}

  4. Question
    Which substances are left out of KcK_\text{c}?
    Answer

    Pure solids and pure liquids (including water as the solvent).

  5. Question
    What does a very large KcK_\text{c} mean?
    Answer

    At equilibrium the mixture is mostly products: the reaction goes almost to completion.

  6. Question
    If Q<KQ < K, which way does the reaction go?
    Answer

    Forward (→), making more products until Q=KQ = K.

  7. Question
    If K for a reaction is 800, what is K for the reverse reaction?
    Answer

    1800=1.25×10−3\dfrac{1}{800} = 1.25 \times 10^{-3}

  8. Question
    What is the only factor that changes the value of KcK_\text{c}?
    Answer

    Temperature.

  9. Question
    What do the letters in an ICE table stand for?
    Answer

    Initial, Change, Equilibrium (concentrations, in M).

  10. Question
    Why does KcK_\text{c} have no unit?
    Answer

    Each concentration is divided by the standard concentration c°=1 Mc° = 1\ \text{M}, so the units cancel.

Quiz

7 questions

  1. Question 1EasyWhich statement describes a system at chemical equilibrium?
    Show answer

    Answer: The forward and reverse rates are equal, and concentrations are constant

    Equilibrium is dynamic: both reactions continue at the same rate, so concentrations stay constant, but they are generally not equal.

  2. Question 2EasyWhat is KcK_\text{c} for 2 SOX2(g)+OX2(g)⇌2 SOX3(g)\ce{2SO2(g) + O2(g) <=> 2SO3(g)}?
    Show answer

    Answer: [SOX3]2[SOX2]2[OX2]\dfrac{[\ce{SO3}]^2}{[\ce{SO2}]^2[\ce{O2}]}

    Products over reactants, each raised to the power of its coefficient.

  3. Question 3MediumWhat is KcK_\text{c} for CaCOX3(s)⇌CaO(s)+COX2(g)\ce{CaCO3(s) <=> CaO(s) + CO2(g)}?
    Show answer

    Answer: [COX2][\ce{CO2}]

    Pure solids are left out of the expression, so only the gas remains.

  4. Question 4MediumAn equilibrium mixture of NX2OX4⇌2 NOX2\ce{N2O4 <=> 2NO2} contains 0.0450 M NX2OX4\ce{N2O4} and 0.0300 M NOX2\ce{NO2}. What is KcK_\text{c}?
    Show answer

    Answer: 0.0200

    Kc=(0.0300 M)2/(0.0450 M)K_\text{c} = (0.0300\ \text{M})^2 / (0.0450\ \text{M}), with each concentration divided by 1 M: 0.000900 ÷ 0.0450 = 0.0200. The answer 0.667 forgets to square [NO₂].

  5. Question 5MediumFor a reaction, Kc=49.0K_\text{c} = 49.0 and Q=16Q = 16. What happens?
    Show answer

    Answer: The reaction goes forward, making more products

    Q<KQ < K means there are too few products, so the reaction goes forward until Q=KQ = K.

  6. Question 6HardFor 2 SOX2+OX2⇌2 SOX3\ce{2SO2 + O2 <=> 2SO3}, Kc=800K_\text{c} = 800. What is KcK_\text{c} for SOX2+12 OX2⇌SOX3\ce{SO2 + 1/2 O2 <=> SO3}?
    Show answer

    Answer: 28.3

    Halving all the coefficients raises K to the power ½: √800 = 28.3. The answer 0.00125 is K for the reverse reaction.

  7. Question 7Hard0.100 M HX2\ce{H2} and 0.100 M IX2\ce{I2} react: HX2+IX2⇌2 HI\ce{H2 + I2 <=> 2HI}, Kc=49.0K_\text{c} = 49.0. What is [HI] at equilibrium?
    Show answer

    Answer: 0.156 M

    Taking the square root of 49.0 = (2x)² / (0.100 M − x)² gives 7.00 = 2x / (0.100 M − x), so x = 0.0778 M and [HI] = 2x = 0.156 M. 0.0778 M is x itself.

Notes and downloads

References

  1. Brown, T. L.; LeMay, H. E., Jr.; Bursten, B. E.; Murphy, C. J.; Woodward, P. M.; Stoltzfus, M. W. Chemistry: The Central Science, 15th ed.; Pearson, 2022.

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