Equilibrium Constant Calculator

What the equilibrium constant measures

A reversible reaction does not run to the end and stop. Reactants turn into products and products turn back, and the mixture settles where the two rates match. The equilibrium constant, K, says where it settles: a ratio of product amounts to reactant amounts, each raised to the power of its coefficient in the balanced equation.

The size of K tells you the ending story before you mix anything. A K of 64 means the mixture ends up rich in products; a K of 0.0002 means almost nothing happens. The table gives the rough bands.

Two habits to build early. First, K belongs to one balanced equation; change the coefficients and K changes with them. Second, K changes with temperature and with nothing else. A catalyst, a bigger flask or more starting material shifts how fast you arrive, not where you arrive.

Writing Kc from a balanced equation

For a reaction written as aA + bB going to cC + dD, the equilibrium expression is Kc = [C]^c x [D]^d / ([A]^a x [B]^b), where the square brackets are concentrations in moles per liter (M) at equilibrium. Coefficients in the equation become exponents in the expression. Pure solids and pure liquids are left out entirely, because their amounts do not change the balance.

Worked example. Hydrogen and iodine form hydrogen iodide: H2 + I2 going to 2 HI. Suppose the equilibrium mixture holds 0.20 M of H2, 0.20 M of I2 and 1.60 M of HI. The numerator is 1.60 x 1.60 = 2.56 and the denominator is 0.20 x 0.20 = 0.040, so Kc = 2.56 / 0.040 = 64. Products are favored.

Three properties of K follow straight from the expression.

Reverse the reaction and K flips: the reverse has K = 1 / 64 = 0.0156.

Double every coefficient and K is raised to that power: 64^2 = 4,096. This is why the K you quote must always be tied to the way the equation was written.

Add two steps and their K values multiply. A two-step route with K values of 4 and 9 has an overall K of 36.

If the equation involves gases and you are working in pressures, the same shape gives Kp with partial pressures in atmospheres instead of concentrations. The next section converts between them.

Kp, Kc and the (RT)^delta n bridge

Kp and Kc describe the same balance in different units, and they connect through Kp = Kc x (RT)^delta n. Here R is 0.08206 L atm per mol per K, T is the kelvin temperature, and delta n is the change in moles of gas: coefficients of gaseous products minus coefficients of gaseous reactants.

The ammonia row shows the arithmetic. RT = 0.08206 x 700 = 57.44. Since delta n is -2, the factor is 1 / (57.44 x 57.44) = 0.000303, and Kp = 0.50 x 0.000303 = 0.000152. Fewer gas molecules on the product side push Kp far below Kc at this temperature.

When delta n is zero, as in the hydrogen iodide reaction, Kp equals Kc and no conversion is needed. When delta n is positive, more gas appears on the product side and Kp comes out larger.

Keep the units straight inside the factor: the 0.08206 value pairs with pressures in atmospheres and volumes in liters. If your pressures are in bar or your volume in cubic meters, use the matching R or convert first, because a mixed R is the most common source of Kp errors off by factors of ten or more.

Solving for x with an ICE table

Often you know K and the starting amounts, and you want the equilibrium mixture. Set up an ICE table: Initial, Change, Equilibrium. For a reaction written as A going to B + C starting with only A at concentration c0, the changes are -x, +x and +x, so K = x^2 / (c0 - x). That is a quadratic, and the calculator solves it directly.

Worked example. K = 0.010 and the starting concentration is 0.10 M. The equation becomes x^2 / (0.10 - x) = 0.010, which rearranges to x^2 + 0.010x - 0.0010 = 0. The quadratic formula gives x = 0.0270 M, so at equilibrium [A] = 0.0730 M and [B] = [C] = 0.0270 M. That is a 27 percent conversion, which matches what a K near 0.01 suggests.

A weak acid, the same shape. Acetic acid has Ka = 0.000018. In a 0.10 M solution, x^2 / (0.10 - x) = 0.000018. Multiplying 0.000018 x 0.10 = 0.0000018, and the square root of that, 0.00134, is a good first guess. The full quadratic gives x = 0.00133 M, so about 1.3 percent of the acid is ionized and the pH lands near 2.88. The shortcut of taking the square root works whenever x is small next to c0, and the calculator shows you the exact answer so you can check the shortcut.

Two ways this goes wrong: if K is large, x approaches c0 and the square root shortcut collapses, so use the quadratic; and if product is already present at the start, the change row is not plus and minus x, so rebuild the ICE table first.

K and free energy: the -RT ln K link

Thermodynamics ties K to the standard free energy change through delta G standard = -RT x ln K, with R = 8.314 J per mol per K and T in kelvin. The link runs both ways: given K you get the free energy change, and given a free energy change you can recover K.

Worked example. Kc = 64 at 298 K. The natural log of 64 is 4.159, so delta G = -8.314 x 298 x 4.159 / 1000 = -10.3 kJ per mol. The negative sign says the reaction releases free energy as it approaches equilibrium from standard conditions, which fits a K above 1.

Read the table as a trade of scale: every factor of 100 in K moves delta G by about 11.4 kJ per mol at room temperature, and the sign flips with the direction. Ten kJ per mol sounds small next to the hundreds carried by bond energies, yet it is enough to swing a yield from a few percent to nearly complete, which is why the free energy scale is drawn in single kJ.

The standard in delta G standard matters. It refers to every species at 1 M or 1 atm, not to your actual flask. Free energy at nonstandard conditions is delta G = delta G standard + RT x ln Q, which is exactly the comparison the next section makes.

Q versus K: which way the reaction moves

The same product-over-reactant expression built from whatever sits in the beaker right now is called the reaction quotient, Q. It uses the formula of K with current numbers instead of equilibrium numbers. Comparing Q with K predicts the direction of travel.

Q below K means too few products, so the forward reaction runs and products build up. Q above K means the mixture has overshot, so the reverse reaction runs. Q equal to K means the mixture is already at equilibrium and the net change is zero.

Worked example. Take the hydrogen iodide reaction with Kc = 64, but mix 0.20 M of H2, 0.10 M of I2 and 0.50 M of HI. The numerator is 0.50 x 0.50 = 0.25, the denominator is 0.20 x 0.10 = 0.020, so Q = 0.25 / 0.020 = 12.5. Since 12.5 is below 64, the mixture makes more HI until the ratio climbs to 64.

This comparison is the practical use of K in industry and the lab. Sample the mixture, compute Q, and you know whether to wait, add reactant, or pull product out. Removing a product lowers Q and drives the reaction forward again, which is the everyday version of Le Chatelier's principle that the OpenStax equilibrium chapter develops with worked ICE tables.

Temperature, pressure and the usual traps

Four traps account for most wrong K answers.

Temperature is the only thing that changes K. Heating an exothermic reaction lowers K; heating an endothermic one raises it. The van't Hoff equation quantifies the shift, and for small temperature changes K moves by a predictable factor. Concentrations, pressure and catalysts never move K itself; they only move Q and the speed.

Solids and pure liquids never appear in K. Calcium carbonate decomposing to calcium oxide and carbon dioxide has K = P of CO2, one term only. Adding more solid changes nothing at equilibrium.

K has no units in the standard convention. Textbook values are dimensionless because each concentration is divided by the standard 1 M. When a problem hands you K with units attached, treat the number as the value to use and drop the units.

Match Kp or Kc to the data you have. Concentrations go with Kc, partial pressures in atmospheres go with Kp, and the RT bridge converts between them. The NIST Chemistry WebBook tabulates Kp and thermodynamic values for hundreds of gas reactions at stated temperatures.

Where equilibrium constants show up outside class

Equilibrium constants decide real things in a house and a body.

Pool and aquarium water. Chlorine disinfection, the carbon dioxide bicarbonate buffer and the calcium carbonate scale balance are all equilibrium systems. Test kits read the shifted position, and dosing moves Q, not K. The reason a pool resists pH change is the same K that the ICE table above solves.

Blood buffering. The carbonic acid pair in blood has Ka near 0.00000042, which holds blood pH in a narrow band around 7.4. Breathing changes how much carbon dioxide the system holds, which moves Q, and the buffer's K decides how much the pH can swing. The USGS water science school explains the pH side of that chemistry.

Baking soda and vinegar. The acid base reaction behind the classroom volcano has an enormous K, which is why it goes to completion and why no ICE table is needed for it. When a K is that large, equilibrium problems turn into limiting reactant problems.

Limescale and water softening. The calcium carbonate that crusts a kettle sits at a solubility equilibrium with a small K, so heating water that holds bicarbonate pushes the balance toward solid scale. Water softeners work by swapping ions so the K of the unwanted solid is never reached.

In every case the number in the table was measured once, at a stated temperature, and everything you see afterwards is the mixture finding its way back to that ratio.