How this instrument works
Kp is the equilibrium constant for a gas-phase reaction written in terms of partial pressures rather than concentrations. For a reaction where products and reactants are gases, Kp equals the product of each product's partial pressure raised to its stoichiometric coefficient, divided by the product of each reactant's partial pressure raised to its coefficient: Kp = (product pressures)^coefficients / (reactant pressures)^coefficients. It is the direct gas-phase counterpart of Kc, the same law-of-mass-action expression built from molar concentrations instead.
Kp is only meaningful once a reaction has actually reached equilibrium — the partial pressures you plug in must be the values measured (or calculated) at that settled state, not at some arbitrary point along the way. This is the detail that distinguishes it from a reaction quotient Q, which uses the identical mathematical form but accepts partial pressures from any moment, equilibrium or not, to tell you which direction a reaction still needs to shift. Kp is Q evaluated specifically at equilibrium; the two share a formula but answer different questions.
Chemists prefer Kp over Kc for gas-phase reactions because partial pressure is what's actually measured with a manometer or pressure gauge in many lab and industrial setups, without needing to first convert to molar concentration via the ideal gas law. Kp and Kc are related by Kp = Kc(RT)^Δn, where Δn is the change in moles of gas from reactants to products — they agree numerically only when Δn is zero, and diverge more as a reaction changes the total number of gas moles.
- Enter the first product's equilibrium partial pressure into Product 1 partial pressure (atm) and its balanced-equation coefficient into Product 1 coefficient.
- If a second product exists, fill in Product 2 partial pressure (atm, 1 if none) and Product 2 coefficient (0 if none); otherwise leave the pressure at 1 and coefficient at 0 so it drops out of the calculation.
- Enter the first reactant's equilibrium partial pressure into Reactant 1 partial pressure (atm) and its coefficient into Reactant 1 coefficient.
- Fill in Reactant 2 partial pressure (atm, 1 if none) and Reactant 2 coefficient (0 if none) the same way if a second reactant is involved.
- Read Equilibrium constant Kp beneath the inputs — it is unitless in the sense that pressures are treated as numbers relative to the standard-state 1 atm reference.
Worked example — a product whose coefficient is 2
Suppose a manometer reading at equilibrium shows a product sitting at 0.5 atm, and that product carries a coefficient of 2 in the balanced equation. Enter 0.5 into Product 1 partial pressure and 2 into Product 1 coefficient; leave Product 2 partial pressure at 1 and Product 2 coefficient at 0, since there's no second product to track. Enter 0.2 into Reactant 1 partial pressure with Reactant 1 coefficient of 1, and leave Reactant 2 partial pressure at 1 and Reactant 2 coefficient at 0. Kp comes out to 1.25.
Squaring the product's pressure first gives 0.5 x 0.5 = 0.25 (Product 2's unused slot contributes only 1^0 = 1, so it never touches the result); the reactant's pressure stays at its own first-power coefficient, so the denominator is just 0.2. Dividing the two, 0.25 / 0.2, lands on 1.25. Reading that number takes context — a Kp above 1 only says this particular equilibrium mixture has more product pressure than reactant pressure once each is raised to its coefficient, not that the reaction is complete or fast.
Questions
What's the difference between Kp and Kc?
Kp is built from partial pressures of gases; Kc is built from molar concentrations, and can apply to gas, liquid, or aqueous reactions. Both share the identical law-of-mass-action structure — products over reactants, each raised to its stoichiometric coefficient — but for a gas-phase reaction they're related by Kp = Kc(RT)^Δn, where Δn is the change in moles of gas. They agree numerically only when a reaction produces the same number of gas moles it consumes (Δn = 0).
What's the difference between Kp and a reaction quotient Qp?
The formula is identical; the timing is what differs. Kp is calculated strictly from partial pressures measured once a reaction has settled into true equilibrium. Qp uses the same expression but accepts partial pressures from any moment — before, during, or after equilibrium — to tell you which direction the reaction still needs to shift: Qp < Kp means the reaction will proceed toward products, Qp > Kp means it will shift toward reactants, and Qp = Kp means it has already arrived.
Why does this instrument treat missing products or reactants as pressure 1, coefficient 0?
Because raising any pressure to the power 0 gives 1, so a coefficient of 0 makes that term vanish from the multiplication regardless of what pressure sits beside it — setting the unused pressure field to 1 is simply the tidiest 'do nothing' value. This lets one instrument handle reactions with one or two products and one or two reactants without needing separate calculators for each stoichiometry.
Does a large Kp mean the reaction happens fast?
No — Kp describes the position of equilibrium, not the speed of getting there. A large Kp means that at equilibrium, products dominate over reactants; it says nothing about how quickly the reaction reaches that state, which is governed by kinetics (reaction rate and activation energy) rather than by the equilibrium constant.
Can Kp be used for reactions involving solids or liquids mixed with gases?
Yes, but only the gas-phase species appear in the expression. Pure solids and pure liquids have activities defined as 1 regardless of how much is present, so they're omitted from Kp entirely — only the partial pressures of gaseous reactants and products, each raised to its coefficient, go into the calculation.