Reaction Quotient Calculator
Calculate the reaction quotient Q and determine which direction a reaction will shift to reach equilibrium. Compare Q to Keq with step-by-step solutions.
Reaction Quotient Calculator
Calculator
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Formula: Q = [C]^c [D]^d / [A]^a [B]^b
Additional inputs: Keq.
Worked example โ Q = 12.5, Reaction shifts in reverse
Formula
Q = [C]^c [D]^d / [A]^a [B]^b
The reaction quotient Q uses the same expression as Keq but with current (non-equilibrium) concentrations. If Q < Keq, the reaction shifts forward. If Q > Keq, it shifts in reverse. If Q = Keq, the system is at equilibrium.
Worked Examples
Example 1: Comparing Q to Keq
Problem:For N2 + 3H2 -> 2NH3, Keq = 0.500. Current: [N2] = 0.200, [H2] = 0.100, [NH3] = 0.050. Find Q and direction.
Solution:Q = [NH3]^2 / ([N2] * [H2]^3) Q = (0.050)^2 / (0.200 * (0.100)^3) Q = 0.0025 / 0.0002 = 12.5 Q (12.5) > Keq (0.500) => Reverse shift
Result:Q = 12.5, Reaction shifts in reverse
Example 2: System at Equilibrium Check
Problem:For H2 + I2 -> 2HI, Keq = 54.3. [H2] = 0.01, [I2] = 0.01, [HI] = 0.0737.
Solution:Q = [HI]^2 / ([H2]*[I2]) Q = (0.0737)^2 / (0.01 * 0.01) Q = 0.005432 / 0.0001 = 54.3 Q = Keq => At equilibrium
Result:Q = 54.3 = Keq, System is at equilibrium
Frequently Asked Questions
What is the reaction quotient Q?
The reaction quotient Q is a measure of the relative amounts of products and reactants present in a reaction mixture at any point in time. It has the same mathematical form as the equilibrium constant Keq, but uses current concentrations rather than equilibrium concentrations. For the reaction aA + bB -> cC + dD, Q = [C]^c[D]^d / [A]^a[B]^b. Comparing Q to Keq predicts which direction the reaction will shift to reach equilibrium. Q is dimensionless when activities are used but may have units when concentrations are used directly.
How do Q and Keq determine reaction direction?
When Q < Keq, there are relatively too many reactants and too few products compared to equilibrium, so the reaction proceeds forward (toward products). When Q > Keq, there are too many products, so the reaction shifts in reverse (toward reactants). When Q = Keq, the system is at equilibrium and no net reaction occurs. The farther Q is from Keq, the greater the driving force for the reaction to shift. This principle is known as Le Chatelier principle applied quantitatively to concentration changes.
Why are pure solids and liquids excluded from Q?
Pure solids and pure liquids have constant concentrations (or more precisely, constant activities equal to 1) regardless of the amount present. Since their concentrations do not change during the reaction, they do not affect the position of equilibrium and are omitted from the Q expression. For example, in CaCO3(s) -> CaO(s) + CO2(g), Q = P(CO2) only. Water as a solvent in aqueous reactions is also typically excluded because its concentration (55.5 M) remains essentially constant even as small amounts react.
What is the relationship between Q, Keq, and Gibbs free energy?
The Gibbs free energy change under non-standard conditions is related to Q by the equation delta-G = delta-G-standard + RT ln(Q). Since delta-G-standard = -RT ln(Keq), this simplifies to delta-G = RT ln(Q/Keq). When Q < Keq, delta-G is negative (spontaneous forward reaction). When Q > Keq, delta-G is positive (spontaneous reverse reaction). When Q = Keq, delta-G = 0 (equilibrium). This equation connects thermodynamics to the kinetic approach of predicting reaction direction.
What is the difference between Qc and Qp?
Qc is built from molar concentrations of dissolved or gaseous species, while Qp uses partial pressures of the gases and is the natural choice for gas-phase equilibria. They are related through the ideal gas law: Kp = Kc x (RT) raised to delta n, where delta n is the moles of gaseous product minus the moles of gaseous reactant. When delta n is zero the two are numerically equal. Always compare Qc with Kc and Qp with Kp, never mix the two.
How does adding an inert gas affect the reaction quotient?
At constant volume, adding argon or another non-reacting gas raises the total pressure but leaves every partial pressure and every concentration unchanged, so Q does not move and there is no shift. If instead the total pressure is held constant, the vessel must expand, every partial pressure falls, and Q changes unless delta n is zero. In that case the system shifts toward whichever side has more moles of gas, exactly as if it had been diluted.
What happens to Q when an aqueous equilibrium is diluted?
Diluting divides every concentration by the same factor, so Q changes by that factor raised to the power of delta n, the moles of dissolved product minus moles of dissolved reactant. If the product side has more dissolved particles, dilution lowers Q below Keq and the reaction shifts forward. This is why weak acids ionise more completely in dilute solution, and why a dilution step alone can move a system that was previously at equilibrium.
How does removing a product drive a reaction forward?
Taking product out of the mixture lowers the numerator of the Q expression, pushing Q below Keq, so the system responds by making more product. Industrial processes exploit this continuously: condensing out a liquid product, distilling off a volatile one, or precipitating an insoluble salt keeps Q permanently below Keq, so a reaction with an unfavourable equilibrium constant can still be driven to high conversion by removing what it makes.
How is the reaction quotient used to predict precipitation?
For a sparingly soluble salt the quotient is the ion product Qsp, formed from the current ion concentrations raised to their stoichiometric powers, and it is compared with the solubility product Ksp. If Qsp is less than Ksp the solution is unsaturated and any solid dissolves; if Qsp equals Ksp the solution is exactly saturated; if Qsp exceeds Ksp the solution is supersaturated and a precipitate should form, though nucleation can delay it.
Does the size of Q tell you how fast equilibrium will be reached?
No. Q and Keq are thermodynamic quantities: their comparison tells you the direction of net change and, through the free energy relationship, the driving force, but nothing about the speed. A mixture with Q far below Keq can sit unchanged for years if the activation barrier is high, which is why hydrogen and oxygen coexist at room temperature. A catalyst speeds the approach to equilibrium but changes neither Q nor Keq nor the final composition.
References
Background & Theory
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Reviewed for accuracy by Manoj Kumar, Mathematics Educator ยท Editorial policy
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