Explain equilibrium with two continuing reactions

Constant concentration does not mean no reaction

Imagine two forms of a fictional substance, A and B, interconverting in a closed vessel: A ⇌ B. At a given instant, some A becomes B while some B becomes A. If both transfers occur at the same rate, the net amount of each form stays constant. Individual transformations can continue even though a concentration measurement does not change.

The widget fixes total concentration at [A] + [B] = 1 mol/L. Brackets denote concentration. Choosing a different initial B redistributes that fixed total; it is not an injection of additional B into a running experiment. The vessel’s volume and temperature are fixed, and the model assumes no other reactions.

Calculate both directions before taking their difference

This model assigns forward rate rf = kf[A] and reverse rate rr = kr[B]. The rate constants are kf = 0.03 s⁻¹ and kr = 0.01 s⁻¹. Multiplying either by concentration in mol/L gives a rate in mol/(L s). The net formation rate of B is rf − rr; net formation of A has the opposite sign.

Start with [B] = 0 and therefore [A] = 1 mol/L. Forward rate is 0.03 mol/(L s), reverse rate is zero, and B initially increases. As B accumulates, its reverse rate grows while the forward rate falls because A is being depleted.

Start instead with [B] = 1 and [A] = 0. Forward rate is zero and reverse rate is 0.01 mol/(L s), so B decreases. This initial reverse direction is visible without calculating a concentration ratio. Both histories approach the same composition because their total, temperature and assigned rate constants are the same.

Solve the balance rather than assuming equal amounts

At equilibrium, 0.03[A] = 0.01[B]. Therefore [B] = 3[A]. Combine this with [A] + [B] = 1: four times [A] equals 1, so [A] = 0.25 and [B] = 0.75 mol/L.

Now calculate the rates: 0.03 × 0.25 = 0.0075 and 0.01 × 0.75 = 0.0075 mol/(L s). They are equal and nonzero. The concentrations are unequal. A system can favor one form at equilibrium while continuously exchanging material in both directions.

Select initial B = 0.75 mol/L and move the time control. The concentrations remain fixed from the start, while the rate chart continues to show the two positive rates. This is the clearest counterexample to the claim that equilibrium means everything stops.

Compare a current ratio with the equilibrium ratio

For this ideal 1:1 system, the reaction quotient is Q = [B]/[A]. It describes the current state. The equilibrium constant is K = 3 for the assigned conditions. Q below 3 corresponds here to net A → B; Q above 3 corresponds to net B → A. Q equal to 3 gives balanced rates.

For example, equal concentrations of 0.5 mol/L give Q = 1, forward rate 0.015 and reverse rate 0.005 mol/(L s). The mixture is not at equilibrium: B increases at a net 0.010 mol/(L s). Equal amounts are insufficient evidence of equal rates.

At the all-B starting point, dividing by [A] = 0 is undefined. The display reports that missing finite ratio rather than silently substituting zero. The rates still give an unambiguous initial reverse direction. At subsequent positive times, A is present and Q becomes finite.

Concentration units cancel in this particular 1:1 ratio. General thermodynamic equilibrium constants use dimensionless activities, and other stoichiometries require different exponents and expressions. The ratio kf/kr follows directly for the two assigned first-order directions here; do not apply that shortcut to arbitrary overall reaction mechanisms.

Read a late-time curve honestly

The concentration solution used by the widget is [B]t = 0.75 + ([B]0 − 0.75) exp(−0.04t), with t in seconds and concentrations in mol/L. exp is the exponential function; its argument is dimensionless. The deviation from 0.75 shrinks continuously. You can understand the guided investigation through the rate balance without deriving this equation.

From an all-A start, B at 120 s is approximately 0.743828 mol/L; from an all-B start it is approximately 0.752057 mol/L. Neither is exactly 0.75. The reference row is the calculated equilibrium state, not an extra measurement at 120 s. Rounding or a visually flat curve must not manufacture an exact equilibrium claim.

These results are synthetic predictions. A real test needs measured concentrations, stable conditions and sufficient precision to detect changes. Constant observations alone also need interpretation: a system may be changing too slowly to resolve. This closed two-state model does not represent an open-system steady state, a temperature change or competing reactions.

Transfer: a different vessel contains total A + B = 2 mol/L under the same ideal rate constants. Calculate its equilibrium concentrations and both rates. Explain why the equilibrium ratio can remain the same while the concentrations and rates change.

Check the equilibrium transfer

B = 3A and A + B = 2 give A = 0.5 and B = 1.5 mol/L. Each rate is 0.015 mol/(L s), twice the original equilibrium rate. The ratio B/A remains 3. This is a written extension beyond the widget's fixed total, not a control setting that adds material.

Reference explanations: OpenStax chemical equilibria and equilibrium constants.