Measure reaction rates over a chosen interval

Measure a change and specify its time interval

Suppose a reactant concentration falls from 1 to 0.5 mol/L over ten seconds. The amount consumed per unit volume is 0.5 mol/L. Dividing by ten seconds gives an average disappearance rate of 0.05 mol/(L s). A concentration and a rate answer different questions: how much is present, and how quickly it changes.

In this investigation, fictional A becomes B in a 1:1 reaction at constant volume. A starts at 1 mol/L and B at zero. The model assigns a first-order law with a ten-second half-life. It generates exact synthetic values; the table is not a record of measurements. Selecting a later interval changes which portion of the same reaction you inspect, rather than restarting the reaction.

Read signs from the graph

Write average disappearance rate = ([A]start − [A]end)/(tend − tstart). Brackets mean concentration in mol/L; t means elapsed time in seconds. The denominator is the interval duration, not the final clock reading. A falling reactant curve has negative signed slope, Δ[A]/Δt. Its disappearance rate is the positive opposite of that slope.

The solid curve is A and the dashed curve is B. Two dots mark the selected endpoints; the straight dotted segment connects them. Its slope summarizes the whole interval. The curved trace need not have that slope at every point. The rate chart separately shows the instantaneous start rate, interval average and instantaneous end rate, all on one fixed scale.

From 0 to 10 s, A decreases from 1 to 0.5 mol/L: average disappearance rate is 0.05 mol/(L s). From 10 to 20 s, it decreases from 0.5 to 0.25: the rate is 0.025 mol/(L s). Equal durations do not guarantee equal rates. In this particular model, less reactant remains and the rate declines.

Do not confuse the window with the whole reaction

Select start 10 s and duration 20 s. The endpoint is 30 s, where A is 0.125 mol/L. The interval consumes 0.5 − 0.125 = 0.375 mol/L, so its average rate is 0.01875 mol/(L s). Total B at 30 s is 0.875 mol/L, because 0.5 mol/L had already formed before this observation window began. Dividing total product by the selected duration would mix incompatible boundaries.

For this 1:1, constant-volume reaction, B’s formation rate equals A’s disappearance rate. That numerical equality is not universal. In a constant-volume reaction 2 A → B, consuming 0.2 mol/L A creates only 0.1 mol/L B. A stoichiometrically normalized reaction rate divides each species’ disappearance or formation rate by its coefficient. Do not change the widget’s 1:1 conservation ledger to represent that different reaction without changing its model.

Interpret an instantaneous rate carefully

An instantaneous rate describes a local slope at a particular time; the widget calculates it from its stipulated equation. A measured pair of finite-time concentrations estimates an average instead. Shortening an interval can make its average closer to a local rate, but measurement noise can then become large relative to the small concentration difference. A finer interval does not automatically produce more trustworthy experimental evidence.

Product amount continues to rise here while reaction rate falls. A slow final portion can coexist with a large accumulated yield. Neither observation establishes reaction enthalpy, activation energy or equilibrium behavior.

Transfer: calculate the average disappearance rate from 20 to 40 s using A = 0.25 and 0.0625 mol/L. Then calculate the increase in B over this interval and distinguish it from total B at 40 s.

Check the interval-rate transfer

The consumed concentration is 0.1875 mol/L. Divide by 20 s to obtain 0.009375 mol/(L s). B increases by 0.1875 mol/L during the interval but totals 0.9375 mol/L at 40 s. A signed concentration slope would be negative; the disappearance rate is positive.

OpenStax reaction rates provides the species-rate and stoichiometric conventions.