One reaction, three rates?

2N₂O₅ → 4NO₂ + O₂ in a vessel with three concentration curves; a tangent shows each species' d[ ]/dt and the same rate after dividing by coefficients.

Move the tangent to the moment when the instantaneous rate equals the average rate over the first 100 min.

Equating −d[N₂O₅]/dt with d[NO₂]/dt without dividing by coefficients.

• Rate is the change in concentration of a reactant or product per unit time. Concentration is in mol L⁻¹ and time in s, min or h, so rate has units such as mol L⁻¹ s⁻¹; for gases, pressure in atm can replace concentration, giving atm s⁻¹.
• For R → P, rate = −Δ[R]/Δt = +Δ[P]/Δt. The minus sign makes the rate positive, because [R] falls.
• Average rate is the change over a finite interval. It hides the fact that the rate itself changes during that interval, usually falling as reactants are used up.
• Instantaneous rate is the limit as Δt → 0: r = −d[R]/dt = d[P]/dt. On a concentration-time graph it is the slope of the tangent at that moment.
• When coefficients differ, divide each rate of change by its coefficient so that every species gives the same number. For 2HI → H₂ + I₂: rate = −½ d[HI]/dt = d[H₂]/dt = d[I₂]/dt.
• For 2N₂O₅ → 4NO₂ + O₂, rate = −½ Δ[N₂O₅]/Δt = ¼ Δ[NO₂]/Δt = Δ[O₂]/Δt. If [N₂O₅] falls from 2.33 to 2.08 mol L⁻¹ in 184 min, the reaction rate is ½ × 0.25/184 = 6.79 × 10⁻⁴ mol L⁻¹ min⁻¹, and NO₂ forms at four times that, 2.72 × 10⁻³ mol L⁻¹ min⁻¹.
• Rates span a huge range: ionic precipitation (such as AgCl from Ag⁺ and Cl⁻) is almost instantaneous, the inversion of cane sugar and hydrolysis of starch go at a moderate pace, and the rusting of iron in moist air is very slow.

Teaching this idea? Put the simulation on your own page or LMS with this code.