chemical kinetics set 2

• 1) we use a graph of conc. vs. time
• 2) we find 2000 s and mark it
• 3) draw a tangent line
• 4) pick 2 points on the tangent line
• 5) derive a slope equation such that, for example: Δ[reactant] / Δ t

the rate of a rxn is based on the species that has a coefficient of one.

• Say we have NO's instantaneous rate of consumption to be -1.2 x 10^-6 M/s.
• We cannot use NO bc its coefficient is 2.  We can it convert though, to O₂'s rate.

For example, if we have an instantaneous rate (t = 2000s) of NO (consumption), we can translate that to O₂'s rate:

instantaneous rate of consumption of O2 = - (Δ[NO] / Δt) = - (-1.2 x 10^-6 M/s) = 6.0 x 10^-7 M/s

rate = (Δ[N₂] / Δt) = (Δ[CO₂] / Δt) = - (Δ[CO] / Δt) = - (Δ[NO] / Δt)

(Δ[N₂] / Δt) = - (Δ[CO] / Δ t)

The rate of formation of N₂ is one half the rate of consumption of CO

(Δ[CO₂] / Δt) = - (Δ[NO] / Δt)

The rate of formation of CO₂ is the same as the rate of consumption of NO

for the average rate of N₂O₅ consumption:

Δ[N₂O₅] / Δt

*for Δt, use a specific time interval (for avg. rate)

For example:

average rate = (Δ[N₂O₅]_final - Δ[N₂O₅]_inital) / (t_final - t_inital)

this will give us the rate for the rxn (for the time interval stated) also because the coefficient for N₂O₅ is one.