Measured kinetics and mechanisms
Введено| English | Значение |
|---|---|
| mechanism/ˈmekənɪzəm/ | механизм реакции |
| tangent/ˈtændʒənt/ | касательная |
| half-life/hɑːf laɪf/ | период полураспада |
| rate law/reɪt lɔː/ | закон скорости |
What changes the speed?
- Two reactants can appear in the overall equation but have different effects on rate. The balanced equation alone does not supply their orders.
- Compare controlled measurements at the same temperature before proposing a mechanism 反应机理.
Infer each order separately
- Doubling A while holding B fixed doubles rate: first order in A. Doubling B while holding A fixed quadruples rate: second order in B.
- The law is r = k[A][B]² and overall order is three. Rearrange to calculate k and derive its units from the measured quantities.
Doubling A quadruples rate at fixed B. What is the order in A?
Difficulty 1/5 · 100 coins + a small bonus on first completion today
A factor of four equals two squared, so the order is two.
Read a concentration trace
- A straight falling concentration–time line supports zero order. Constant successive half-lives support first order at fixed conditions.
- Read a tangent 切线 for an instantaneous rate; a chord gives an interval average. One measured half-life 半衰期 does not establish that later half-lives are equal.
A gradient of −6000 K and R = 8.31 J per mole per kelvin gives Ea = −R × gradient = 49,860 J per mole = 49.9 kJ per mole.
A first-order reactant halves every 40 s. Starting at 0.160 mol per cubic decimetre, what is its concentration after 120 s?
Difficulty 2/5 · 140 coins + a small bonus on first completion today
Three half-lives give 0.160 → 0.080 → 0.040 → 0.020.
Name the line whose slope gives instantaneous rate at a point on a curved concentration–time graph.
Difficulty 1/5 · 100 coins + a small bonus on first completion today
A tangent measures the slope at that instant; a chord averages over an interval.
Test a mechanism against evidence
- The proposed steps must sum to the overall equation and agree with the rate law 速率定律. Intermediates cancel between their formation and consumption steps.
- S_N1 has slow ionisation in the simple model; S_N2 involves nucleophile attack in the slow concerted step. Agreement supports a mechanism without proving it unique.
A rate law consistent with one proposed mechanism proves that no other mechanism is possible.
Difficulty 1/5 · 100 coins + a small bonus on first completion today
Different sequences can yield the same law. Stoichiometric and rate consistency support rather than uniquely prove a mechanism.
Use temperature data correctly
- With supplied $\ln k=\ln A-E_a/(RT)$, the gradient of ln k against 1/T equals $-E_a/R$. Use kelvin and consistent energy units.
- A solid catalyst offers surface sites; a poison can block them. A catalyst changes rate but leaves the equilibrium constant unchanged at fixed temperature.
“The overall coefficient is always the reaction order” is false. Orders are measured, and a mechanism can contain more than one step.
Separate controlled data, a supported rate law and a possible mechanism. Each is a different claim.
Which two checks are needed before using an Arrhenius gradient?
Difficulty 2/5 · 140 coins + a small bonus on first completion today
Kelvin and the actual axis scaling/energy units are necessary. Catalysts do not change K at fixed temperature.