Higher Tier, Chemistry-only: choose a reaction pathway from several criteria
مُقدَّم| English | المعنى |
|---|---|
| reaction pathway/rɪˈækʃn ˈpæθweɪ/ | مسار التفاعل |
| process criterion/ˈprəʊses kraɪˈtɪərɪən/ | معيار العملية |
What would explain this observation?
- Higher Tier: A high atom-economy pathway may be slow or give a low recovered yield. A justified industrial choice compares the relevant evidence and states the decision priority.
- Start with a prediction. State the quantities or features you would compare, then decide what evidence could distinguish two explanations.
Build the model
- Use supplied data about atom economy, yield, reaction rate, equilibrium position and usefulness of by-products. Atom economy describes stoichiometric allocation of starting material; yield describes desired product obtained against its theoretical maximum. Rate affects output over time, while equilibrium may limit conversion. Useful saleable by-products may reduce waste or improve economics, but their market and processing requirements need evidence.
- reaction pathway 反应路线: A chosen sequence or route of reactions for producing a substance; process criterion 工艺评价标准: A stated factor used to compare production methods.
What makes a pathway choice justified?
Difficulty 1/5 · 100 coins + a small bonus on first completion today
A justified choice weighs the supplied criteria - yield, atom economy, rate, cost, hazards - against the stated priority. One number alone can be misleading, and assuming a market for every by-product invents evidence.
Match each technical term to its precise meaning.
Difficulty 2/5 · 140 coins + a small bonus on first completion today
Use the definitions to distinguish related quantities and processes.
Choose evidence that can test it
- Imagine route A with atom economy 90%, yield 50% and a slow rate, and route B with atom economy 70%, yield 90% and a faster rate under supplied comparable conditions. Neither percentage alone decides every use. If equal starting mass could theoretically enter product according to these fractions, recovered desired mass per 100 g starting material is 45 g for A and 63 g for B under this simplified comparison.
- State what the data hold constant: feedstock mass, purity, conditions, reaction time or cost. Rank criteria against the question’s objective and acknowledge one disadvantage of the chosen route. Do not equate fast reaction with low energy cost unless the conditions support that inference. A favourable equilibrium position is different from a catalyst speeding the approach to it.
Which two habits make the investigation or model in this case more defensible?
Difficulty 2/5 · 140 coins + a small bonus on first completion today
Stating the conditions and assumptions lets others repeat the work exactly, and recording the supporting evidence lets them check the conclusion. Hiding disagreeing results removes the evidence that could challenge a wrong conclusion, and one measurement can never establish a wider claim.
Work from known quantities
- State the known values and their units. Choose the relation because its assumptions fit this case, then rearrange before substitution.
- Known: a supplied route directs 80% of 200 g reactant mass into theoretical desired product, giving 160 g. At 75% recovery, actual desired mass is 120 g. This product-of-fractions illustration assumes the atom-economy fraction applies to the stated total feed and no reactant excess is included; other industrial datasets may require a different basis.
Under the stated mass basis, a route has 80% atom economy and 75% yield from 200 g total reactant feed. Find recovered desired-product mass. Use the same sequence: known quantities → model → relation → substitution → unit and interpretation.
Under the stated mass basis, a route has 80% atom economy and 75% yield from 200 g total reactant feed. Find recovered desired-product mass.
Difficulty 2/5 · 140 coins + a small bonus on first completion today
The result is 120 g. Known: 200 g feed; 80% atom economy; 75% yield. Theoretical desired product = 0.80 x 200 = 160 g. Recovered = 0.75 x 160 = 120 g. The fractions multiply because they apply in sequence.
Check the conclusion and its limits
- Higher atom economy alone does not prove an entire process is sustainable. Toxicity, energy, separation and resource availability can matter when supplied. Keep yield and atom economy denominators separate, and avoid inventing cost or environmental facts absent from the evidence.
- Return to the original observation. Explain what the result supports, which conditions it assumes, and one way to test a competing explanation.
A catalyst necessarily changes equilibrium position to improve the final product fraction. This claim is false: A catalyst speeds up reaching equilibrium but cannot shift it: the equilibrium composition at fixed conditions is unchanged. Only changing temperature, pressure or concentration moves the position.
Higher Tier, Chemistry-only: choose a reaction pathway from several criteria: Imagine route A with atom economy 90%, yield 50% and a slow rate, and route B with atom economy 70%, yield 90% and a faster rate under supplied comparable conditions. Neither percentage alone decides every use. If equal starting mass could theoretically enter product according to these fractions, recovered desired mass per 100 g starting material is 45 g for A and 63 g for B under this simplified comparison.
A catalyst necessarily changes equilibrium position to improve the final product fraction.
Difficulty 1/5 · 100 coins + a small bonus on first completion today
A catalyst speeds up reaching equilibrium but cannot shift it: the equilibrium composition at fixed conditions is unchanged. Only changing temperature, pressure or concentration moves the position.
A chosen sequence or route of reactions for producing a substance: write the technical term.
Difficulty 1/5 · 100 coins + a small bonus on first completion today
reaction pathway means A chosen sequence or route of reactions for producing a substance.