Higher Tier: calculate tangent gradients and molar rates
導入済み| English | 意味 |
|---|---|
| instantaneous rate/ˌɪnstənˈteɪnɪəs reɪt/ | 瞬間速度 |
| molar rate/ˈməʊlə reɪt/ | モル速度 |
What would explain this observation?
- Higher Tier: A curve can flatten during a reaction. Its mean rate over a long interval can therefore differ from its rate at one particular instant.
- Start with a prediction. State the quantities or features you would compare, then decide what evidence could distinguish two explanations.
Build the model
- An instantaneous rate 瞬时速率 is estimated by the gradient of a tangent to the quantity–time curve at the selected time. Gradient=vertical change/horizontal change using two well-separated points on the tangent itself. For a product-volume graph this gives $\dfrac{\text{cm}^3}{\text{s}}$. Higher-tier quantity can also be in moles, giving $\dfrac{\text{mol}}{\text{s}}$. The units come from the actual axes, not from a remembered numerical example.
- instantaneous rate: Rate at a stated time estimated by the gradient of a tangent; molar rate 摩尔速率: Amount of reactant used or product formed in moles per unit time.
Where should the two points for a tangent-gradient calculation be selected?
難易度 1/5 · 100 コイン + 今日初回完了時に小ボーナス付与
The gradient belongs to the TANGENT, so both points must be read off the tangent line, widely spaced for precision. Points on the original curve give a chord - an average rate - not the instantaneous rate.
Match each technical term to its precise meaning.
難易度 2/5 · 140 コイン + 今日初回完了時に小ボーナス付与
Use the definitions to distinguish related quantities and processes.
Choose evidence that can test it
- First mark the requested time, draw a tangent following the smooth curve’s local direction, then select two readable points on that line. They need not be measured data points or lie on the curve elsewhere. A larger triangle reduces the relative effect of reading uncertainty. Tangent estimates can differ slightly between reasonable drawings. A chord linking two curve points measures an interval mean and must not be presented as the tangent gradient.
- Show each coordinate pair, differences, division and final unit. For reactant remaining, the tangent gradient is negative; report its positive magnitude when asked for rate of reactant consumption. Convert time or amount units before dividing: milliseconds and seconds are not interchangeable, nor are millimoles and moles. This numerical tangent calculation and the molar-rate requirement are the embedded Higher clauses, while common-tier students still draw and interpret tangent steepness.
Which two habits make the investigation or model in this case more defensible?
難易度 2/5 · 140 コイン + 今日初回完了時に小ボーナス付与
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: points read from a supplied tangent are (10 s,18 cm³) and (50 s,42 cm³). Gradient=(42−18)/(50−10)=24/40=0.60 $\dfrac{\text{cm}^3}{\text{s}}$ at the tangent’s stated contact time. In a separate amount measurement, 0.012 mol product forms in 30 s; mean molar rate=0.012/30=0.00040 $\dfrac{\text{mol}}{\text{s}}$. This second number is a mean, not automatically an instantaneous rate.
A supplied tangent passes through (10 s,16 cm³) and (50 s,44 cm³). Find its gradient. Use the same sequence: known quantities → model → relation → substitution → unit and interpretation.
A supplied tangent passes through (10 s,16 cm³) and (50 s,44 cm³). Find its gradient.
難易度 2/5 · 140 コイン + 今日初回完了時に小ボーナス付与
The result is 0.7 cm³ per s. Known: tangent passes through (10 s, 16 cm^3) and (50 s, 44 cm^3). Gradient = (44 - 16)/(50 - 10) = 28/40 = 0.7 cm^3 per second at the tangent's contact time.
Check the conclusion and its limits
- Do not take points from the original curve when a tangent calculation is requested. A negative gradient of reactant remaining does not mean particles react backwards. Do not convert cm³ to moles without supplied conditions or an appropriate relation. The tangent contact time must be distinguished from the endpoints used for the gradient triangle.
- Return to the original observation. Explain what the result supports, which conditions it assumes, and one way to test a competing explanation.
The gradient of any long chord is necessarily the instantaneous rate at its midpoint. This claim is false: A chord averages the rate over its span; the curve bends, so that average matches the instantaneous rate at its midpoint only by accident. The tangent, touching at one time, is the correct graphical model of instantaneous rate.
Higher Tier: calculate tangent gradients and molar rates: First mark the requested time, draw a tangent following the smooth curve’s local direction, then select two readable points on that line. They need not be measured data points or lie on the curve elsewhere. A larger triangle reduces the relative effect of reading uncertainty. Tangent estimates can differ slightly between reasonable drawings. A chord linking two curve points measures an interval mean and must not be presented as the tangent gradient.
The gradient of any long chord is necessarily the instantaneous rate at its midpoint.
難易度 1/5 · 100 コイン + 今日初回完了時に小ボーナス付与
A chord averages the rate over its span; the curve bends, so that average matches the instantaneous rate at its midpoint only by accident. The tangent, touching at one time, is the correct graphical model of instantaneous rate.
Rate at a stated time estimated by the gradient of a tangent: write the technical term.
難易度 1/5 · 100 コイン + 今日初回完了時に小ボーナス付与
instantaneous rate means Rate at a stated time estimated by the gradient of a tangent.