Mean rates: use the change over the chosen interval
導入済み| English | 意味 |
|---|---|
| mean reaction rate/miːn rɪˈækʃn reɪt/ | 平均反応速度 |
| tangent/ˈtændʒənt/ | 接線 |
What would explain this observation?
- Two reactions can produce the same final gas volume but reach it at different times. The final amount and the speed of formation answer different questions.
- Start with a prediction. State the quantities or features you would compare, then decide what evidence could distinguish two explanations.
Build the model
- Mean rate is quantity of reactant used divided by time taken, or quantity of product formed divided by time taken. Common-tier quantities include mass in grams and gas volume in cm³, giving $\dfrac{\text{g}}{\text{s}}$ or $\dfrac{\text{cm}^3}{\text{s}}$. If a graph already shows an accumulated quantity, use the change between the two stated times, not the final reading alone. A product curve normally rises and becomes less steep as reactants are used up.
- mean reaction rate 平均反应速率: Quantity of reactant used or product formed divided by the stated time interval; tangent 切线: A straight line matching a curve’s local direction at a chosen point.
Gas volume rises from 10 to 30 cm³ between 20 and 60 s. Which calculation gives the interval mean rate?
難易度 1/5 · 100 コイン + 今日初回完了時に小ボーナス付与
Mean rate over an interval is the CHANGE in quantity divided by the CHANGE in time: (30 - 10)/(60 - 20). Using 30 alone ignores the starting value, and inverting the ratio gives seconds per cm^3 - not a 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
- Plot time on the horizontal axis and measured quantity on the vertical axis with labelled units and sensible scales. A steeper product-time curve means faster formation; a horizontal plateau means no further measured product forms. A tangent touches the local curve direction at the chosen point and its steepness indicates rate there. Drawing and interpreting tangents is both-tier; calculating their numerical gradients is in the Higher lesson. A straight chord between distant points gives an interval mean, not the local rate.
- Collect readings at regular intervals with a consistent start. Check whether the graph is product formed, reactant used up or reactant remaining. Remaining reactant falls; its negative slope represents a positive rate of consumption when expressed as the amount used. Compare curves at the same time and under stated conditions. Do not assume a plateau proves every reactant is exhausted: a limiting reactant, incomplete collection or another stopped process needs consideration.
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: gas readings are 12 cm³ at 20 s and 36 cm³ at 60 s. Quantity formed during that interval=36−12=24 cm³; time=60−20=40 s. Mean rate=24/40=0.60 $\dfrac{\text{cm}^3}{\text{s}}$. A separate mass-loss example records 0.80 g escaping gas over 40 s, giving 0.020 $\dfrac{\text{g}}{\text{s}}$ if that gas loss is the measured reaction quantity.
Gas volume increases from 8 to 32 cm³ between 10 and 50 s. Find mean rate. Use the same sequence: known quantities → model → relation → substitution → unit and interpretation.
Gas volume increases from 8 to 32 cm³ between 10 and 50 s. Find mean rate.
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 0.6 cm³ per s. Known: volume rises from 8 to 32 cm^3 between 10 and 50 s. Change in volume = 24 cm^3; change in time = 40 s. Mean rate = 24/40 = 0.6 cm^3 per second over that interval.
Check the conclusion and its limits
- Do not divide 36 by 40 for the interval calculation or call a final volume a rate. A tangent line is a graphical model of the local direction, not a new experimental trace. Mol/s calculations are Higher-only. Curves with the same plateau can have different rates, and a larger plateau alone does not prove a faster initial reaction.
- Return to the original observation. Explain what the result supports, which conditions it assumes, and one way to test a competing explanation.
Two curves with the same final gas volume must have identical reaction rates. This claim is false: The final volume is fixed by the amounts reacting; the rate is how fast the curve gets there. A fast and a slow reaction with the same reactant amounts share a plateau but have completely different curves.
Mean rates: use the change over the chosen interval: Plot time on the horizontal axis and measured quantity on the vertical axis with labelled units and sensible scales. A steeper product-time curve means faster formation; a horizontal plateau means no further measured product forms. A tangent touches the local curve direction at the chosen point and its steepness indicates rate there. Drawing and interpreting tangents is both-tier; calculating their numerical gradients is in the Higher lesson. A straight chord between distant points gives an interval mean, not the local rate.
Two curves with the same final gas volume must have identical reaction rates.
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 final volume is fixed by the amounts reacting; the rate is how fast the curve gets there. A fast and a slow reaction with the same reactant amounts share a plateau but have completely different curves.
Quantity of reactant used or product formed divided by the stated time interval: 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
mean reaction rate means Quantity of reactant used or product formed divided by the stated time interval.