Monitoring Chemical Reactions (OCR GCSE Chemistry A (Gateway)): Flashcards

Concentration (mol/dm³) = moles of solute (mol) ÷ volume of solution (dm³).

A solution has 0.5 mol of solute in 2.5 dm³. Its concentration is 0.2 mol/dm³.

0.5 mol ÷ 2.5 dm³ = 0.2 mol/dm³.

True.

Moles allow chemists to directly compare amounts of different substances and relate them to reaction equations, making mol/dm³ the more chemically useful unit.

Moles = concentration × volume = 0.2 mol/dm³ × 2.5 dm³ = 0.5 mol.

Rearrange the concentration formula: moles = concentration × volume.

When calculating concentration in mol/dm³, convert volume in cm³ to dm³ by dividing by 1000.

For example, 500 cm³ ÷ 1000 = 0.5 dm³.

Molar mass of NaOH = 23 + 16 + 1 = 40 g/mol.

Moles of NaOH = 80 ÷ 40 = 2 mol.

Volume = 500 ÷ 1000 = 0.5 dm³. Concentration = 2 ÷ 0.5 = 4 mol/dm³.

False.

Volume (dm³) = moles ÷ concentration. Dividing concentration by moles gives the reciprocal of volume, not volume itself.

The concentration formula uses moles, not grams. Converting mass to moles first using the molar mass ensures the correct unit (mol/dm³) is obtained.

This is the first step in a structured calculation.

A titration is a practical technique used to analyse the concentration of a solution by reacting it with another solution of known concentration.

It identifies the exact volume needed for complete reaction.

An indicator shows the endpoint of the titration through a sharp colour change.

This tells you when the acid and alkali have exactly neutralised each other.

False.

Universal indicator is a mixture of indicators that produces many subtle colour changes, making it impossible to identify a single sharp endpoint.

Litmus does not give a sharp enough colour change at the endpoint.

A suitable titration indicator must change colour suddenly at a precise point to identify neutralisation accurately.

In a titration, the solution with unknown concentration is placed in the conical flask and the solution of known concentration is added from the burette.

A pipette is used to measure an accurate fixed volume into the flask.

True.

Phenolphthalein gives a clear colour change at the endpoint, making it well suited for acid-base titrations.

Concordant results are two titre values that are within 0.1 cm³ of each other.

They are used to calculate a reliable mean titre, improving the accuracy of the final concentration value.

Burette readings in a titration are recorded to the nearest 0.05 cm³.

This level of precision minimises reading error and is the standard required in GCSE practicals.

A white tile provides a plain background that makes it easier to see the colour change of the indicator at the endpoint.

This helps to identify neutralisation precisely.

False.

Anomalous results are excluded from the mean titre calculation. Only concordant results are used to ensure an accurate and reliable average.

The mean titre is the average volume of solution added from the burette, calculated from concordant results.

It is used in titration calculations to find an unknown concentration or volume.

1. Write a balanced equation for the reaction.

2. Calculate moles of the known solution using moles = concentration × volume.

3. Use the molar ratio to find moles of the unknown solution.

4. Calculate concentration of the unknown: moles ÷ volume.

In a titration calculation, moles of the known solution = concentration × volume (in dm³).

True.

When the molar ratio is 1:1, equal moles of acid and alkali react at neutralisation, so the moles of each substance are equal at the endpoint.

Moles of NaOH = 0.0121 dm³ × 0.100 mol/dm³ = 1.21 × 10^-3 mol.

Moles of HCl = 1.21 × 10^-3 mol (1:1 ratio).

Concentration of HCl = 1.21 × 10^-3 ÷ 0.025 dm³ = 0.0484 mol/dm³.

Once moles of the unknown solution are known, its concentration = moles ÷ volume (in dm³).

This is the final step in a titration calculation.

The balanced equation gives the molar ratio between the acid and alkali.

This ratio is needed to convert the moles of the known solution into the moles of the unknown solution.

False.

Volumes must be converted to dm³ first by dividing by 1000. Using cm³ directly would give an answer 1000 times too small.

The molar ratio tells you how many moles of the unknown react with each mole of the known solution.

For example, in a 1:2 ratio, the moles of the unknown are twice those of the known.

The molar gas volume is the volume occupied by one mole of any gas at a given temperature and pressure. At RTP it is 24 dm³ (or 24,000 cm³).

False.

By Avogadro's Law, equal amounts of any gas occupy the same volume at the same temperature and pressure. Both would occupy 24 dm³ at RTP.

Avogadro's Law states that at the same temperature and pressure, equal amounts (in moles) of any gas occupy the same volume.

RTP (room temperature and pressure) is 20°C and 1 atm. At these conditions, the molar gas volume is exactly 24 dm³ mol⁻¹, so the value can be used directly in volume calculations without correction.

Volume of gas (dm³) = moles × molar gas volume

At RTP, the molar gas volume is 24 dm³.

Divide the volume by the molar gas volume:

Moles = Volume ÷ 24 (if volume is in dm³)

Moles = Volume ÷ 24,000 (if volume is in cm³)

By Avogadro's Law, equal moles of gases occupy equal volumes at the same conditions. So the mole ratio from the equation equals the volume ratio of the gases.

True.

You divide mass by molar mass to get moles, then multiply by the molar gas volume (24 dm³ at RTP) to get the volume.

The mole ratio of O₂ to C₃H₈ is 5 : 1, so volumes are in the same ratio.

Volume of O₂ = 5 × 200 cm³ = 1000 cm³

Yield is the amount of product obtained from a chemical reaction. In practice, the yield is always less than the maximum theoretically possible.

Any three from:

• Reactants may remain in equipment and not fully react

• The reaction may be reversible, so products reform reactants

• Product is lost during separation or purification (e.g. filtration, distillation)

• Side reactions consume some reactant

• Product is lost during transfer between containers

Actual yield is the mass of product obtained in an experiment. Theoretical yield is the maximum mass calculated from the balanced equation, assuming 100% conversion.

False.

The actual yield can never exceed the theoretical yield. If it appears to, the calculation or measurement contains an error.

Percentage yield = ( actual yield ÷ theoretical yield ) × 100

Percentage yield = (1.6 ÷ 2.0) × 100

= 80%

A higher percentage yield means more product is obtained from the same amount of reactants, which reduces costs, increases profit and reduces waste.

True.

In a reversible reaction, products continuously reform reactants, so complete conversion to products is impossible.

The percentage yield indicates how efficiently each pathway converts reactants into the desired product. A pathway with a higher percentage yield is generally more economical and produces less waste.

Atom economy is the percentage of the total mass of reactants that ends up in the useful product(s) of a reaction. A higher atom economy means a more sustainable and efficient process.

Atom economy = ( total M_r of useful products ÷ total M_r of all products ) × 100

A high atom economy means more reactant mass becomes useful product and less becomes waste. This reduces raw material costs, disposal costs and environmental impact, making the process more sustainable and economically attractive.

True.

If waste by-products can be sold or reused, the overall process becomes more economical even if the atom economy figure is low.

The only way to improve atom economy is to change the chemicals used (i.e. change the reaction). Unlike percentage yield, it cannot be improved by running the reaction more completely or reducing product loss.

Atom economy measures how much of the reactant mass becomes useful product in theory. Percentage yield measures how much of the theoretical product is actually obtained in practice. Both are needed because a reaction can have high atom economy but low yield, or vice versa.

True.

If there is only one product, all the reactant atoms end up in that product, so 100% of the reactant mass becomes the useful product.

A sustainable process meets current needs without depleting resources for the future. In chemistry, this means using reactions with high atom economies that minimise waste and raw material consumption.

CO is also formed as a by-product. The atoms in CO are not in the useful product (H₂), so not all reactant mass becomes useful product. This means the atom economy is less than 100%.

A reaction pathway is a sequence of reactions used to produce a desired product. Different pathways may use different reactants, conditions or numbers of steps to make the same compound.

1. Atom economy — how much reactant mass becomes useful product

2. Percentage yield — how much of the theoretical product is obtained

3. Rate of reaction — how quickly the process runs

4. Position of equilibrium — whether conditions can be adjusted to favour products

True.

Low atom economy means much of the reactant mass becomes waste by-product. Disposing of this waste requires chemicals, equipment, transport and space, making the process expensive and unsustainable.

The atom economy of a given reaction cannot be changed without changing the chemistry. However, the effective economy can be improved by selling or reusing by-products, which reduces waste and adds revenue.

In reversible reactions, the position of the equilibrium may need to be shifted towards the products by altering reaction conditions to increase yield.

A high percentage yield ensures more product is made from a given amount of reactant, reducing waste and cost. A fast reaction rate means more product is made per unit time, increasing productivity and reducing running costs.

True.

Expensive raw materials raise production costs, and waste disposal requires chemicals, equipment, space and transport. Minimising both is a key driver in pathway selection.

The lower atom economy pathway might have a higher percentage yield, a faster rate, or produce by-products that can be sold or reused. Overall economic viability depends on all these factors together, not atom economy alone.

If the equilibrium lies towards reactants, very little product forms. Conditions (temperature, pressure, catalysts) can be adjusted to shift the equilibrium position towards products, increasing the yield and making the pathway more economically attractive.