6.1 Measuring the rate of reaction
6 exam questions worth 12 marks. Write your answer, open the mark scheme, then give yourself the marks you earned. Your answers are saved on this device.
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In a reaction, 48 cm3 of gas was produced in 40 seconds.
Calculate the mean rate of reaction. Give the unit.
Mark scheme
- 48 ÷ 40 (1)
- = 1.2 (1)
- cm3/s (1)
Model answer
mean rate = 48 ÷ 40 = 1.2 cm3/s.
Describe two ways to measure the rate of a reaction that gives off a gas.
Mark scheme
- measure the volume of gas with a gas syringe / upturned measuring cylinder over time (1)
- measure the loss of mass on a balance over time (1)
Model answer
Collect the gas in a gas syringe and record the volume every 10 seconds, or put the flask on a balance and record the loss in mass every 10 seconds.
Explain why the graph of volume of gas against time becomes flat.
Mark scheme
- one reactant is used up (1)
- so the reaction stops / no more gas is made (1)
Model answer
One of the reactants has been used up, so the reaction stops and no more gas is produced.
Why is the graph steepest at the start of the reaction?
Mark scheme
most reactant particles / highest concentration, so the rate is fastest
Model answer
At the start there are the most reactant particles, so collisions are most frequent and the rate is fastest.
A tangent drawn to a curve at 30 s passes through (10 s, 12 cm3) and (50 s, 52 cm3).
Calculate the rate of reaction at 30 s.
Mark scheme
- (52 − 12) ÷ (50 − 10) / 40 ÷ 40 (1)
- = 1.0 cm3/s (1)
Model answer
gradient = (52 − 12) ÷ (50 − 10) = 40 ÷ 40 = 1.0 cm3/s.
In the loss-of-mass method, a cotton wool plug is put in the neck of the flask. Explain why.
Mark scheme
- lets the gas escape (so mass falls) (1)
- stops acid spraying out / liquid being lost (1)
Model answer
The cotton wool lets the gas escape so the mass falls, but stops any acid spray escaping, which would also lower the mass.
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