Sodium thiosulfate, Na2S2O3, can be used to investigate the effect of reaction conditions on the rate of reaction - Scottish Highers Chemistry - Question 1 - 2019

To find the volumes of water used in each experiment to ensure constant total volume:

• For experiment A: 0 cm³ (50 - 50 = 0)
• For experiment B: 10 cm³ (50 - 40 = 10)
• For experiment C: 20 cm³ (50 - 30 = 20)
• For experiment D: 30 cm³ (50 - 20 = 30)
• For experiment E: 40 cm³ (50 - 10 = 40)

Thus the completed table is:

The rate of reaction for experiment C is given as 0.0285 s⁻¹.

To calculate the time, use the formula:

Thus,

Referring to the graph, if the rate doubles from 0.0285 s⁻¹ (experiment C) to approximately 0.0570 s⁻¹, determine the temperature at which these rates occur.

From observation, the initial temperature is around 25°C, and the new temperature required to achieve the doubled rate is about 50°C. Therefore:

$ext{Temperature rise} ext{ = } 50°C - 25°C = 25°C$

1. Particles must have sufficient or enough energy to react.
2. Particles must collide with the correct orientation.

Mark the top of the peak on the potential energy diagram with an X to indicate the position of the activated complex.

Draw a dotted line that shows a lower activation energy pathway compared to the original diagram, indicating that the presence of the catalyst lowers the activation energy required for the reaction.