Core Practical 8: Calculate the enthalpy change for the thermal decomposition of potassium hydrogencarbonate

Question 1

2 mol dm-3 is

Answer 1

an irritant

Question 2

potassium carbonate is

Answer 2

an irritant

Question 3

the method for carrying out this practical is:

Answer 3

• weigh about 3g of potassium carbonate: do this by adding the solid to a test tube, and reweigh once all of the solid has been added to the mixture below
• use a burette to dispense 30cm3 of 2 mol dm-3 hydrochloric acid into a polystyrene cup supported by a beaker
• measure the temperature of the acid and record
• add the solid and stir with the thermometer, and poke this through a polystyrene lid if possible
• record the highest temperature reached
• (reweigh empty test tube)

-now repeat with potassium hydrogencarbonate instead, and record the lowest temperature reached.

Question 4

the two equations for the reactions occurring are:

Answer 4

reaction 1: K2CO3(s) + 2HCl(aq) → 2KCl(aq) + CO2(g) + H2O(l)
reaction 2: KHCO3(s) + HCl(aq) → KCl(aq) + CO2(g) + H2O(l) 2

Question 5

Why is it not possible to measure the enthalpy change for the decomposition of potassium hydrogencarbonate directly?

Answer 5

-Heat energy must be supplied for the compound to undergo thermal decomposition; hence, the temperature change measured is not solely due to the decomposition. 2

Question 6

the equation for the thermal decomposition of potassium hydrogen carbonate is:

Answer 6

reaction 3:

2KHCO3(s) → K2CO3(s) + CO2(g) + H2O(l)

Question 7

Explain why the reactions are conducted in a polystyrene cup rather than a glass beaker.

Answer 7

-Polystyrene is a better insulator than glass. Therefore, less heat energy is lost to/gained from the surroundings, so temperature changes are more accurate.