5.1 - Determination of the Heat of Reaction of Hydrochloric Acid with Sodium Hydroxide (Leaving Cert Chemistry): Flashcards
📚Flashcards
Practise the cards
14 cards from this deck
ShowHide
Practise the cards
14 cards from this deck
Solutions used in HCl-NaOH heat of reaction expt
Solutions used in HCl-NaOH heat of reaction expt
M and M
Formula to calculate heat liberated in neutralisation
Formula to calculate heat liberated in neutralisation
Type of reaction: HCl + NaOH neutralisation
Type of reaction: HCl + NaOH neutralisation
Exothermic
Volume of each solution used in expt
Volume of each solution used in expt
cm³
Why polystyrene cup used in heat of reaction expt
Why polystyrene cup used in heat of reaction expt
Excellent insulator, minimises heat loss
Why lid placed on cup in expt
Why lid placed on cup in expt
Reduce heat loss to surroundings
Temperature rise in sample calculation
Temperature rise in sample calculation
°C
Moles of HCl used in expt ( cm³ of M)
Moles of HCl used in expt ( cm³ of M)
mol
Specific heat capacity of water used in calculation
Specific heat capacity of water used in calculation
J/g°C
Mass of combined solution assumed in calculation
Mass of combined solution assumed in calculation
g
Heat of reaction per mole in sample calculation
Heat of reaction per mole in sample calculation
kJ/mol
Significance of negative sign in heat of reaction
Significance of negative sign in heat of reaction
Exothermic; releases heat to surroundings
Why strong acids/bases give similar heat of neutralisation
Why strong acids/bases give similar heat of neutralisation
Heat mainly from + reaction (same for all)
Temperature condition before mixing HCl and NaOH
Temperature condition before mixing HCl and NaOH
Both at same temperature
