5.1 - Determination of the Heat of Reaction of Hydrochloric Acid with Sodium Hydroxide (Leaving Cert Chemistry): Flashcards

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- Determination of the Heat of Reaction of Hydrochloric Acid with Sodium Hydroxide
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Solutions used in HCl-NaOH heat of reaction expt

11 M HClHCl and 11 M NaOHNaOH

Formula to calculate heat liberated in neutralisation

q=mcΔTq = mc\Delta T

Type of reaction: HCl + NaOH neutralisation

Exothermic

Volume of each solution used in expt

5050 cm³

Why polystyrene cup used in heat of reaction expt

Excellent insulator, minimises heat loss

Why lid placed on cup in expt

Reduce heat loss to surroundings

Temperature rise in sample calculation

66°C

Moles of HCl used in expt (5050 cm³ of 11 M)

0.050.05 mol

Specific heat capacity of water used in calculation

4.24.2 J/g°C

Mass of combined solution assumed in calculation

100100 g

Heat of reaction per mole in sample calculation

−50.4-50.4 kJ/mol

Significance of negative sign in heat of reaction

Exothermic; releases heat to surroundings

Why strong acids/bases give similar heat of neutralisation

Heat mainly from H+H^+ + OH−OH^- reaction (same for all)

Temperature condition before mixing HCl and NaOH

Both at same temperature

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