Calorimetry example answer

25.0 cm³ of 2.00 M sodium hydroxide solution was mixed with 35.0 cm³ of 2.00M hydrochloric acid in an insulated cup. The temperature rose from 18.0 ºC to 29.5 ºC. Given that the specific heat capacity of water is 4.18 g^-1 K^-1, calculate the enthalpy change for this reaction.

The temperature rise is 11.5 ºC and the total volume of solution is 60 cm³. We need to use the total volume as it is both solutions which rise in temperature, not just the one in the cup where the initial temperature was measured.

Heat released = 4.18 × 60 × 11.5 = 2884 J = 2.88 kJ

NaOH(aq) + HCl(aq) ⇒ NaCl(aq) + H₂O(l)

The equation shows that the two solutions react in a 1:1 ratio. This means that excess hydrochloric acid was used (same concentration, but larger volume). Consequently, we need to use the number of moles of sodium hydroxide in our calculation as all of the sodium hydroxide has reacted. Only some of the hydrochloric acid has reacted (not all of the excess used).

Number of moles of NaOH = 2.00 × 25.0/1000 = 0.0500 mol

Enthalpy change is measured in kilojoules per mole, that is kilojoules divided by moles:

DH = - 2.88/0.0500 = - 57.6 kJ mol^-1