So I thought I'd write this question as I wouldn't necessarily know how to do it straight away upon first glance
) A 25.0 cm3 sample of 2.00 mol dm–3 hydrochloric acid was mixed with 50.0 cm3 of a1.00 mol dm–3 solution of sodium hydroxide. Both solutions were initially at 18.0 °C.After mixing, the temperature of the final solution was 26.5 °C.Use this information to calculate a value for the standard enthalpy change for thefollowing reaction.HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)In your calculation, assume that the density of the final solution is 1.00 g cm–3 and thatits specific heat capacity is the same as that of water. (Ignore the heat capacity of thecontainer.)...............................................................
Energy absorbed (q) = Mass of Solution X Heat Capacity X Temperature change
- You use mass of both solutions combines (each cm cubed is roughly a gram) (25+50= 75)
- Temp change = 26.5-18 = 8.5
So 75 x 4.18 X 8.5 = 2665
you use moles of whatever is not in XS - so the lowest value. In this case, both are 0.05 so that's what we use
Enthalpy change = Q/No of Moles
Therefore -2665/0105
= -53.4 Kjmol-1