When 40.0 ml of 0.200 m hcl at 21.5°c is added to 40.0 ml of 0.200 m naoh also at 21.5°c in a coffee-cup calorimeter, the temperature of the resulting solution rises to 22.8°c. assume that the volumes are additive, the specific heat of the solution is 4.18 jg -1°c -1 and that the density of the solution is 1.00 g ml -1 calculate the enthalpy change, δh in kj for the reaction:

The enthalpy change for the reaction is ΔH = - 54.3 kJ/mol

Explanation:

The reaction between HCl and NaOH is a neutralization reaction:

Heat released during neutralization = Heat gained by water

i.e.

where:

m = total mass of solution

ΔT = change in temperature = 22.8 - 21.5 = 1.3 C

c = specific heat = 4.18 J/g C

As per equation (1): qrxn = -434.7 J

The reaction enthalpy ΔH is the heat released per mole of acid (or base)

3 1/4 %

explanation: