Question

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:

Answer 1

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

Step-by-step explanation:

The reaction between HCl and NaOH is a neutralization reaction:

`HCl + NaOH \rightarrow NaCl + H2O`

Heat released during neutralization = Heat gained by water

i.e.
`\qrxn = -\q(solution)-----(1)`

where:

`\q(solution) = mc\Delta T-----(2)`

m = total mass of solution

`m = density*total\ volume = 1.00g/ml*(40.00+40.00)ml = 80.00\ g`

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

c = specific heat = 4.18 J/g C

`\q(solution) = 80.00g*4.18J/gC*1.3C = 434.7 J`

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

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

`Moles\ of\ HCl = V(HCl) * M(HCl) = 0.040 L*0.200moles/L = 0.008\ moles`

`\Delta Hrxn = (q)/(mole)=(-434.7J)/(0.008mole)=-54337 J/mol=-54.3 kJ/mol`