Question

Find the enthalpy of neutralization of HCl and NaOH. 137 cm3 of 2.6 mol dm-3 hydrochloric acid was neutralized by 137 cm3 of 2.6 mol dm-3 NaOH. The temperature rose from 298 K to 325.8 K. The specific heat capacity is the same as water, 4.18 J/K g.

Answer 1

Approximately
`\rm -89 \; kJ\cdot mol^(-1)`.

Assumption: the density of the solution is equal to the density of pure water.

Step-by-step explanation:

The enthalpy of neutralization is defined as the enthalpy change for each moles of water produced. (Clark, Physical & Theoretical Chemistry, Chemistry Libretexts.)

Each mole of
`\rm NaOH` formula units will neutralize one mole of
`\rm HCl` to produce one mole of water.
`\rm HCl` and
`\rm NaOH` are available at equal volume and concentration. In other words, there's an equal number of both reactants. All
`\rm HCl` and
`\rm NaOH` will react to form water.

`V(\mathrm{HCl}) = \rm 137\; cm^(3) = 0.137\;dm^(3)`.

`V(\mathrm{NaOH}) = \rm 137\; cm^(3) = 0.137\;dm^(3)`.

`n = c\cdot V = \rm 0.137\;dm^(3) * 2.6\;mol\cdot dm^(-3) = 0.3652\; mol`.

In other words, there are
`\rm 0.3652\; mol` of
`\rm HCl` and
`\rm NaOH` each. The two will react to produce
`\rm 0.3652\; mol` of water.

How much heat is released?

Assume that the volume of the liquid is equal to the volume of the
`\rm HCl` solution plus the volume of the
`\rm NaOH` solution. That's
`\rm 0.274\;dm^(3)`. Assume that the density of the solution is equal to that of water under room temperature.
`\rho(\text{water}) = \rm 1.000\; kg\cdot dm^(-3)`. The mass of the liquid will be
`m = \rho \cdot V = \rm 0.274\; dm^(3) * 1.000\; kg\cdot dm^(-3) = 0.274\; kg = 274\;g`.

Change in temperature:

`\Delta T = \rm 325.8 - 298 = 27.8\; K`.

Heat that the solution absorbed:

`Q = c\cdot m \cdot \Delta T = \rm 4.18\;J\cdot K^(-1)\cdot g^(-1) * 274\; g* 27.8\;K = 36410.216\; J = 36.410216\; kJ`.

That will also be the amount of heat released from the reaction if there's no energy loss.

`\displaystyle \Delta H(\text{Neutralization}) = \frac{-Q}{n(\text{water produced})} = \rm (36.410216\; kJ)/(0.3652\; mol) \approx 89\; kJ\cdot mol^(-1)`.