Question

Find the enthalpy of neutralization of HCl and NaOH. 87 cm3 of 1.6 mol dm-3 hydrochloric acid was neutralized by 87 cm3 of 1.6 mol dm-3 NaOH. The temperature rose from 298 K to 317.4 K. The specific heat capacity is the same as water, 4.18 J/K g.

Answer 1

Explanation:

The reaction equation for the given reaction will be as follows.

               $HCl + NaOH \rightarrow NaCl + H_{2}O$

Each mole of both HCl and NaOH gives one mole of water.

Also, it is given that 1 liter of NaOH and HCl solution contains 1.6 mol $dm^{-3}$ of NaOH (HCl).

It is known that 1 $cm^{3}$ = 0.001 liter. So, 87 $cm^{3}$ = 0.087 liter.

Hence, number of moles of water obtained from the given reaction are as follows.

                  0.087 liter × 1.6 mol = 0.1392 moles

            No. of moles = $\frac{mass}{molar mass of water}$

                0.1392 moles = $\frac{mass}{18 g/mol}$

                       mass = 2.5056 g

Now, volume of water present before the reaction is $2 \times 0.087$ liter = 0.174 liter or 0.174 Kg (as density is 1 $Kg/cm^{3}$) or 174 g (as 1 kg = 1000 g).

Therefore, total weight of water present = 2.5056 g + 174 g = 176.5056 g

Formula to calculate enthalpy of neutralization is as follows.

                Enthalpy of neutralization = $mS \Delta T$

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where,            m = mass

                       S = specific heat capacity

                   $\Delta T$ = change in temperature

Putting the given values in the formula as follows.

       Enthalpy of neutralization = $mS \Delta T$

                     = $176.5056 g \times 4.18 J/K g \times (317.4 K - 298 K)$              

                                                           = 14333.73 J

   or,                                                    = 14.33 kJ

Thus, we can conclude that the enthalpy of neutralization of given reaction is 14.33 kJ.