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.

Explanation:

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

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 of NaOH (HCl).

It is known that 1 = 0.001 liter. So, 87 = 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 =

                0.1392 moles =

                       mass = 2.5056 g

Now, volume of water present before the reaction is liter = 0.174 liter or 0.174 Kg (as density is 1 ) 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 =

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where,            m = mass

                       S = specific heat capacity

                    = change in temperature

Putting the given values in the formula as follows.

       Enthalpy of neutralization =

                     =              

                                                           = 14333.73 J

   or,                                                    = 14.33 kJ

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

the electrical  conductivity  of asolution  of an electrolyte is measured by determining the resistance of thesolution  between two flat or cylindrical electrodes separated by a fixed distance.