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.

A. -101.37 kJ


B. -7.05 kJ


C. 7055 kJ


D. 10,1365 kJ

Answer 1

Answer : The correct option is, (A) -101.37 KJ

Explanation :

First we have to calculate the moles of HCl and NaOH.

`\text{Moles of HCl}=\text{Concentration of HCl}* \text{Volume of solution}=1.6mole/L* 0.087L=0.1392mole`

`\text{Moles of NaOH}=\text{Concentration of NaOH}* \text{Volume of solution}=1.6mole/L* 0.087L=0.1392mole`

The balanced chemical reaction will be,

`HCl+NaOH\rightarrow NaCl+H_2O`

From the balanced reaction we conclude that,

As, 1 mole of HCl neutralizes by 1 mole of NaOH

So, 0.1392 mole of HCl neutralizes by 0.1392 mole of NaOH

Thus, the number of neutralized moles = 0.1392 mole

Now we have to calculate the mass of water.

As we know that the density of water is 1 g/ml. So, the mass of water will be:

The volume of water =
`87ml+87ml=174ml`

`\text{Mass of water}=\text{Density of water}* \text{Volume of water}=1g/ml* 174ml=174g`

Now we have to calculate the heat absorbed during the reaction.

`q=m* c* (T_(final)-T_(initial))`

where,

q = heat absorbed = ?

`c` = specific heat of water =
`4.18J/g^oC`

m = mass of water = 174 g

`T_(final)` = final temperature of water = 317.4 K

`T_(initial)` = initial temperature of metal = 298 K

Now put all the given values in the above formula, we get:

`q=174g* 4.18J/g^oC* (317.4-298)K`

`q=14110.008J=14.11KJ`

Thus, the heat released during the neutralization = -14.11 KJ

Now we have to calculate the enthalpy of neutralization.

`\Delta H=(q)/(n)`

where,

`\Delta H` = enthalpy of neutralization = ?

q = heat released = -14.11 KJ

n = number of moles used in neutralization = 0.1392 mole

`\Delta H=(-14.11KJ)/(0.1392mole)=-101.37KJ/mole`

Therefore, the enthalpy of neutralization is, -101.37 KJ