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.

A. 44.69 kJ/mol


B. 6123.06 kJ/mol


C. 597.46 kJ/mol


D. 89.39 kJ/mol

Answer 1

Answer : The correct option is, (D) 89.39 KJ/mole

Explanation :

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

`\text{Moles of HCl}=\text{Concentration of HCl}* \text{Volume of solution}=2.6mole/L* 0.137L=0.3562mole`

`\text{Moles of NaOH}=\text{Concentration of NaOH}* \text{Volume of solution}=2.6mole/L* 0.137L=0.3562mole`

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, 0.3562 mole of HCl neutralizes by 0.3562 mole of NaOH

Thus, the number of neutralized moles = 0.3562 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 =
`137ml+137ml=274ml`

`\text{Mass of water}=\text{Density of water}* \text{Volume of water}=1g/ml* 274ml=274g`

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 = 274 g

`T_(final)` = final temperature of water = 325.8 K

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

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

`q=274g* 4.18J/g^oC* (325.8-298)K`

`q=31839.896J=31.84KJ`

Thus, the heat released during the neutralization = -31.84 KJ

Now we have to calculate the enthalpy of neutralization.

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

where,

`\Delta H` = enthalpy of neutralization = ?

q = heat released = -31.84 KJ

n = number of moles used in neutralization = 0.3562 mole

`\Delta H=(-31.84KJ)/(0.3562mole)=-89.39KJ/mole`

The negative sign indicate the heat released during the reaction.

Therefore, the enthalpy of neutralization is, 89.39 KJ/mole