Question

If 15.0 mL of a 1.5M HCl solution at 21.50C is mixed with 25.0mL of a 1.5M NaOH solution at 21.50C that is in a calorimeter, and the final mixed solution temperature rises to 28.50C, what is the ΔHrxn for this process?

Answer 1

Answer : The enthalpy of neutralization for this process is, 52.02 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}=1.5mole/L* 0.0150L=0.0225mole`

`\text{Moles of NaOH}=\text{Concentration of NaOH}* \text{Volume of solution}=1.5mole/L* 0.0250L=0.0375mole`

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, 0.0225 mole of HCl neutralizes by 0.0225 mole of NaOH

Thus, the number of neutralized moles = 0.0225 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 =
`15.0ml+25.0ml=40ml`

`\text{Mass of water}=\text{Density of water}* \text{Volume of water}=1g/ml* 40ml=40g`

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 = 40 g

`T_(final)` = final temperature of water =
`28.50^oC=273+28.50=301.5K`

`T_(initial)` = initial temperature of metal =
`21.50^oC=273+21.50=294.5K`

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

`q=40g* 4.18J/g^oC* (301.5-294.5)K`

`q=1170.4J`

Thus, the heat released during the neutralization = -1170.4 J

Now we have to calculate the enthalpy of neutralization.

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

where,

`\Delta H` = enthalpy of neutralization = ?

q = heat released = -1170.4 J

n = number of moles used in neutralization = 0.0225 mole

`\Delta H=(-1170.4J)/(0.0225mole)=-52017.77J/mole=-52.02kJ/mole`

The negative sign indicate the heat released during the reaction.

Therefore, the enthalpy of neutralization for this process is, 52.02 kJ/mole