You mix 200. mL of 0.400M HCl with 200. mL of 0.400M NaOH in a coffee cup calorimeter. The temperature of the solution goes from 25.10°C to 27.78° C. What is the molar enthalpy of neutralization of the acid in kJ/mol? Assume all densities are 1.00 g/mL and the specific heat capacities are 4.184 J/g*K.

Answer : The enthalpy of neutralization is, 56.012 kJ/mole

Explanation :

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

The balanced chemical reaction will be,

From the balanced reaction we conclude that,

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

So, 0.08 mole of HCl neutralizes by 0.08 mole of NaOH

Thus, the number of neutralized moles = 0.08 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 =

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

where,

q = heat absorbed = ?

= specific heat of water =

m = mass of water = 400 g

= final temperature of water =

= initial temperature of metal =

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

Thus, the heat released during the neutralization = -4480.96 J

Now we have to calculate the enthalpy of neutralization.

where,

= enthalpy of neutralization = ?

q = heat released = -4480.96 J

n = number of moles used in neutralization = 0.08 mole

The negative sign indicate the heat released during the reaction.

Therefore, the enthalpy of neutralization is, 56.012 kJ/mole

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