In a constant‑pressure calorimeter, 70.0 ml of 0.770 m h2so4 is added to 70.0 ml of 0.420 m naoh. the reaction caused the temperature of the solution to rise from 23.14 ∘c to 26.00 ∘c. if the solution has the same density and specific heat as water ( 1.00 g/ml and 4.184 j/(g⋅°c), respectively), what is δh for this reaction (per mole of h2o produced)

Answer : The enthalpy of neutralization is, -113.9 KJ/mole

Explanation :

First we have to calculate the moles of and .

The balanced chemical reaction will be,

From the balanced reaction we conclude that,

As, 2 mole of neutralizes by 1 mole of

As, 0.0294 mole of neutralizes by mole of

Thus, the number of neutralized moles = 0.0147 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 = 140.0 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 = -1.675 KJ

Now we have to calculate the enthalpy of neutralization.

where,

= enthalpy of neutralization = ?

q = heat released = -1.675 KJ

n = number of moles used in neutralization = 0.0147 mole

The negative sign indicate the heat released during the reaction.

Therefore, the enthalpy of neutralization is, -113.9 KJ/mole