Answer : The enthalpy of neutralization is, 56.96 kJ/mole
Explanation :
First we have to calculate the moles of H₂SO₄ and NaOH.
The balanced chemical reaction will be,
From the balanced reaction we conclude that,
As, 2 mole of NaOH neutralizes by 1 mole of H₂SO₄
So, 0.0169 mole of NaOH neutralizes by 0.00845 mole of H₂SO₄
That means, NaOH is a limiting reagent and H₂SO₄ is an excess reagent.
Now we have to calculate the moles of H₂O.
As, 2 mole of NaOH react to give 2 mole of H₂O
So, 0.0169 mole of NaOH react to give 0.0169 mole of H₂O
Now we have to calculate the mass of water.
As we know that the density of water is 1.00 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 = 130 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 = -962.7 J
Now we have to calculate the enthalpy of neutralization.
where,
= enthalpy of neutralization = ?
q = heat released = -962.7 J
n = number of moles used in neutralization = 0.0169 mole
The negative sign indicate the heat released during the reaction.
Therefore, the enthalpy of neutralization is, 56.96 kJ/mole