Final answer:
The enthalpy of neutralization of HCl and NaOH is 8.18 kJ/mol.
Step-by-step explanation:
The enthalpy of neutralization is the heat released or absorbed when an acid and a base react to form one mole of water. In this case, the reaction is between hydrochloric acid (HCl) and sodium hydroxide (NaOH).
From the given information, 87 cm3 of 1.6 mol dm-3 HCl is neutralized by 87 cm3 of 1.6 mol dm-3 NaOH. The temperature rose from 298 K to 317.4 K.
To calculate the enthalpy change, we can use the formula:
Enthalpy change of neutralization = (mole of limiting reactant) x (heat evolved per mole of reaction)
From the balanced equation: HCl(aq) + NaOH(aq) -> NaCl(aq) + H2O(l)
This means that 1 mole of HCl reacts with 1 mole of NaOH to produce 1 mole of water. So, the heat evolved per mole of reaction is equal to the enthalpy change of neutralization.
Now, let's calculate the mole of the limiting reactant:
Given volume of HCl = 87 cm3 = 87 x 10-3 dm3
Given molarity of HCl = 1.6 mol dm-3
Mole of HCl = volume x molarity = (87 x 10-3) x 1.6 = 0.1392 mol
Since the mole of HCl and NaOH are equal, the mole of the limiting reactant is 0.1392 mol.
To determine the heat evolved per reaction, we need to divide the heat evolved by the mole of reaction:
Given temperature change = 317.4 K - 298 K = 19.4 K
Given the specific heat capacity of water = 4.18 J/K g
Assuming the density of the solutions is the same as water, we can use the mass of the solutions:
Mass of solution = volume x density = 87 x 10-3 dm3 x 1 g/cm3 = 87 g
Now, we can calculate the heat evolved:
Heat evolved = mass x specific heat capacity x temperature change = 87 g x 4.18 J/K g x 19.4 K = 8,177.56 J/mol
Convert the heat evolved from Joules to kilojoules:
The heat evolved = 8,177.56 J/mol = 8.18 kJ/mol (rounded to 2 decimal places)
Therefore, the enthalpy of neutralization of HCl and NaOH is 8.18 kJ/mol.