Question

0 m L 150 𝑚 𝐿 of 0.5 N H C l 0.5 𝑁 𝐻 𝐶 𝑙 solution at 25 ∘ C 25 ∘ 𝐶 was mixed with 150 m L 150 𝑚 𝐿 of 0.5 N N a O H 0.5 𝑁 𝑁 𝑎 𝑂 𝐻 solution at same temperature. Calculate the heat of neutralization of H C l 𝐻 𝐶 𝑙 with N a O H 𝑁 𝑎 𝑂 𝐻 , if find temperature was recorded to be 29 ∘ C 29 ∘ 𝐶 . ( ρ H 2 O = 1 g / m L )

Answer 1

-16 kcal/mol

Answer 2

The heat of neutralization is the heat released when one mole of water is formed from the reaction of an acid and a base.

1. Calculate the total volume and mass of the solution: Total volume of solution = Volume of HCl + Volume of NaOH Total volume = 150 mL + 150 mL = 300 mL Given the density of the solution (approximated as water) ρ = 1 g/mL: Total mass (m) = Total volume × density = 300 mL × 1 g/mL = 300 g

2. Calculate the change in temperature (ΔT): Initial temperature = 25 °C Final temperature = 29 °C ΔT = Final temperature - Initial temperature = 29 °C - 25 °C = 4 °C

3. Calculate the heat absorbed by the solution (Q): The specific heat capacity of the solution (c) is assumed to be that of water, which is 1 cal/g°C. Q = m × c × ΔT Q = 300 g × 1 cal/g°C × 4 °C = 1200 calories

4. Calculate the moles of water formed: The reaction is: HCl(aq) + NaOH(aq) → NaCl(aq) + H₂O(l) Since it's a strong acid and strong base, the normality is equal to molarity (n-factor = 1). Moles of HCl = Normality × Volume (in Liters) = 0.5 N × (150/1000) L = 0.5 × 0.150 = 0.075 moles Moles of NaOH = Normality × Volume (in Liters) = 0.5 N × (150/1000) L = 0.5 × 0.150 = 0.075 moles Since both reactants are present in stoichiometric amounts, 0.075 moles of water (H₂O) are formed.

5. Calculate the heat of neutralization per mole: The heat of neutralization (ΔH_neut) is the heat released per mole of water formed. Since heat is released, the value will be negative (exothermic reaction). ΔH_neut = -Q / moles of H₂O formed ΔH_neut = -1200 cal / 0.075 moles ΔH_neut = -16000 cal/mol

6. Convert the heat of neutralization to kcal/mol: ΔH_neut = -16000 cal/mol × (1 kcal / 1000 cal) = -16 kcal/mol

The heat of neutralization of HCl with NaOH is -16 kcal/mol.

The final answer is $\boxed{-16 \text{ kcal/mol}}$.

Explanation: The heat released during the neutralization reaction is absorbed by the solution, causing its temperature to rise. This heat (Q) is calculated using the formula Q = mcΔT, where m is the total mass of the solution, c is its specific heat capacity (assumed to be 1 cal/g°C), and ΔT is the temperature change. The total mass is determined from the combined volume and given density. The moles of water formed are calculated from the initial moles of the limiting reactant (HCl or NaOH, which are equal in this case). Finally, the heat of neutralization is obtained by dividing the total heat released by the moles of water formed, with a negative sign indicating an exothermic process.