Final answer:
The molar enthalpy of neutralization for sodium hydroxide in the described calorimeter experiment is -55.73 kJ/mol, calculated by dividing the heat absorbed by the water during the temperature rise by the moles of NaOH that reacted.
Step-by-step explanation:
Calculating the Molar Enthalpy of Neutralization for Sodium Hydroxide
To calculate the molar enthalpy of neutralization for sodium hydroxide, we will use the information given about the coffee-cup calorimeter experiment. Here are the steps to calculate it:
- First, calculate the number of moles of sodium hydroxide that reacted. We do this by multiplying the volume of the NaOH solution (60.0 mL or 0.060 L) by its concentration (0.700 mol/L). Therefore, the moles of NaOH are 0.060 L × 0.700 mol/L = 0.042 mol.
- Since the temperature increased by 5.6 °C, we assume that the entire solution's heat capacity is the same as water and that the density of the solution is 1.00 g/mL. Thus, the total mass of the solution is the sum of the volumes of the NaOH and H2SO4 solutions, which is 60.0 mL + 40.0 mL = 100.0 mL or 100.0 g of water.
- The amount of heat absorbed by the solution, q, is calculated using q = m × c × ΔT, where
m is the mass of the solution,
c is the specific heat capacity of water (4.18 J/g°C), and
ΔT is the change in temperature.
So q = 100.0 g × 4.18 J/g°C × 5.6 °C = 2340.8 J or 2.3408 kJ. - The molar enthalpy of neutralization (ΔH_neut) is the amount of heat released per mole of NaOH. To find this, we divide the heat absorbed by the number of moles of NaOH. ΔH_neut = -2.3408 kJ / 0.042 mol = -55.73 kJ/mol.
The negative sign indicates that it is an exothermic reaction, which means heat is released during the process of neutralization.