Answer:
a. To calculate the standard enthalpy of neutralization of sulphuric acid and sodium hydroxide, we can use the formula:
ΔH_neutralization = q_calorimeter / n_reactant
where q_calorimeter is the heat absorbed or released by the solution and n_reactant is the number of moles of reactant.
First, we need to calculate the heat absorbed or released by the solution using the equation:
q_calorimeter = mcΔT
where m is the mass of the solution, c is the specific heat capacity of the solution and ΔT is the change in temperature.
In this case, the mass of the solution is 50 cm³ + 50 cm³ = 100 cm³. The specific heat capacity of the solution is approximately 4.18 J/g°C. The change in temperature is 6.8°C.
q_calorimeter = (100 cm³) (4.18 J/g°C) (6.8°C) = 28.624 J
Then we need to calculate the number of moles of reactant used. The acid is sulfuric acid and the base is sodium hydroxide.
H2SO4 + 2NaOH -> Na2SO4 + 2H2O
The balanced equation shows that 1 mole of H2SO4 reacts with 2 moles of NaOH.
n_reactant = (50 cm³)(0.50 M) + (50 cm³)(0.10 M) = 2.5 moles
Therefore, the standard enthalpy of neutralization of sulphuric acid and sodium hydroxide is:
ΔH_neutralization = q_calorimeter / n_reactant = (28.624 J) / (2.5 moles) = -11.4496 kJ/mol
b. The value obtained in (a) is less than the expected value because heat exchange occurs between the reaction mixture and the calorimeter, which causes a loss of energy. Additionally, the solution may not have been perfectly insulated and heat may have been lost to the surroundings, also causing a loss of energy. Furthermore, the reaction may not have been performed under standard conditions, such as at a specific temperature or pressure.