Final answer:
By applying the calorimetry formula q = mcΔT to the given data, it is calculated that the total heat released from the neutralization reaction is 2.1 kJ, which helps to find the correct answer related to the enthalpy of neutralization for the given reaction. The correct answer is D. ΔH of 2H₂H₂O(I) → 2H⁺(aq.) + 2OH⁻(aq) is 114 kJ
Step-by-step explanation:
To solve the student's question involving thermochemistry and stoichiometry, one must understand the principles of calorimetry and molar enthalpy changes. The change in temperature caused by a chemical reaction, given the heat capacity of the calorimeter, can be used to calculate the amount of heat absorbed or released during the reaction. In this instance, we are asked to determine the enthalpy changes related to different neutralization reactions.
We first calculate the amount of heat (q) released using the formula q = mcΔT, where 'm' is the mass of solution, 'c' is the heat capacity, and ΔT is the change in temperature. Quantities given are already in appropriate units, so we can directly apply them: q = 1.5 kJ/°C * 1.4°C = 2.1 kJ. This is the total heat released when 100 ml of 0.5 Normal (0.25 mole/L) H₂SO₄ reacts completely with 200 ml of 0.2 Mole NH₄OH. Since normality is twice the molarity for H₂SO₄ (due to it being a diprotic acid), the mole ratio of H₂SO₄ to NH₄OH is 1:1. Given this, we can calculate the molar enthalpy change.
The correct answer is provided by applying this logic and calculation steps to the potential answer choices. Remember that the enthalpy of neutralization typically involves the formation of water from an acid and a base. The given examples in the question relate to different neutralization and dissociation reactions that may be considered for solving the problem. However, with the given data, the student's question can be directly answered.