Chemistry - Thermochemistry Question with Solution | TestHub
A volume of 100 ml of 0.5 N-H2SO4 solution is neutralized with 200 ml of 0.2 M-NH4OH in a constant pressure calorimeter which resulted 1.4°C rise in temperature. The heat capacity of the calorimeter system is 1.5 kJ/°C. Some useful thermochemical equations are:
HCl + NaOH → NaCl + H2O + 57 kJ
CH3COOH + NH4OH → CH3COONH4 + H2O + 48.1 kJ
Which of the following statements are correct? (one or more than one correct)
Options:(select one or more)
Answer:
Solution:
The available H⁺ is calculated as 100 mL × 0.5 N = 0.05 eq.
For NH₄OH, we have 0.2 M × 0.2 L = 0.04 mol.
Since 0.04 mol < 0.05 eq, NH₄OH is the limiting reactant.
The heat released is C Δ T = 1.5 × 1.4 = 2.1 kJ for 0.04 mol.
Therefore, the heat released per mole is 2.1/0.04 = 52.5 kJ/mol.
So, Δ Hneut = -52.5 kJ/mol (statement 'a' is true).
A strong acid with a strong base yields -57 kJ/mol. The difference in heat released indicates the energy absorbed by the ionization of NH₄OH: 57 - 52.5 = 4.5 kJ/mol (statement 'b' is true).
For the reaction CH₃COOH + NH₄OH, we have:
-48.1 = Δ Hion(CH₃COOH) + 4.5 - 57
Solving for Δ Hion(CH₃COOH):
Δ Hion(CH₃COOH) = -48.1 - 4.5 + 57 = 4.4 kJ/mol.
This value is 4.4 kJ/mol, not 4.6 kJ/mol (statement 'c' is false).
The reaction H₂O → H⁺ + OH⁻ is the reverse of neutralization, which requires +57 kJ per mole. For 2 mol, the energy required is 2 × 57 = +114 kJ (statement 'd' is true).
The correct statements are (a), (b), and (d).
