Chemistry - Thermochemistry Question with Solution | TestHub
Column I contains some chemical reactions and Column II contains some ΔH values (in kJ). Match the correct ΔH in Column II for the chemical reaction in Column I, with the help of the following thermochemical equations given:
H+(aq) + OH-(aq) → H2O(l); ΔH = -57.3 kJ
ΔHSolution of HA(g) = -70.7 kJ/mol
ΔHSolution of BOH(g) = 20 kJ/mol
ΔHIonization of HA = 15 kJ/mol and BOH is a strong base.
Column I | Column II |
|---|---|
P) HA(aq) + BOH(aq) → BA(aq) + H2O | i) -42.3 |
Q) HA(g) + BOH(g) → BA(aq) + H2O | ii) -93 |
R) HA(g) → H+(aq) + A-(aq) | iii) -55.7 |
S) B+(aq) + OH-(aq) → BOH(aq) | iv) 0 |
(Column I = Chemical reactions; Column II = ΔH values in kJ)
Options:
Answer:
Solution:
R: The ionization of gaseous HA followed by dissolution is represented by the reaction:
HA(g) tarrow H⁺(aq) + A⁻(aq)
The enthalpy change for this process is Δ H = 15 + (-70.7) = -55.7 kJ, which corresponds to R-iii.
P: In solution, HA must first ionize (+15 kJ), then H⁺ neutralizes OH⁻ (-57.3 kJ). Since BOH is a strong base, the overall enthalpy change is Δ H = 15 - 57.3 = -42.3 kJ, which corresponds to P-i.
Q: Starting from gases, the total enthalpy change is Δ H = (-55.7) + 20 + (-57.3) = -93 kJ. Here, +20 kJ is the enthalpy of dissolution for gaseous BOH, and its ionization contributes nothing because it is a strong base. This corresponds to Q-ii.
S: BOH is a strong base, meaning BOH(aq) is already fully ionized. Therefore, the reaction B⁺(aq) + OH⁻(aq) tarrow BOH(aq) has an approximate enthalpy change of Δ H ≈ 0, which corresponds to S-iv.
Answer: P-i; Q-ii; R-iii; S-iv.
