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 = -57.0 kJ
ΔHSolution of HX(g) = -60.0 kJ/mol
ΔHSolution of MOH(g) = +10.0 kJ/mol
ΔHIonization of HX = 13.0 kJ/mol and MOH is a strong base.
Column I | Column II |
|---|---|
P) HX(aq) + MOH(aq) → MX(aq) + H2O | i) -47 |
Q) HX(g) + MOH(g) → MX(aq) + H2O | ii) -94 |
R) HX(g) → H+(aq) + X-(aq) | iii) -44 |
S) M+(aq) + OH-(aq) → MOH(aq) | iv) 0 |
(Column I = Chemical reactions; Column II = ΔH values in kJ)
Options:
Answer:
Solution:
R: The reaction HX(g) → H+(aq) + X-(aq) represents ionization plus dissolution. The enthalpy change is ΔH = 13 + (-60) = -47 kJ, which corresponds to R-i.
P: In solution, a weak acid HX first ionizes, requiring +13 kJ. Then, H+ neutralizes OH-, releasing -57 kJ. Thus, the total enthalpy change is ΔH = 13 - 57 = -44 kJ, which corresponds to P-iii.
Q: Starting from the gaseous reactants, the total enthalpy change is ΔH = (-47) + 10 + (-57) = -94 kJ. Here, +10 kJ is for the dissolution of gaseous MOH, and the strong base contributes no further ionization enthalpy. This corresponds to Q-ii.
S: Since MOH is a strong base, MOH(aq) is already fully ionized. Therefore, the reaction M+(aq) + OH-(aq) → MOH(aq) has ΔH ≈ 0. This corresponds to S-iv.
Answer: P-iii; Q-ii; R-i; S-iv.
